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Theorem List for Metamath Proof Explorer - 25401-25500   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremisosctrlem2 25401 Lemma for isosctr 25403. 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 ∈ ℂ ∧ (abs‘𝐴) = 1 ∧ ¬ 1 = 𝐴) → (ℑ‘(log‘(1 − 𝐴))) = (ℑ‘(log‘(-𝐴 / (1 − 𝐴)))))

Theoremisosctrlem3 25402* Lemma for isosctr 25403. Corresponds to the case where one vertex is at 0. (Contributed by Saveliy Skresanov, 1-Jan-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))       (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0 ∧ 𝐴𝐵) ∧ (abs‘𝐴) = (abs‘𝐵)) → (-𝐴𝐹(𝐵𝐴)) = ((𝐴𝐵)𝐹-𝐵))

Theoremisosctr 25403* Isosceles triangle theorem. This is Metamath 100 proof #65. (Contributed by Saveliy Skresanov, 1-Jan-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))       (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴𝐶𝐵𝐶𝐴𝐵) ∧ (abs‘(𝐴𝐶)) = (abs‘(𝐵𝐶))) → ((𝐶𝐴)𝐹(𝐵𝐴)) = ((𝐴𝐵)𝐹(𝐶𝐵)))

Theoremssscongptld 25404* 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 25398 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 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝐸 ∈ ℂ)    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)    &   (𝜑𝐷𝐸)    &   (𝜑𝐸𝐺)    &   (𝜑 → (abs‘(𝐴𝐵)) = (abs‘(𝐷𝐸)))    &   (𝜑 → (abs‘(𝐵𝐶)) = (abs‘(𝐸𝐺)))    &   (𝜑 → (abs‘(𝐶𝐴)) = (abs‘(𝐺𝐷)))       (𝜑 → (cos‘((𝐴𝐵)𝐹(𝐶𝐵))) = (cos‘((𝐷𝐸)𝐹(𝐺𝐸))))

Theoremaffineequiv 25405 Equivalence between two ways of expressing 𝐵 as an affine combination of 𝐴 and 𝐶. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → (𝐵 = ((𝐷 · 𝐴) + ((1 − 𝐷) · 𝐶)) ↔ (𝐶𝐵) = (𝐷 · (𝐶𝐴))))

Theoremaffineequiv2 25406 Equivalence between two ways of expressing 𝐵 as an affine combination of 𝐴 and 𝐶. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → (𝐵 = ((𝐷 · 𝐴) + ((1 − 𝐷) · 𝐶)) ↔ (𝐵𝐴) = ((1 − 𝐷) · (𝐶𝐴))))

Theoremaffineequiv3 25407 Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶. (Contributed by AV, 22-Jan-2023.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ (𝐴𝐵) = (𝐷 · (𝐶𝐵))))

Theoremaffineequiv4 25408 Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶. (Contributed by AV, 22-Jan-2023.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ 𝐴 = ((𝐷 · (𝐶𝐵)) + 𝐵)))

Theoremaffineequivne 25409 Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶 if 𝐵 and 𝐶 are not equal. (Contributed by AV, 22-Jan-2023.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝐵𝐶)       (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ 𝐷 = ((𝐴𝐵) / (𝐶𝐵))))

Theoremangpieqvdlem 25410 Equivalence used in the proof of angpieqvd 25413. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐴𝐶)       (𝜑 → (-((𝐶𝐵) / (𝐴𝐵)) ∈ ℝ+ ↔ ((𝐶𝐵) / (𝐶𝐴)) ∈ (0(,)1)))

Theoremangpieqvdlem2 25411* Equivalence used in angpieqvd 25413. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)       (𝜑 → (-((𝐶𝐵) / (𝐴𝐵)) ∈ ℝ+ ↔ ((𝐴𝐵)𝐹(𝐶𝐵)) = π))

Theoremangpined 25412* If the angle at ABC is π, then A is not equal to C. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)       (𝜑 → (((𝐴𝐵)𝐹(𝐶𝐵)) = π → 𝐴𝐶))

Theoremangpieqvd 25413* The angle ABC is π 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 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)       (𝜑 → (((𝐴𝐵)𝐹(𝐶𝐵)) = π ↔ ∃𝑤 ∈ (0(,)1)𝐵 = ((𝑤 · 𝐴) + ((1 − 𝑤) · 𝐶))))

