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Theorem List for Metamath Proof Explorer - 23801-23900   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremlawcos 23801* Law of cosines (also known as the Al-Kashi theorem or the generalized Pythagorean theorem, or the cosine formula or cosine rule). Given three distinct points A, B, and C, prove a relationship between their segment lengths. This theorem is expressed using the complex number plane as a plane, where  F is the signed angle construct (as used in ang180 23799),  X is the distance of line segment BC,  Y is the distance of line segment AC,  Z is the distance of line segment AB, and  O is the signed angle m/_ BCA on the complex plane. We translate triangle ABC to move C to the origin (C-C), B to U=(B-C), and A to V=(A-C), then use lemma lawcoslem1 23800 to prove this algebraically simpler case. The metamath convention is to use a signed angle; in this case the sign doesn't matter because we use the cosine of the angle (see cosneg 14256). The Pythagorean theorem pythag 23802 is a special case of the law of cosines. The theorem's expression and approach were suggested by Mario Carneiro. This is Metamath 100 proof #94. (Contributed by David A. Wheeler, 12-Jun-2015.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  X  =  ( abs `  ( B  -  C ) )   &    |-  Y  =  ( abs `  ( A  -  C ) )   &    |-  Z  =  ( abs `  ( A  -  B ) )   &    |-  O  =  ( ( B  -  C ) F ( A  -  C ) )   =>    |-  ( ( ( A  e.  CC  /\  B  e.  CC  /\  C  e.  CC )  /\  ( A  =/=  C  /\  B  =/=  C ) )  ->  ( Z ^ 2 )  =  ( ( ( X ^ 2 )  +  ( Y ^
 2 ) )  -  ( 2  x.  (
 ( X  x.  Y )  x.  ( cos `  O ) ) ) ) )
 
Theorempythag 23802* Pythagorean theorem. Given three distinct points A, B, and C that form a right triangle (with the right angle at C), prove a relationship between their segment lengths. This theorem is expressed using the complex number plane as a plane, where  F is the signed angle construct (as used in ang180 23799),  X is the distance of line segment BC,  Y is the distance of line segment AC,  Z is the distance of line segment AB (the hypotenuse), and  O is the signed right angle m/_ BCA. We use the law of cosines lawcos 23801 to prove this, along with simple trigonometry facts like coshalfpi 23480 and cosneg 14256. (Contributed by David A. Wheeler, 13-Jun-2015.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  X  =  ( abs `  ( B  -  C ) )   &    |-  Y  =  ( abs `  ( A  -  C ) )   &    |-  Z  =  ( abs `  ( A  -  B ) )   &    |-  O  =  ( ( B  -  C ) F ( A  -  C ) )   =>    |-  ( ( ( A  e.  CC  /\  B  e.  CC  /\  C  e.  CC )  /\  ( A  =/=  C  /\  B  =/=  C )  /\  O  e.  { ( pi  / 
 2 ) ,  -u ( pi  /  2 ) }
 )  ->  ( Z ^ 2 )  =  ( ( X ^
 2 )  +  ( Y ^ 2 ) ) )
 
Theoremisosctrlem1 23803 Lemma for isosctr 23806. (Contributed by Saveliy Skresanov, 30-Dec-2016.)
 |-  ( ( A  e.  CC  /\  ( abs `  A )  =  1  /\  -.  1  =  A ) 
 ->  ( Im `  ( log `  ( 1  -  A ) ) )  =/=  pi )
 
Theoremisosctrlem2 23804 Lemma for isosctr 23806. Corresponds to the case where one vertex is at 0, another at 1 and the third lies on the unit circle. (Contributed by Saveliy Skresanov, 31-Dec-2016.)
 |-  ( ( A  e.  CC  /\  ( abs `  A )  =  1  /\  -.  1  =  A ) 
 ->  ( Im `  ( log `  ( 1  -  A ) ) )  =  ( Im `  ( log `  ( -u A  /  ( 1  -  A ) ) ) ) )
 
Theoremisosctrlem3 23805* Lemma for isosctr 23806. Corresponds to the case where one vertex is at 0. (Contributed by Saveliy Skresanov, 1-Jan-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   =>    |-  ( ( ( A  e.  CC  /\  B  e.  CC )  /\  ( A  =/=  0  /\  B  =/=  0  /\  A  =/=  B )  /\  ( abs `  A )  =  ( abs `  B )
 )  ->  ( -u A F ( B  -  A ) )  =  ( ( A  -  B ) F -u B ) )
 
Theoremisosctr 23806* Isosceles triangle theorem. This is Metamath 100 proof #65. (Contributed by Saveliy Skresanov, 1-Jan-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   =>    |-  ( ( ( A  e.  CC  /\  B  e.  CC  /\  C  e.  CC )  /\  ( A  =/=  C  /\  B  =/=  C  /\  A  =/=  B )  /\  ( abs `  ( A  -  C ) )  =  ( abs `  ( B  -  C ) ) ) 
 ->  ( ( C  -  A ) F ( B  -  A ) )  =  ( ( A  -  B ) F ( C  -  B ) ) )
 
Theoremssscongptld 23807* If two triangles have equal sides, one angle in one triangle has the same cosine as the corresponding angle in the other triangle. This is a partial form of the SSS congruence theorem.

This theorem is proven by using lawcos 23801 on both triangles to express one side in terms of the other two, and then equating these expressions and reducing this algebraically to get an equality of cosines of angles. (Contributed by David Moews, 28-Feb-2017.)

 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  E  e.  CC )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  A  =/=  B )   &    |-  ( ph  ->  B  =/=  C )   &    |-  ( ph  ->  D  =/=  E )   &    |-  ( ph  ->  E  =/=  G )   &    |-  ( ph  ->  ( abs `  ( A  -  B ) )  =  ( abs `  ( D  -  E ) ) )   &    |-  ( ph  ->  ( abs `  ( B  -  C ) )  =  ( abs `  ( E  -  G ) ) )   &    |-  ( ph  ->  ( abs `  ( C  -  A ) )  =  ( abs `  ( G  -  D ) ) )   =>    |-  ( ph  ->  ( cos `  ( ( A  -  B ) F ( C  -  B ) ) )  =  ( cos `  (
 ( D  -  E ) F ( G  -  E ) ) ) )
 
Theoremaffineequiv 23808 Equivalence between two ways of expressing  B as an affine combination of  A and  C. (Contributed by David Moews, 28-Feb-2017.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   =>    |-  ( ph  ->  ( B  =  ( ( D  x.  A )  +  ( ( 1  -  D )  x.  C ) )  <->  ( C  -  B )  =  ( D  x.  ( C  -  A ) ) ) )
 
Theoremaffineequiv2 23809 Equivalence between two ways of expressing  B as an affine combination of  A and  C. (Contributed by David Moews, 28-Feb-2017.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   =>    |-  ( ph  ->  ( B  =  ( ( D  x.  A )  +  ( ( 1  -  D )  x.  C ) )  <->  ( B  -  A )  =  (
 ( 1  -  D )  x.  ( C  -  A ) ) ) )
 
Theoremangpieqvdlem 23810 Equivalence used in the proof of angpieqvd 23813. (Contributed by David Moews, 28-Feb-2017.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  A  =/=  B )   &    |-  ( ph  ->  A  =/=  C )   =>    |-  ( ph  ->  (
 -u ( ( C  -  B )  /  ( A  -  B ) )  e.  RR+  <->  ( ( C  -  B )  /  ( C  -  A ) )  e.  (
 0 (,) 1 ) ) )
 
Theoremangpieqvdlem2 23811* Equivalence used in angpieqvd 23813. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  A  =/=  B )   &    |-  ( ph  ->  B  =/=  C )   =>    |-  ( ph  ->  ( -u ( ( C  -  B )  /  ( A  -  B ) )  e.  RR+  <->  ( ( A  -  B ) F ( C  -  B ) )  =  pi ) )
 
Theoremangpined 23812* If the angle at ABC is  pi, then A is not equal to C. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  A  =/=  B )   &    |-  ( ph  ->  B  =/=  C )   =>    |-  ( ph  ->  (
 ( ( A  -  B ) F ( C  -  B ) )  =  pi  ->  A  =/=  C ) )
 
Theoremangpieqvd 23813* The angle ABC is  pi iff B is a nontrivial convex combination of A and C, i.e., iff B is in the interior of the segment AC. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  A  =/=  B )   &    |-  ( ph  ->  B  =/=  C )   =>    |-  ( ph  ->  (
 ( ( A  -  B ) F ( C  -  B ) )  =  pi  <->  E. w  e.  (
 0 (,) 1 ) B  =  ( ( w  x.  A )  +  ( ( 1  -  w )  x.  C ) ) ) )
 
Theoremchordthmlem 23814* If M is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 23807 to observe that, since AMQ and BMQ have equal sides, the angles QMB and QMA must be equal. Since they are supplementary, both must be right. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  M  =  ( ( A  +  B )  / 
 2 ) )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( B  -  Q ) ) )   &    |-  ( ph  ->  A  =/=  B )   &    |-  ( ph  ->  Q  =/=  M )   =>    |-  ( ph  ->  (
 ( Q  -  M ) F ( B  -  M ) )  e. 
 { ( pi  / 
 2 ) ,  -u ( pi  /  2 ) }
 )
 
Theoremchordthmlem2 23815* If M is the midpoint of AB, AQ = BQ, and P is on the line AB, then QMP is a right angle. This is proven by reduction to the special case chordthmlem 23814, where P = B, and using angrtmuld 23793 to observe that QMP is right iff QMB is. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  RR )   &    |-  ( ph  ->  M  =  ( ( A  +  B )  /  2 ) )   &    |-  ( ph  ->  P  =  ( ( X  x.  A )  +  (
 ( 1  -  X )  x.  B ) ) )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( B  -  Q ) ) )   &    |-  ( ph  ->  P  =/=  M )   &    |-  ( ph  ->  Q  =/=  M )   =>    |-  ( ph  ->  (
 ( Q  -  M ) F ( P  -  M ) )  e. 
 { ( pi  / 
 2 ) ,  -u ( pi  /  2 ) }
 )
 
Theoremchordthmlem3 23816 If M is the midpoint of AB, AQ = BQ, and P is on the line AB, then PQ 2 = QM 2  + PM 2 . This follows from chordthmlem2 23815 and the Pythagorean theorem (pythag 23802) in the case where P and Q are unequal to M. If either P or Q equals M, the result is trivial. (Contributed by David Moews, 28-Feb-2017.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  RR )   &    |-  ( ph  ->  M  =  ( ( A  +  B )  / 
 2 ) )   &    |-  ( ph  ->  P  =  ( ( X  x.  A )  +  ( (
 1  -  X )  x.  B ) ) )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( B  -  Q ) ) )   =>    |-  ( ph  ->  (
 ( abs `  ( P  -  Q ) ) ^
 2 )  =  ( ( ( abs `  ( Q  -  M ) ) ^ 2 )  +  ( ( abs `  ( P  -  M ) ) ^ 2 ) ) )
 
Theoremchordthmlem4 23817 If P is on the segment AB and M is the midpoint of AB, then PA  x. PB = BM 2  - PM 2 . If all lengths are reexpressed as fractions of AB, this reduces to the identity  X  x.  (
1  -  X )  =  ( 1  / 
2 ) 2  -  ( ( 1  /  2 )  -  X ) 2 . (Contributed by David Moews, 28-Feb-2017.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  X  e.  (
 0 [,] 1 ) )   &    |-  ( ph  ->  M  =  ( ( A  +  B )  /  2
 ) )   &    |-  ( ph  ->  P  =  ( ( X  x.  A )  +  ( ( 1  -  X )  x.  B ) ) )   =>    |-  ( ph  ->  ( ( abs `  ( P  -  A ) )  x.  ( abs `  ( P  -  B ) ) )  =  ( ( ( abs `  ( B  -  M ) ) ^ 2 )  -  ( ( abs `  ( P  -  M ) ) ^ 2 ) ) )
 
Theoremchordthmlem5 23818 If P is on the segment AB and AQ = BQ, then PA  x. PB = BQ 2  - PQ 2 . This follows from two uses of chordthmlem3 23816 to show that PQ 2 = QM 2  + PM 2 and BQ 2 = QM 2  + BM 2 , so BQ 2  - PQ 2 = (QM 2  + BM 2 )  - (QM 2  + PM 2 ) = BM 2  - PM 2 , which equals PA  x. PB by chordthmlem4 23817. (Contributed by David Moews, 28-Feb-2017.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  ( 0 [,] 1
 ) )   &    |-  ( ph  ->  P  =  ( ( X  x.  A )  +  ( ( 1  -  X )  x.  B ) ) )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( B  -  Q ) ) )   =>    |-  ( ph  ->  (
 ( abs `  ( P  -  A ) )  x.  ( abs `  ( P  -  B ) ) )  =  ( ( ( abs `  ( B  -  Q ) ) ^ 2 )  -  ( ( abs `  ( P  -  Q ) ) ^ 2 ) ) )
 
Theoremchordthm 23819* The intersecting chords theorem. If points A, B, C, and D lie on a circle (with center Q, say), and the point P is on the interior of the segments AB and CD, then the two products of lengths PA  x. PB and PC  x. PD are equal. The Euclidean plane is identified with the complex plane, and the fact that P is on AB and on CD is expressed by the hypothesis that the angles APB and CPD are equal to  pi. The result is proven by using chordthmlem5 23818 twice to show that PA  x. PB and PC  x. PD both equal BQ 2  - PQ 2 . This is similar to the proof of the theorem given in Euclid's Elements, where it is Proposition III.35. This is Metamath 100 proof #55. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  P  e.  CC )   &    |-  ( ph  ->  A  =/=  P )   &    |-  ( ph  ->  B  =/=  P )   &    |-  ( ph  ->  C  =/=  P )   &    |-  ( ph  ->  D  =/=  P )   &    |-  ( ph  ->  ( ( A  -  P ) F ( B  -  P ) )  =  pi )   &    |-  ( ph  ->  ( ( C  -  P ) F ( D  -  P ) )  =  pi )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( B  -  Q ) ) )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( C  -  Q ) ) )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( D  -  Q ) ) )   =>    |-  ( ph  ->  (
 ( abs `  ( P  -  A ) )  x.  ( abs `  ( P  -  B ) ) )  =  ( ( abs `  ( P  -  C ) )  x.  ( abs `  ( P  -  D ) ) ) )
 