Theoremchordthmlem 25414* If M is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 25404 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 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑀 = ((𝐴 + 𝐵) / 2))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))    &   (𝜑𝐴𝐵)    &   (𝜑𝑄𝑀)       (𝜑 → ((𝑄𝑀)𝐹(𝐵𝑀)) ∈ {(π / 2), -(π / 2)})

Theoremchordthmlem2 25415* 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 25414, where P = B, and using angrtmuld 25390 to observe that QMP is right iff QMB is. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℝ)    &   (𝜑𝑀 = ((𝐴 + 𝐵) / 2))    &   (𝜑𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))    &   (𝜑𝑃𝑀)    &   (𝜑𝑄𝑀)       (𝜑 → ((𝑄𝑀)𝐹(𝑃𝑀)) ∈ {(π / 2), -(π / 2)})

Theoremchordthmlem3 25416 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 25415 and the Pythagorean theorem (pythag 25399) 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.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℝ)    &   (𝜑𝑀 = ((𝐴 + 𝐵) / 2))    &   (𝜑𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))       (𝜑 → ((abs‘(𝑃𝑄))↑2) = (((abs‘(𝑄𝑀))↑2) + ((abs‘(𝑃𝑀))↑2)))

Theoremchordthmlem4 25417 If P is on the segment AB and M is the midpoint of AB, then PA · PB = BM 2 PM 2 . If all lengths are reexpressed as fractions of AB, this reduces to the identity 𝑋 · (1 − 𝑋) = (1 / 2) 2 − ((1 / 2) − 𝑋) 2 . (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑋 ∈ (0[,]1))    &   (𝜑𝑀 = ((𝐴 + 𝐵) / 2))    &   (𝜑𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))       (𝜑 → ((abs‘(𝑃𝐴)) · (abs‘(𝑃𝐵))) = (((abs‘(𝐵𝑀))↑2) − ((abs‘(𝑃𝑀))↑2)))

Theoremchordthmlem5 25418 If P is on the segment AB and AQ = BQ, then PA · PB = BQ 2 PQ 2 . This follows from two uses of chordthmlem3 25416 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 · PB by chordthmlem4 25417. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ (0[,]1))    &   (𝜑𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))       (𝜑 → ((abs‘(𝑃𝐴)) · (abs‘(𝑃𝐵))) = (((abs‘(𝐵𝑄))↑2) − ((abs‘(𝑃𝑄))↑2)))

Theoremchordthm 25419* 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 · PB and PC · 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 π. The result is proven by using chordthmlem5 25418 twice to show that PA · PB and PC · 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 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑃 ∈ ℂ)    &   (𝜑𝐴𝑃)    &   (𝜑𝐵𝑃)    &   (𝜑𝐶𝑃)    &   (𝜑𝐷𝑃)    &   (𝜑 → ((𝐴𝑃)𝐹(𝐵𝑃)) = π)    &   (𝜑 → ((𝐶𝑃)𝐹(𝐷𝑃)) = π)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐶𝑄)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐷𝑄)))       (𝜑 → ((abs‘(𝑃𝐴)) · (abs‘(𝑃𝐵))) = ((abs‘(𝑃𝐶)) · (abs‘(𝑃𝐷))))

Theoremheron 25420* 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 · 𝑌 · abs(sin𝑂), is equal to the square root of 𝑆 · (𝑆𝑋) · (𝑆𝑌) · (𝑆𝑍), 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 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   𝑋 = (abs‘(𝐵𝐶))    &   𝑌 = (abs‘(𝐴𝐶))    &   𝑍 = (abs‘(𝐴𝐵))    &   𝑂 = ((𝐵𝐶)𝐹(𝐴𝐶))    &   𝑆 = (((𝑋 + 𝑌) + 𝑍) / 2)    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐶)    &   (𝜑𝐵𝐶)       (𝜑 → (((1 / 2) · (𝑋 · 𝑌)) · (abs‘(sin‘𝑂))) = (√‘((𝑆 · (𝑆𝑋)) · ((𝑆𝑌) · (𝑆𝑍)))))