Theoremheron 23820* Heron's formula gives the area of a triangle given only the side lengths. If points A, B, C form a triangle, then the area of the triangle, represented here as 
( 1  /  2
)  x.  X  x.  Y  x.  abs ( sin O ), is equal to the square root of  S  x.  ( S  -  X )  x.  ( S  -  Y
)  x.  ( S  -  Z ), where  S  =  ( X  +  Y  +  Z )  /  2 is half the perimeter of the triangle. Based on work by Jon Pennant. This is Metamath 100 proof #57. (Contributed by Mario Carneiro, 10-Mar-2019.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  X  =  ( abs `  ( B  -  C ) )   &    |-  Y  =  ( abs `  ( A  -  C ) )   &    |-  Z  =  ( abs `  ( A  -  B ) )   &    |-  O  =  ( ( B  -  C ) F ( A  -  C ) )   &    |-  S  =  ( ( ( X  +  Y )  +  Z )  /  2
 )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  A  =/=  C )   &    |-  ( ph  ->  B  =/=  C )   =>    |-  ( ph  ->  ( ( ( 1  / 
 2 )  x.  ( X  x.  Y ) )  x.  ( abs `  ( sin `  O ) ) )  =  ( sqr `  ( ( S  x.  ( S  -  X ) )  x.  (
 ( S  -  Y )  x.  ( S  -  Z ) ) ) ) )
 
14.3.7  Solutions of quadratic, cubic, and quartic equations
 
Theoremquad2 23821 The quadratic equation, without specifying the particular branch  D to the square root. (Contributed by Mario Carneiro, 23-Apr-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  A  =/=  0 )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  ( D ^ 2 )  =  ( ( B ^
 2 )  -  (
 4  x.  ( A  x.  C ) ) ) )   =>    |-  ( ph  ->  (
 ( ( A  x.  ( X ^ 2 ) )  +  ( ( B  x.  X )  +  C ) )  =  0  <->  ( X  =  ( ( -u B  +  D )  /  (
 2  x.  A ) )  \/  X  =  ( ( -u B  -  D )  /  (
 2  x.  A ) ) ) ) )
 
Theoremquad 23822 The quadratic equation. (Contributed by Mario Carneiro, 23-Apr-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  A  =/=  0 )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  D  =  ( ( B ^ 2
 )  -  ( 4  x.  ( A  x.  C ) ) ) )   =>    |-  ( ph  ->  (
 ( ( A  x.  ( X ^ 2 ) )  +  ( ( B  x.  X )  +  C ) )  =  0  <->  ( X  =  ( ( -u B  +  ( sqr `  D ) )  /  (
 2  x.  A ) )  \/  X  =  ( ( -u B  -  ( sqr `  D ) )  /  (
 2  x.  A ) ) ) ) )
 
Theorem1cubrlem 23823 The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
 |-  ( ( -u 1  ^c  ( 2  /  3 ) )  =  ( ( -u 1  +  ( _i  x.  ( sqr `  3
 ) ) )  / 
 2 )  /\  (
 ( -u 1  ^c 
 ( 2  /  3
 ) ) ^ 2
 )  =  ( (
 -u 1  -  ( _i  x.  ( sqr `  3
 ) ) )  / 
 2 ) )
 
Theorem1cubr 23824 The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
 |-  R  =  { 1 ,  ( ( -u 1  +  ( _i  x.  ( sqr `  3
 ) ) )  / 
 2 ) ,  (
 ( -u 1  -  ( _i  x.  ( sqr `  3
 ) ) )  / 
 2 ) }   =>    |-  ( A  e.  R 
 <->  ( A  e.  CC  /\  ( A ^ 3
 )  =  1 ) )
 
Theoremdcubic1lem 23825 Lemma for dcubic1 23827 and dcubic2 23826: simplify the cubic equation under the substitution  X  =  U  -  M  /  U. (Contributed by Mario Carneiro, 26-Apr-2015.)
 |-  ( ph  ->  P  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  e.  CC )   &    |-  ( ph  ->  ( T ^ 3 )  =  ( G  -  N ) )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  ( G ^ 2 )  =  ( ( N ^
 2 )  +  ( M ^ 3 ) ) )   &    |-  ( ph  ->  M  =  ( P  / 
 3 ) )   &    |-  ( ph  ->  N  =  ( Q  /  2 ) )   &    |-  ( ph  ->  T  =/=  0 )   &    |-  ( ph  ->  U  e.  CC )   &    |-  ( ph  ->  U  =/=  0 )   &    |-  ( ph  ->  X  =  ( U  -  ( M  /  U ) ) )   =>    |-  ( ph  ->  (
 ( ( X ^
 3 )  +  (
 ( P  x.  X )  +  Q )
 )  =  0  <->  ( ( ( U ^ 3 ) ^ 2 )  +  ( ( Q  x.  ( U ^ 3 ) )  -  ( M ^ 3 ) ) )  =  0 ) )
 
Theoremdcubic2 23826* Reverse direction of dcubic 23828. Given a solution  U to the "substitution" quadratic equation  X  =  U  -  M  /  U, show that  X is in the desired form. (Contributed by Mario Carneiro, 25-Apr-2015.)
 |-  ( ph  ->  P  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  e.  CC )   &    |-  ( ph  ->  ( T ^ 3 )  =  ( G  -  N ) )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  ( G ^ 2 )  =  ( ( N ^
 2 )  +  ( M ^ 3 ) ) )   &    |-  ( ph  ->  M  =  ( P  / 
 3 ) )   &    |-  ( ph  ->  N  =  ( Q  /  2 ) )   &    |-  ( ph  ->  T  =/=  0 )   &    |-  ( ph  ->  U  e.  CC )   &    |-  ( ph  ->  U  =/=  0 )   &    |-  ( ph  ->  X  =  ( U  -  ( M  /  U ) ) )   &    |-  ( ph  ->  ( ( X ^ 3
 )  +  ( ( P  x.  X )  +  Q ) )  =  0 )   =>    |-  ( ph  ->  E. r  e.  CC  (
 ( r ^ 3
 )  =  1  /\  X  =  ( (
 r  x.  T )  -  ( M  /  ( r  x.  T ) ) ) ) )
 
Theoremdcubic1 23827 Forward direction of dcubic 23828: the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015.)
 |-  ( ph  ->  P  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  e.  CC )   &    |-  ( ph  ->  ( T ^ 3 )  =  ( G  -  N ) )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  ( G ^ 2 )  =  ( ( N ^
 2 )  +  ( M ^ 3 ) ) )   &    |-  ( ph  ->  M  =  ( P  / 
 3 ) )   &    |-  ( ph  ->  N  =  ( Q  /  2 ) )   &    |-  ( ph  ->  T  =/=  0 )   &    |-  ( ph  ->  X  =  ( T  -  ( M 
 /  T ) ) )   =>    |-  ( ph  ->  (
 ( X ^ 3
 )  +  ( ( P  x.  X )  +  Q ) )  =  0 )
 
Theoremdcubic 23828* Solutions to the depressed cubic, a special case of cubic 23831. (The definitions of  M ,  N ,  G ,  T here differ from mcubic 23829 by scale factors of  -u 9,  5 4,  5 4 and  -u 2
7 respectively, to simplify the algebra and presentation.) (Contributed by Mario Carneiro, 26-Apr-2015.)
 |-  ( ph  ->  P  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  e.  CC )   &    |-  ( ph  ->  ( T ^ 3 )  =  ( G  -  N ) )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  ( G ^ 2 )  =  ( ( N ^
 2 )  +  ( M ^ 3 ) ) )   &    |-  ( ph  ->  M  =  ( P  / 
 3 ) )   &    |-  ( ph  ->  N  =  ( Q  /  2 ) )   &    |-  ( ph  ->  T  =/=  0 )   =>    |-  ( ph  ->  ( ( ( X ^
 3 )  +  (
 ( P  x.  X )  +  Q )
 )  =  0  <->  E. r  e.  CC  ( ( r ^
 3 )  =  1 
 /\  X  =  ( ( r  x.  T )  -  ( M  /  ( r  x.  T ) ) ) ) ) )
 
Theoremmcubic 23829* Solutions to a monic cubic equation, a special case of cubic 23831. (Contributed by Mario Carneiro, 24-Apr-2015.)
 |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  e.  CC )   &    |-  ( ph  ->  ( T ^
 3 )  =  ( ( N  +  G )  /  2 ) )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  ( G ^ 2 )  =  ( ( N ^ 2 )  -  ( 4  x.  ( M ^ 3 ) ) ) )   &    |-  ( ph  ->  M  =  ( ( B ^ 2 )  -  ( 3  x.  C ) ) )   &    |-  ( ph  ->  N  =  ( ( ( 2  x.  ( B ^ 3
 ) )  -  (
 9  x.  ( B  x.  C ) ) )  +  (; 2 7  x.  D ) ) )   &    |-  ( ph  ->  T  =/=  0
 )   =>    |-  ( ph  ->  (
 ( ( ( X ^ 3 )  +  ( B  x.  ( X ^ 2 ) ) )  +  ( ( C  x.  X )  +  D ) )  =  0  <->  E. r  e.  CC  ( ( r ^
 3 )  =  1 
 /\  X  =  -u ( ( ( B  +  ( r  x.  T ) )  +  ( M  /  (
 r  x.  T ) ) )  /  3
 ) ) ) )
 
Theoremcubic2 23830* The solution to the general cubic equation, for arbitrary choices  G and  T of the square and cube roots. (Contributed by Mario Carneiro, 23-Apr-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  A  =/=  0 )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  e.  CC )   &    |-  ( ph  ->  ( T ^ 3 )  =  ( ( N  +  G )  / 
 2 ) )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  ( G ^ 2 )  =  ( ( N ^
 2 )  -  (
 4  x.  ( M ^ 3 ) ) ) )   &    |-  ( ph  ->  M  =  ( ( B ^ 2 )  -  ( 3  x.  ( A  x.  C ) ) ) )   &    |-  ( ph  ->  N  =  ( ( ( 2  x.  ( B ^ 3 ) )  -  ( ( 9  x.  A )  x.  ( B  x.  C ) ) )  +  (; 2 7  x.  ( ( A ^ 2 )  x.  D ) ) ) )   &    |-  ( ph  ->  T  =/=  0 )   =>    |-  ( ph  ->  ( ( ( ( A  x.  ( X ^
 3 ) )  +  ( B  x.  ( X ^ 2 ) ) )  +  ( ( C  x.  X )  +  D ) )  =  0  <->  E. r  e.  CC  ( ( r ^
 3 )  =  1 
 /\  X  =  -u ( ( ( B  +  ( r  x.  T ) )  +  ( M  /  (
 r  x.  T ) ) )  /  (
 3  x.  A ) ) ) ) )
 
Theoremcubic 23831* The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4040 to convert the existential quantifier to a triple disjunction. This is Metamath 100 proof #37. (Contributed by Mario Carneiro, 26-Apr-2015.)
 |-  R  =  { 1 ,  ( ( -u 1  +  ( _i  x.  ( sqr `  3
 ) ) )  / 
 2 ) ,  (
 ( -u 1  -  ( _i  x.  ( sqr `  3
 ) ) )  / 
 2 ) }   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  A  =/=  0 )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  =  ( ( ( N  +  ( sqr `  G )
 )  /  2 )  ^c  ( 1  /  3 ) ) )   &    |-  ( ph  ->  G  =  ( ( N ^ 2 )  -  ( 4  x.  ( M ^ 3 ) ) ) )   &    |-  ( ph  ->  M  =  ( ( B ^ 2 )  -  ( 3  x.  ( A  x.  C ) ) ) )   &    |-  ( ph  ->  N  =  ( ( ( 2  x.  ( B ^ 3 ) )  -  ( ( 9  x.  A )  x.  ( B  x.  C ) ) )  +  (; 2 7  x.  ( ( A ^ 2 )  x.  D ) ) ) )   &    |-  ( ph  ->  M  =/=  0 )   =>    |-  ( ph  ->  ( ( ( ( A  x.  ( X ^
 3 ) )  +  ( B  x.  ( X ^ 2 ) ) )  +  ( ( C  x.  X )  +  D ) )  =  0  <->  E. r  e.  R  X  =  -u ( ( ( B  +  (
 r  x.  T ) )  +  ( M 
 /  ( r  x.  T ) ) ) 
 /  ( 3  x.  A ) ) ) )
 
Theorembinom4 23832 Work out a quartic binomial. (You would think that by this point it would be faster to use binom 13943, but it turns out to be just as much work to put it into this form after clearing all the sums and calculating binomial coefficients.) (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ( A  e.  CC  /\  B  e.  CC )  ->  ( ( A  +  B ) ^
 4 )  =  ( ( ( A ^
 4 )  +  (
 4  x.  ( ( A ^ 3 )  x.  B ) ) )  +  ( ( 6  x.  ( ( A ^ 2 )  x.  ( B ^
 2 ) ) )  +  ( ( 4  x.  ( A  x.  ( B ^ 3 ) ) )  +  ( B ^ 4 ) ) ) ) )
 