14.3.7  Solutions of quadratic, cubic, and quartic equations

Theoremquad2 25421 The quadratic equation, without specifying the particular branch 𝐷 to the square root. (Contributed by Mario Carneiro, 23-Apr-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑 → (𝐷↑2) = ((𝐵↑2) − (4 · (𝐴 · 𝐶))))       (𝜑 → (((𝐴 · (𝑋↑2)) + ((𝐵 · 𝑋) + 𝐶)) = 0 ↔ (𝑋 = ((-𝐵 + 𝐷) / (2 · 𝐴)) ∨ 𝑋 = ((-𝐵𝐷) / (2 · 𝐴)))))

(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝐷 = ((𝐵↑2) − (4 · (𝐴 · 𝐶))))       (𝜑 → (((𝐴 · (𝑋↑2)) + ((𝐵 · 𝑋) + 𝐶)) = 0 ↔ (𝑋 = ((-𝐵 + (√‘𝐷)) / (2 · 𝐴)) ∨ 𝑋 = ((-𝐵 − (√‘𝐷)) / (2 · 𝐴)))))

Theorem1cubrlem 25423 The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
((-1↑𝑐(2 / 3)) = ((-1 + (i · (√‘3))) / 2) ∧ ((-1↑𝑐(2 / 3))↑2) = ((-1 − (i · (√‘3))) / 2))

Theorem1cubr 25424 The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
𝑅 = {1, ((-1 + (i · (√‘3))) / 2), ((-1 − (i · (√‘3))) / 2)}       (𝐴𝑅 ↔ (𝐴 ∈ ℂ ∧ (𝐴↑3) = 1))

Theoremdcubic1lem 25425 Lemma for dcubic1 25427 and dcubic2 25426: simplify the cubic equation under the substitution 𝑋 = 𝑈𝑀 / 𝑈. (Contributed by Mario Carneiro, 26-Apr-2015.)
(𝜑𝑃 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 ∈ ℂ)    &   (𝜑 → (𝑇↑3) = (𝐺𝑁))    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   (𝜑𝑀 = (𝑃 / 3))    &   (𝜑𝑁 = (𝑄 / 2))    &   (𝜑𝑇 ≠ 0)    &   (𝜑𝑈 ∈ ℂ)    &   (𝜑𝑈 ≠ 0)    &   (𝜑𝑋 = (𝑈 − (𝑀 / 𝑈)))       (𝜑 → (((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0 ↔ (((𝑈↑3)↑2) + ((𝑄 · (𝑈↑3)) − (𝑀↑3))) = 0))

Theoremdcubic2 25426* Reverse direction of dcubic 25428. Given a solution 𝑈 to the "substitution" quadratic equation 𝑋 = 𝑈𝑀 / 𝑈, show that 𝑋 is in the desired form. (Contributed by Mario Carneiro, 25-Apr-2015.)
(𝜑𝑃 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 ∈ ℂ)    &   (𝜑 → (𝑇↑3) = (𝐺𝑁))    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   (𝜑𝑀 = (𝑃 / 3))    &   (𝜑𝑁 = (𝑄 / 2))    &   (𝜑𝑇 ≠ 0)    &   (𝜑𝑈 ∈ ℂ)    &   (𝜑𝑈 ≠ 0)    &   (𝜑𝑋 = (𝑈 − (𝑀 / 𝑈)))    &   (𝜑 → ((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0)       (𝜑 → ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = ((𝑟 · 𝑇) − (𝑀 / (𝑟 · 𝑇)))))

Theoremdcubic1 25427 Forward direction of dcubic 25428: the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015.)
(𝜑𝑃 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 ∈ ℂ)    &   (𝜑 → (𝑇↑3) = (𝐺𝑁))    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   (𝜑𝑀 = (𝑃 / 3))    &   (𝜑𝑁 = (𝑄 / 2))    &   (𝜑𝑇 ≠ 0)    &   (𝜑𝑋 = (𝑇 − (𝑀 / 𝑇)))       (𝜑 → ((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0)

Theoremdcubic 25428* Solutions to the depressed cubic, a special case of cubic 25431. (The definitions of 𝑀, 𝑁, 𝐺, 𝑇 here differ from mcubic 25429 by scale factors of -9, 54, 54 and -27 respectively, to simplify the algebra and presentation.) (Contributed by Mario Carneiro, 26-Apr-2015.)
(𝜑𝑃 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 ∈ ℂ)    &   (𝜑 → (𝑇↑3) = (𝐺𝑁))    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   (𝜑𝑀 = (𝑃 / 3))    &   (𝜑𝑁 = (𝑄 / 2))    &   (𝜑𝑇 ≠ 0)       (𝜑 → (((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = ((𝑟 · 𝑇) − (𝑀 / (𝑟 · 𝑇))))))

Theoremmcubic 25429* Solutions to a monic cubic equation, a special case of cubic 25431. (Contributed by Mario Carneiro, 24-Apr-2015.)
(𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 ∈ ℂ)    &   (𝜑 → (𝑇↑3) = ((𝑁 + 𝐺) / 2))    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑 → (𝐺↑2) = ((𝑁↑2) − (4 · (𝑀↑3))))    &   (𝜑𝑀 = ((𝐵↑2) − (3 · 𝐶)))    &   (𝜑𝑁 = (((2 · (𝐵↑3)) − (9 · (𝐵 · 𝐶))) + (27 · 𝐷)))    &   (𝜑𝑇 ≠ 0)       (𝜑 → ((((𝑋↑3) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / 3))))

Theoremcubic2 25430* 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.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 ∈ ℂ)    &   (𝜑 → (𝑇↑3) = ((𝑁 + 𝐺) / 2))    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑 → (𝐺↑2) = ((𝑁↑2) − (4 · (𝑀↑3))))    &   (𝜑𝑀 = ((𝐵↑2) − (3 · (𝐴 · 𝐶))))    &   (𝜑𝑁 = (((2 · (𝐵↑3)) − ((9 · 𝐴) · (𝐵 · 𝐶))) + (27 · ((𝐴↑2) · 𝐷))))    &   (𝜑𝑇 ≠ 0)       (𝜑 → ((((𝐴 · (𝑋↑3)) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / (3 · 𝐴)))))

Theoremcubic 25431* The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4626 to convert the existential quantifier to a triple disjunction. This is Metamath 100 proof #37. (Contributed by Mario Carneiro, 26-Apr-2015.)
𝑅 = {1, ((-1 + (i · (√‘3))) / 2), ((-1 − (i · (√‘3))) / 2)}    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 = (((𝑁 + (√‘𝐺)) / 2)↑𝑐(1 / 3)))    &   (𝜑𝐺 = ((𝑁↑2) − (4 · (𝑀↑3))))    &   (𝜑𝑀 = ((𝐵↑2) − (3 · (𝐴 · 𝐶))))    &   (𝜑𝑁 = (((2 · (𝐵↑3)) − ((9 · 𝐴) · (𝐵 · 𝐶))) + (27 · ((𝐴↑2) · 𝐷))))    &   (𝜑𝑀 ≠ 0)       (𝜑 → ((((𝐴 · (𝑋↑3)) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟𝑅 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / (3 · 𝐴))))

Theorembinom4 25432 Work out a quartic binomial. (You would think that by this point it would be faster to use binom 15181, 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 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴 + 𝐵)↑4) = (((𝐴↑4) + (4 · ((𝐴↑3) · 𝐵))) + ((6 · ((𝐴↑2) · (𝐵↑2))) + ((4 · (𝐴 · (𝐵↑3))) + (𝐵↑4)))))

Theoremdquartlem1 25433 Lemma for dquart 25435. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑆 ∈ ℂ)    &   (𝜑𝑀 = ((2 · 𝑆)↑2))    &   (𝜑𝑀 ≠ 0)    &   (𝜑𝐼 ∈ ℂ)    &   (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))       (𝜑 → ((((𝑋↑2) + ((𝑀 + 𝐵) / 2)) + ((((𝑀 / 2) · 𝑋) − (𝐶 / 4)) / 𝑆)) = 0 ↔ (𝑋 = (-𝑆 + 𝐼) ∨ 𝑋 = (-𝑆𝐼))))