Theoremdquartlem1 23833 Lemma for dquart 23835. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  S  e.  CC )   &    |-  ( ph  ->  M  =  ( ( 2  x.  S ) ^
 2 ) )   &    |-  ( ph  ->  M  =/=  0
 )   &    |-  ( ph  ->  I  e.  CC )   &    |-  ( ph  ->  ( I ^ 2 )  =  ( ( -u ( S ^ 2 )  -  ( B  / 
 2 ) )  +  ( ( C  / 
 4 )  /  S ) ) )   =>    |-  ( ph  ->  ( ( ( ( X ^ 2 )  +  ( ( M  +  B )  /  2
 ) )  +  (
 ( ( ( M 
 /  2 )  x.  X )  -  ( C  /  4 ) ) 
 /  S ) )  =  0  <->  ( X  =  ( -u S  +  I
 )  \/  X  =  ( -u S  -  I
 ) ) ) )
 
Theoremdquartlem2 23834 Lemma for dquart 23835. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  S  e.  CC )   &    |-  ( ph  ->  M  =  ( ( 2  x.  S ) ^
 2 ) )   &    |-  ( ph  ->  M  =/=  0
 )   &    |-  ( ph  ->  I  e.  CC )   &    |-  ( ph  ->  ( I ^ 2 )  =  ( ( -u ( S ^ 2 )  -  ( B  / 
 2 ) )  +  ( ( C  / 
 4 )  /  S ) ) )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  (
 ( ( M ^
 3 )  +  (
 ( 2  x.  B )  x.  ( M ^
 2 ) ) )  +  ( ( ( ( B ^ 2
 )  -  ( 4  x.  D ) )  x.  M )  +  -u ( C ^ 2
 ) ) )  =  0 )   =>    |-  ( ph  ->  (
 ( ( ( M  +  B )  / 
 2 ) ^ 2
 )  -  ( ( ( C ^ 2
 )  /  4 )  /  M ) )  =  D )
 
Theoremdquart 23835 Solve a depressed quartic equation. To eliminate  S, which is the square root of a solution  M to the resolvent cubic equation, apply cubic 23831 or one of its variants. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  S  e.  CC )   &    |-  ( ph  ->  M  =  ( ( 2  x.  S ) ^
 2 ) )   &    |-  ( ph  ->  M  =/=  0
 )   &    |-  ( ph  ->  I  e.  CC )   &    |-  ( ph  ->  ( I ^ 2 )  =  ( ( -u ( S ^ 2 )  -  ( B  / 
 2 ) )  +  ( ( C  / 
 4 )  /  S ) ) )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  (
 ( ( M ^
 3 )  +  (
 ( 2  x.  B )  x.  ( M ^
 2 ) ) )  +  ( ( ( ( B ^ 2
 )  -  ( 4  x.  D ) )  x.  M )  +  -u ( C ^ 2
 ) ) )  =  0 )   &    |-  ( ph  ->  J  e.  CC )   &    |-  ( ph  ->  ( J ^
 2 )  =  ( ( -u ( S ^
 2 )  -  ( B  /  2 ) )  -  ( ( C 
 /  4 )  /  S ) ) )   =>    |-  ( ph  ->  ( (
 ( ( X ^
 4 )  +  ( B  x.  ( X ^
 2 ) ) )  +  ( ( C  x.  X )  +  D ) )  =  0  <->  ( ( X  =  ( -u S  +  I )  \/  X  =  ( -u S  -  I
 ) )  \/  ( X  =  ( S  +  J )  \/  X  =  ( S  -  J ) ) ) ) )
 
Theoremquart1cl 23836 Closure lemmas for quart 23843. (Contributed by Mario Carneiro, 7-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   =>    |-  ( ph  ->  ( P  e.  CC  /\  Q  e.  CC  /\  R  e.  CC ) )
 
Theoremquart1lem 23837 Lemma for quart1 23838. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  Y  =  ( X  +  ( A 
 /  4 ) ) )   =>    |-  ( ph  ->  D  =  ( ( ( ( A ^ 4 ) 
 / ;; 2 5 6 )  +  ( P  x.  ( ( A 
 /  4 ) ^
 2 ) ) )  +  ( ( Q  x.  ( A  / 
 4 ) )  +  R ) ) )
 
Theoremquart1 23838 Depress a quartic equation. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  Y  =  ( X  +  ( A 
 /  4 ) ) )   =>    |-  ( ph  ->  (
 ( ( X ^
 4 )  +  ( A  x.  ( X ^
 3 ) ) )  +  ( ( B  x.  ( X ^
 2 ) )  +  ( ( C  x.  X )  +  D ) ) )  =  ( ( ( Y ^ 4 )  +  ( P  x.  ( Y ^ 2 ) ) )  +  ( ( Q  x.  Y )  +  R ) ) )
 
Theoremquartlem1 23839 Lemma for quart 23843. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  P  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  R  e.  CC )   &    |-  ( ph  ->  U  =  ( ( P ^
 2 )  +  (; 1 2  x.  R ) ) )   &    |-  ( ph  ->  V  =  ( ( -u ( 2  x.  ( P ^ 3 ) )  -  (; 2 7  x.  ( Q ^ 2 ) ) )  +  (; 7 2  x.  ( P  x.  R ) ) ) )   =>    |-  ( ph  ->  ( U  =  ( (
 ( 2  x.  P ) ^ 2 )  -  ( 3  x.  (
 ( P ^ 2
 )  -  ( 4  x.  R ) ) ) )  /\  V  =  ( ( ( 2  x.  ( ( 2  x.  P ) ^
 3 ) )  -  ( 9  x.  (
 ( 2  x.  P )  x.  ( ( P ^ 2 )  -  ( 4  x.  R ) ) ) ) )  +  (; 2 7  x.  -u ( Q ^ 2 ) ) ) ) )
 
Theoremquartlem2 23840 Closure lemmas for quart 23843. (Contributed by Mario Carneiro, 7-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  E  =  -u ( A  /  4
 ) )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   &    |-  ( ph  ->  U  =  ( ( P ^ 2 )  +  (; 1 2  x.  R ) ) )   &    |-  ( ph  ->  V  =  ( ( -u ( 2  x.  ( P ^ 3 ) )  -  (; 2 7  x.  ( Q ^ 2 ) ) )  +  (; 7 2  x.  ( P  x.  R ) ) ) )   &    |-  ( ph  ->  W  =  ( sqr `  (
 ( V ^ 2
 )  -  ( 4  x.  ( U ^
 3 ) ) ) ) )   =>    |-  ( ph  ->  ( U  e.  CC  /\  V  e.  CC  /\  W  e.  CC ) )
 
Theoremquartlem3 23841 Closure lemmas for quart 23843. (Contributed by Mario Carneiro, 7-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  E  =  -u ( A  /  4
 ) )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   &    |-  ( ph  ->  U  =  ( ( P ^ 2 )  +  (; 1 2  x.  R ) ) )   &    |-  ( ph  ->  V  =  ( ( -u ( 2  x.  ( P ^ 3 ) )  -  (; 2 7  x.  ( Q ^ 2 ) ) )  +  (; 7 2  x.  ( P  x.  R ) ) ) )   &    |-  ( ph  ->  W  =  ( sqr `  (
 ( V ^ 2
 )  -  ( 4  x.  ( U ^
 3 ) ) ) ) )   &    |-  ( ph  ->  S  =  ( ( sqr `  M )  /  2
 ) )   &    |-  ( ph  ->  M  =  -u ( ( ( ( 2  x.  P )  +  T )  +  ( U  /  T ) )  /  3
 ) )   &    |-  ( ph  ->  T  =  ( ( ( V  +  W ) 
 /  2 )  ^c  ( 1  / 
 3 ) ) )   &    |-  ( ph  ->  T  =/=  0 )   =>    |-  ( ph  ->  ( S  e.  CC  /\  M  e.  CC  /\  T  e.  CC ) )
 
Theoremquartlem4 23842 Closure lemmas for quart 23843. (Contributed by Mario Carneiro, 7-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  E  =  -u ( A  /  4
 ) )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   &    |-  ( ph  ->  U  =  ( ( P ^ 2 )  +  (; 1 2  x.  R ) ) )   &    |-  ( ph  ->  V  =  ( ( -u ( 2  x.  ( P ^ 3 ) )  -  (; 2 7  x.  ( Q ^ 2 ) ) )  +  (; 7 2  x.  ( P  x.  R ) ) ) )   &    |-  ( ph  ->  W  =  ( sqr `  (
 ( V ^ 2
 )  -  ( 4  x.  ( U ^
 3 ) ) ) ) )   &    |-  ( ph  ->  S  =  ( ( sqr `  M )  /  2
 ) )   &    |-  ( ph  ->  M  =  -u ( ( ( ( 2  x.  P )  +  T )  +  ( U  /  T ) )  /  3
 ) )   &    |-  ( ph  ->  T  =  ( ( ( V  +  W ) 
 /  2 )  ^c  ( 1  / 
 3 ) ) )   &    |-  ( ph  ->  T  =/=  0 )   &    |-  ( ph  ->  M  =/=  0 )   &    |-  ( ph  ->  I  =  ( sqr `  ( ( -u ( S ^ 2
 )  -  ( P 
 /  2 ) )  +  ( ( Q 
 /  4 )  /  S ) ) ) )   &    |-  ( ph  ->  J  =  ( sqr `  (
 ( -u ( S ^
 2 )  -  ( P  /  2 ) )  -  ( ( Q 
 /  4 )  /  S ) ) ) )   =>    |-  ( ph  ->  ( S  =/=  0  /\  I  e.  CC  /\  J  e.  CC ) )
 
Theoremquart 23843 The quartic equation, writing out all roots using square and cube root functions so that only direct substitutions remain, and we can actually claim to have a "quartic equation". Naturally, this theorem is ridiculously long (see quartfull 29938) if all the substitutions are performed. This is Metamath 100 proof #46. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  E  =  -u ( A  /  4
 ) )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   &    |-  ( ph  ->  U  =  ( ( P ^ 2 )  +  (; 1 2  x.  R ) ) )   &    |-  ( ph  ->  V  =  ( ( -u ( 2  x.  ( P ^ 3 ) )  -  (; 2 7  x.  ( Q ^ 2 ) ) )  +  (; 7 2  x.  ( P  x.  R ) ) ) )   &    |-  ( ph  ->  W  =  ( sqr `  (
 ( V ^ 2
 )  -  ( 4  x.  ( U ^
 3 ) ) ) ) )   &    |-  ( ph  ->  S  =  ( ( sqr `  M )  /  2
 ) )   &    |-  ( ph  ->  M  =  -u ( ( ( ( 2  x.  P )  +  T )  +  ( U  /  T ) )  /  3
 ) )   &    |-  ( ph  ->  T  =  ( ( ( V  +  W ) 
 /  2 )  ^c  ( 1  / 
 3 ) ) )   &    |-  ( ph  ->  T  =/=  0 )   &    |-  ( ph  ->  M  =/=  0 )   &    |-  ( ph  ->  I  =  ( sqr `  ( ( -u ( S ^ 2
 )  -  ( P 
 /  2 ) )  +  ( ( Q 
 /  4 )  /  S ) ) ) )   &    |-  ( ph  ->  J  =  ( sqr `  (
 ( -u ( S ^
 2 )  -  ( P  /  2 ) )  -  ( ( Q 
 /  4 )  /  S ) ) ) )   =>    |-  ( ph  ->  (
 ( ( ( X ^ 4 )  +  ( A  x.  ( X ^ 3 ) ) )  +  ( ( B  x.  ( X ^ 2 ) )  +  ( ( C  x.  X )  +  D ) ) )  =  0  <->  ( ( X  =  ( ( E  -  S )  +  I )  \/  X  =  ( ( E  -  S )  -  I
 ) )  \/  ( X  =  ( ( E  +  S )  +  J )  \/  X  =  ( ( E  +  S )  -  J ) ) ) ) )
 
14.3.8  Inverse trigonometric functions
 
Syntaxcasin 23844 The arcsine function.
 class arcsin
 
Syntaxcacos 23845 The arccosine function.
 class arccos
 
Syntaxcatan 23846 The arctangent function.
 class arctan
 
Definitiondf-asin 23847 Define the arcsine function. Because  sin is not a one-to-one function, the literal inverse  `' sin is not a function. Rather than attempt to find the right domain on which to restrict  sin in order to get a total function, we just define it in terms of  log, which we already know is total (except at  0). There are branch points at  -u 1 and  1 (at which the function is defined), and branch cuts along the real line not between  -u
1 and  1, which is to say  ( -oo ,  -u 1 )  u.  (
1 , +oo ). (Contributed by Mario Carneiro, 31-Mar-2015.)
 |- arcsin  =  ( x  e.  CC  |->  ( -u _i  x.  ( log `  ( ( _i 
 x.  x )  +  ( sqr `  ( 1  -  ( x ^ 2
 ) ) ) ) ) ) )
 
Definitiondf-acos 23848 Define the arccosine function. See also remarks for df-asin 23847. Since we define arccos in terms of arcsin, it shares the same branch points and cuts, namely  ( -oo ,  -u
1 )  u.  (
1 , +oo ). (Contributed by Mario Carneiro, 31-Mar-2015.)
 |- arccos  =  ( x  e.  CC  |->  ( ( pi  / 
 2 )  -  (arcsin `  x ) ) )
 
Definitiondf-atan 23849 Define the arctangent function. See also remarks for df-asin 23847. Unlike arcsin and arccos, this function is not defined everywhere, because  tan ( z )  =/=  pm _i for all  z  e.  CC. For all other  z, there is a formula for arctan ( z ) in terms of  log, and we take that as the definition. Branch points are at  pm _i; branch cuts are on the pure imaginary axis not between  -u _i and  _i, which is to say  { z  e.  CC  |  ( _i  x.  z )  e.  ( -oo ,  -u
1 )  u.  (
1 , +oo ) }. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |- arctan  =  ( x  e.  ( CC  \  { -u _i ,  _i } )  |->  ( ( _i  /  2
 )  x.  ( ( log `  ( 1  -  ( _i  x.  x ) ) )  -  ( log `  ( 1  +  ( _i  x.  x ) ) ) ) ) )
 