Theoremdquartlem2 25434 Lemma for dquart 25435. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑆 ∈ ℂ)    &   (𝜑𝑀 = ((2 · 𝑆)↑2))    &   (𝜑𝑀 ≠ 0)    &   (𝜑𝐼 ∈ ℂ)    &   (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑 → (((𝑀↑3) + ((2 · 𝐵) · (𝑀↑2))) + ((((𝐵↑2) − (4 · 𝐷)) · 𝑀) + -(𝐶↑2))) = 0)       (𝜑 → ((((𝑀 + 𝐵) / 2)↑2) − (((𝐶↑2) / 4) / 𝑀)) = 𝐷)

Theoremdquart 25435 Solve a depressed quartic equation. To eliminate 𝑆, which is the square root of a solution 𝑀 to the resolvent cubic equation, apply cubic 25431 or one of its variants. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑆 ∈ ℂ)    &   (𝜑𝑀 = ((2 · 𝑆)↑2))    &   (𝜑𝑀 ≠ 0)    &   (𝜑𝐼 ∈ ℂ)    &   (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑 → (((𝑀↑3) + ((2 · 𝐵) · (𝑀↑2))) + ((((𝐵↑2) − (4 · 𝐷)) · 𝑀) + -(𝐶↑2))) = 0)    &   (𝜑𝐽 ∈ ℂ)    &   (𝜑 → (𝐽↑2) = ((-(𝑆↑2) − (𝐵 / 2)) − ((𝐶 / 4) / 𝑆)))       (𝜑 → ((((𝑋↑4) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ((𝑋 = (-𝑆 + 𝐼) ∨ 𝑋 = (-𝑆𝐼)) ∨ (𝑋 = (𝑆 + 𝐽) ∨ 𝑋 = (𝑆𝐽)))))

Theoremquart1cl 25436 Closure lemmas for quart 25443. (Contributed by Mario Carneiro, 7-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))       (𝜑 → (𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ ∧ 𝑅 ∈ ℂ))

Theoremquart1lem 25437 Lemma for quart1 25438. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑌 = (𝑋 + (𝐴 / 4)))       (𝜑𝐷 = ((((𝐴↑4) / 256) + (𝑃 · ((𝐴 / 4)↑2))) + ((𝑄 · (𝐴 / 4)) + 𝑅)))

Theoremquart1 25438 Depress a quartic equation. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑌 = (𝑋 + (𝐴 / 4)))       (𝜑 → (((𝑋↑4) + (𝐴 · (𝑋↑3))) + ((𝐵 · (𝑋↑2)) + ((𝐶 · 𝑋) + 𝐷))) = (((𝑌↑4) + (𝑃 · (𝑌↑2))) + ((𝑄 · 𝑌) + 𝑅)))

Theoremquartlem1 25439 Lemma for quart 25443. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝑃 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑅 ∈ ℂ)    &   (𝜑𝑈 = ((𝑃↑2) + (12 · 𝑅)))    &   (𝜑𝑉 = ((-(2 · (𝑃↑3)) − (27 · (𝑄↑2))) + (72 · (𝑃 · 𝑅))))       (𝜑 → (𝑈 = (((2 · 𝑃)↑2) − (3 · ((𝑃↑2) − (4 · 𝑅)))) ∧ 𝑉 = (((2 · ((2 · 𝑃)↑3)) − (9 · ((2 · 𝑃) · ((𝑃↑2) − (4 · 𝑅))))) + (27 · -(𝑄↑2)))))

Theoremquartlem2 25440 Closure lemmas for quart 25443. (Contributed by Mario Carneiro, 7-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝐸 = -(𝐴 / 4))    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))    &   (𝜑𝑈 = ((𝑃↑2) + (12 · 𝑅)))    &   (𝜑𝑉 = ((-(2 · (𝑃↑3)) − (27 · (𝑄↑2))) + (72 · (𝑃 · 𝑅))))    &   (𝜑𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))       (𝜑 → (𝑈 ∈ ℂ ∧ 𝑉 ∈ ℂ ∧ 𝑊 ∈ ℂ))

Theoremquartlem3 25441 Closure lemmas for quart 25443. (Contributed by Mario Carneiro, 7-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝐸 = -(𝐴 / 4))    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))    &   (𝜑𝑈 = ((𝑃↑2) + (12 · 𝑅)))    &   (𝜑𝑉 = ((-(2 · (𝑃↑3)) − (27 · (𝑄↑2))) + (72 · (𝑃 · 𝑅))))    &   (𝜑𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   (𝜑𝑆 = ((√‘𝑀) / 2))    &   (𝜑𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   (𝜑𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   (𝜑𝑇 ≠ 0)       (𝜑 → (𝑆 ∈ ℂ ∧ 𝑀 ∈ ℂ ∧ 𝑇 ∈ ℂ))