Theoremasinlem 23850 The argument to the logarithm in df-asin 23847 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  CC  ->  ( ( _i  x.  A )  +  ( sqr `  ( 1  -  ( A ^ 2 ) ) ) )  =/=  0 )
 
Theoremasinlem2 23851 The argument to the logarithm in df-asin 23847 has the property that replacing  A with  -u A in the expression gives the reciprocal. (Contributed by Mario Carneiro, 1-Apr-2015.)
 |-  ( A  e.  CC  ->  ( ( ( _i 
 x.  A )  +  ( sqr `  ( 1  -  ( A ^ 2
 ) ) ) )  x.  ( ( _i 
 x.  -u A )  +  ( sqr `  ( 1  -  ( -u A ^ 2
 ) ) ) ) )  =  1 )
 
Theoremasinlem3a 23852 Lemma for asinlem3 23853. (Contributed by Mario Carneiro, 1-Apr-2015.)
 |-  ( ( A  e.  CC  /\  ( Im `  A )  <_  0 ) 
 ->  0  <_  ( Re
 `  ( ( _i 
 x.  A )  +  ( sqr `  ( 1  -  ( A ^ 2
 ) ) ) ) ) )
 
Theoremasinlem3 23853 The argument to the logarithm in df-asin 23847 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.)
 |-  ( A  e.  CC  ->  0  <_  ( Re `  ( ( _i  x.  A )  +  ( sqr `  ( 1  -  ( A ^ 2 ) ) ) ) ) )
 
Theoremasinf 23854 Domain and range of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |- arcsin : CC --> CC
 
Theoremasincl 23855 Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  CC  ->  (arcsin `  A )  e.  CC )
 
Theoremacosf 23856 Domain and range of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |- arccos : CC --> CC
 
Theoremacoscl 23857 Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  CC  ->  (arccos `  A )  e.  CC )
 
Theorematandm 23858 Since the property is a little lengthy, we abbreviate  A  e.  CC  /\  A  =/=  -u _i  /\  A  =/=  _i as  A  e.  dom arctan. This is the necessary precondition for the definition of arctan to make sense. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  dom arctan  <->  ( A  e.  CC  /\  A  =/=  -u _i  /\  A  =/=  _i ) )
 
Theorematandm2 23859 This form of atandm 23858 is a bit more useful for showing that the logarithms in df-atan 23849 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  dom arctan  <->  ( A  e.  CC  /\  ( 1  -  ( _i  x.  A ) )  =/=  0  /\  ( 1  +  ( _i  x.  A ) )  =/=  0 ) )
 
Theorematandm3 23860 A compact form of atandm 23858. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  dom arctan  <->  ( A  e.  CC  /\  ( A ^
 2 )  =/=  -u 1
 ) )
 
Theorematandm4 23861 A compact form of atandm 23858. (Contributed by Mario Carneiro, 3-Apr-2015.)
 |-  ( A  e.  dom arctan  <->  ( A  e.  CC  /\  ( 1  +  ( A ^ 2
 ) )  =/=  0
 ) )
 
Theorematanf 23862 Domain and range of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |- arctan : ( CC  \  { -u _i ,  _i } ) --> CC
 
Theorematancl 23863 Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  dom arctan  ->  (arctan `  A )  e. 
 CC )
 
Theoremasinval 23864 Value of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  CC  ->  (arcsin `  A )  =  ( -u _i  x.  ( log `  ( ( _i 
 x.  A )  +  ( sqr `  ( 1  -  ( A ^ 2
 ) ) ) ) ) ) )
 
Theoremacosval 23865 Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  CC  ->  (arccos `  A )  =  ( ( pi  / 
 2 )  -  (arcsin `  A ) ) )
 
Theorematanval 23866 Value of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  dom arctan  ->  (arctan `  A )  =  ( ( _i  / 
 2 )  x.  (
 ( log `  ( 1  -  ( _i  x.  A ) ) )  -  ( log `  ( 1  +  ( _i  x.  A ) ) ) ) ) )
 
Theorematanre 23867 A real number is in the domain of the arctangent function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  RR  ->  A  e.  dom arctan )
 
Theoremasinneg 23868 The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.)
 |-  ( A  e.  CC  ->  (arcsin `  -u A )  =  -u (arcsin `  A ) )
 
Theoremacosneg 23869 The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  CC  ->  (arccos `  -u A )  =  ( pi  -  (arccos `  A ) ) )
 
Theoremefiasin 23870 The exponential of the arcsine function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  CC  ->  ( exp `  ( _i  x.  (arcsin `  A ) ) )  =  ( ( _i  x.  A )  +  ( sqr `  ( 1  -  ( A ^ 2 ) ) ) ) )
 
Theoremsinasin 23871 The arcsine function is an inverse to  sin. This is the main property that justifies the notation arcsin or  sin
^ -u 1. Because  sin is not an injection, the other converse identity asinsin 23874 is only true under limited circumstances. (Contributed by Mario Carneiro, 1-Apr-2015.)
 |-  ( A  e.  CC  ->  ( sin `  (arcsin `  A ) )  =  A )
 
Theoremcosacos 23872 The arccosine function is an inverse to  cos. (Contributed by Mario Carneiro, 1-Apr-2015.)
 |-  ( A  e.  CC  ->  ( cos `  (arccos `  A ) )  =  A )
 
Theoremasinsinlem 23873 Lemma for asinsin 23874. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( ( A  e.  CC  /\  ( Re `  A )  e.  ( -u ( pi  /  2
 ) (,) ( pi  / 
 2 ) ) ) 
 ->  0  <  ( Re
 `  ( exp `  ( _i  x.  A ) ) ) )
 
Theoremasinsin 23874 The arcsine function composed with 
sin is equal to the identity. This plus sinasin 23871 allow us to view  sin and arcsin as inverse operations to each other. For ease of use, we have not defined precisely the correct domain of correctness of this identity; in addition to the main region described here it is also true for some points on the branch cuts, namely when  A  =  ( pi 
/  2 )  -  _i y for nonnegative real  y and also symmetrically at  A  =  _i y  -  ( pi  / 
2 ). In particular, when restricted to reals this identity extends to the closed interval  [ -u (
pi  /  2 ) ,  ( pi  / 
2 ) ], not just the open interval (see reasinsin 23878). (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( ( A  e.  CC  /\  ( Re `  A )  e.  ( -u ( pi  /  2
 ) (,) ( pi  / 
 2 ) ) ) 
 ->  (arcsin `  ( sin `  A ) )  =  A )
 
Theoremacoscos 23875 The arccosine function is an inverse to  cos. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( ( A  e.  CC  /\  ( Re `  A )  e.  (
 0 (,) pi ) ) 
 ->  (arccos `  ( cos `  A ) )  =  A )
 
Theoremasin1 23876 The arcsine of  1 is  pi  / 
2. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  (arcsin `  1 )  =  ( pi  /  2
 )
 
Theoremacos1 23877 The arcsine of  1 is  pi  / 
2. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  (arccos `  1 )  =  0
 
Theoremreasinsin 23878 The arcsine function composed with 
sin is equal to the identity. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  ( -u ( pi  /  2
 ) [,] ( pi  / 
 2 ) )  ->  (arcsin `  ( sin `  A ) )  =  A )
 
Theoremasinsinb 23879 Relationship between sine and arcsine. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( ( A  e.  CC  /\  B  e.  CC  /\  ( Re `  B )  e.  ( -u ( pi  /  2 ) (,) ( pi  /  2
 ) ) )  ->  ( (arcsin `  A )  =  B  <->  ( sin `  B )  =  A )
 )
 
Theoremacoscosb 23880 Relationship between sine and arcsine. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( ( A  e.  CC  /\  B  e.  CC  /\  ( Re `  B )  e.  ( 0 (,) pi ) )  ->  ( (arccos `  A )  =  B  <->  ( cos `  B )  =  A )
 )
 
Theoremasinbnd 23881 The arcsine function has range within a vertical strip of the complex plane with real part between  -u pi  /  2 and  pi  /  2. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  CC  ->  ( Re `  (arcsin `  A ) )  e.  ( -u ( pi  / 
 2 ) [,] ( pi  /  2 ) ) )
 
Theoremacosbnd 23882 The arccosine function has range within a vertical strip of the complex plane with real part between  0 and  pi. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  CC  ->  ( Re `  (arccos `  A ) )  e.  ( 0 [,] pi ) )
 
Theoremasinrebnd 23883 Bounds on the arcsine function. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  ( -u 1 [,] 1 ) 
 ->  (arcsin `  A )  e.  ( -u ( pi  / 
 2 ) [,] ( pi  /  2 ) ) )
 
Theoremasinrecl 23884 The arcsine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  ( -u 1 [,] 1 ) 
 ->  (arcsin `  A )  e.  RR )
 
Theoremacosrecl 23885 The arccosine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  ( -u 1 [,] 1 ) 
 ->  (arccos `  A )  e.  RR )
 
Theoremcosasin 23886 The cosine of the arcsine of  A is  sqr ( 1  -  A ^ 2 ). (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  CC  ->  ( cos `  (arcsin `  A ) )  =  ( sqr `  (
 1  -  ( A ^ 2 ) ) ) )
 
Theoremsinacos 23887 The sine of the arccosine of  A is  sqr ( 1  -  A ^ 2 ). (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  CC  ->  ( sin `  (arccos `  A ) )  =  ( sqr `  (
 1  -  ( A ^ 2 ) ) ) )
 
Theorematandmneg 23888 The domain of the arctangent function is closed under negatives. (Contributed by Mario Carneiro, 3-Apr-2015.)
 |-  ( A  e.  dom arctan  ->  -u A  e.  dom arctan )
 
Theorematanneg 23889 The arctangent function is odd. (Contributed by Mario Carneiro, 3-Apr-2015.)
 |-  ( A  e.  dom arctan  ->  (arctan `  -u A )  =  -u (arctan `  A )
 )
 
Theorematan0 23890 The arctangent of zero is zero. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  (arctan `  0 )  =  0
 
Theorematandmcj 23891 The arctangent function distributes under conjugation. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  dom arctan  ->  ( * `  A )  e.  dom arctan )
 
Theorematancj 23892 The arctangent function distributes under conjugation. (The condition that  Re ( A )  =/=  0 is necessary because the branch cuts are chosen so that the negative imaginary line "agrees with" neighboring values with negative real part, while the positive imaginary line agrees with values with positive real part. This makes atanneg 23889 true unconditionally but messes up conjugation symmetry, and it is impossible to have both in a single-valued function. The claim is true on the imaginary line between  -u 1 and  1, though.) (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( ( A  e.  CC  /\  ( Re `  A )  =/=  0
 )  ->  ( A  e.  dom arctan  /\  ( * `  (arctan `  A ) )  =  (arctan `  ( * `  A ) ) ) )
 
Theorematanrecl 23893 The arctangent function is real for all real inputs. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  RR  ->  (arctan `  A )  e.  RR )
 
Theoremefiatan 23894 Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  dom arctan  ->  ( exp `  ( _i  x.  (arctan `  A )
 ) )  =  ( ( sqr `  (
 1  +  ( _i 
 x.  A ) ) )  /  ( sqr `  ( 1  -  ( _i  x.  A ) ) ) ) )
 
Theorematanlogaddlem 23895 Lemma for atanlogadd 23896. (Contributed by Mario Carneiro, 3-Apr-2015.)
 |-  ( ( A  e.  dom arctan  /\  0  <_  ( Re `  A ) )  ->  ( ( log `  (
 1  +  ( _i 
 x.  A ) ) )  +  ( log `  ( 1  -  ( _i  x.  A ) ) ) )  e.  ran  log )
 
Theorematanlogadd 23896 The rule  sqr ( z w )  =  ( sqr z ) ( sqr w ) is not always true on the complex numbers, but it is true when the arguments of  z and  w sum to within the interval  ( -u pi ,  pi ], so there are some cases such as this one with  z  =  1  +  _i A and  w  =  1  -  _i A which are true unconditionally. This result can also be stated as " sqr ( 1  +  z )  +  sqr ( 1  -  z
) is analytic". (Contributed by Mario Carneiro, 3-Apr-2015.)
 |-  ( A  e.  dom arctan  ->  ( ( log `  (
 1  +  ( _i 
 x.  A ) ) )  +  ( log `  ( 1  -  ( _i  x.  A ) ) ) )  e.  ran  log )
 
Theorematanlogsublem 23897 Lemma for atanlogsub 23898. (Contributed by Mario Carneiro, 4-Apr-2015.)
 |-  ( ( A  e.  dom arctan  /\  0  <  ( Re
 `  A ) ) 
 ->  ( Im `  (
 ( log `  ( 1  +  ( _i  x.  A ) ) )  -  ( log `  ( 1  -  ( _i  x.  A ) ) ) ) )  e.  ( -u pi (,) pi ) )
 
Theorematanlogsub 23898 A variation on atanlogadd 23896, to show that  sqr ( 1  +  _i z )  /  sqr ( 1  -  _i z )  =  sqr ( ( 1  +  _i z )  /  ( 1  -  _i z ) ) under more limited conditions. (Contributed by Mario Carneiro, 4-Apr-2015.)
 |-  ( ( A  e.  dom arctan  /\  ( Re `  A )  =/=  0 )  ->  ( ( log `  (
 1  +  ( _i 
 x.  A ) ) )  -  ( log `  ( 1  -  ( _i  x.  A ) ) ) )  e.  ran  log )
 
Theoremefiatan2 23899 Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 3-Apr-2015.)
 |-  ( A  e.  dom arctan  ->  ( exp `  ( _i  x.  (arctan `  A )
 ) )  =  ( ( 1  +  ( _i  x.  A ) ) 
 /  ( sqr `  (
 1  +  ( A ^ 2 ) ) ) ) )
 
Theorem2efiatan 23900 Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  dom arctan  ->  ( exp `  ( 2  x.  ( _i  x.  (arctan `  A ) ) ) )  =  ( ( ( 2  x.  _i )  /  ( A  +  _i ) )  -  1
 ) )
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