Theoremquartlem4 25442 Closure lemmas for quart 25443. (Contributed by Mario Carneiro, 7-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝐸 = -(𝐴 / 4))    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))    &   (𝜑𝑈 = ((𝑃↑2) + (12 · 𝑅)))    &   (𝜑𝑉 = ((-(2 · (𝑃↑3)) − (27 · (𝑄↑2))) + (72 · (𝑃 · 𝑅))))    &   (𝜑𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   (𝜑𝑆 = ((√‘𝑀) / 2))    &   (𝜑𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   (𝜑𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   (𝜑𝑇 ≠ 0)    &   (𝜑𝑀 ≠ 0)    &   (𝜑𝐼 = (√‘((-(𝑆↑2) − (𝑃 / 2)) + ((𝑄 / 4) / 𝑆))))    &   (𝜑𝐽 = (√‘((-(𝑆↑2) − (𝑃 / 2)) − ((𝑄 / 4) / 𝑆))))       (𝜑 → (𝑆 ≠ 0 ∧ 𝐼 ∈ ℂ ∧ 𝐽 ∈ ℂ))

Theoremquart 25443 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 32437) if all the substitutions are performed. This is Metamath 100 proof #46. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝐸 = -(𝐴 / 4))    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))    &   (𝜑𝑈 = ((𝑃↑2) + (12 · 𝑅)))    &   (𝜑𝑉 = ((-(2 · (𝑃↑3)) − (27 · (𝑄↑2))) + (72 · (𝑃 · 𝑅))))    &   (𝜑𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   (𝜑𝑆 = ((√‘𝑀) / 2))    &   (𝜑𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   (𝜑𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   (𝜑𝑇 ≠ 0)    &   (𝜑𝑀 ≠ 0)    &   (𝜑𝐼 = (√‘((-(𝑆↑2) − (𝑃 / 2)) + ((𝑄 / 4) / 𝑆))))    &   (𝜑𝐽 = (√‘((-(𝑆↑2) − (𝑃 / 2)) − ((𝑄 / 4) / 𝑆))))       (𝜑 → ((((𝑋↑4) + (𝐴 · (𝑋↑3))) + ((𝐵 · (𝑋↑2)) + ((𝐶 · 𝑋) + 𝐷))) = 0 ↔ ((𝑋 = ((𝐸𝑆) + 𝐼) ∨ 𝑋 = ((𝐸𝑆) − 𝐼)) ∨ (𝑋 = ((𝐸 + 𝑆) + 𝐽) ∨ 𝑋 = ((𝐸 + 𝑆) − 𝐽)))))

14.3.8  Inverse trigonometric functions

Syntaxcasin 25444 The arcsine function.
class arcsin

Syntaxcacos 25445 The arccosine function.
class arccos

Syntaxcatan 25446 The arctangent function.
class arctan

Definitiondf-asin 25447 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 -1 and 1 (at which the function is defined), and branch cuts along the real line not between -1 and 1, which is to say (-∞, -1) ∪ (1, +∞). (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin = (𝑥 ∈ ℂ ↦ (-i · (log‘((i · 𝑥) + (√‘(1 − (𝑥↑2)))))))

Definitiondf-acos 25448 Define the arccosine function. See also remarks for df-asin 25447. Since we define arccos in terms of arcsin, it shares the same branch points and cuts, namely (-∞, -1) ∪ (1, +∞). (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos = (𝑥 ∈ ℂ ↦ ((π / 2) − (arcsin‘𝑥)))

Definitiondf-atan 25449 Define the arctangent function. See also remarks for df-asin 25447. Unlike arcsin and arccos, this function is not defined everywhere, because tan(𝑧) ≠ ±i for all 𝑧 ∈ ℂ. 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 ±i; branch cuts are on the pure imaginary axis not between -i and i, which is to say {𝑧 ∈ ℂ ∣ (i · 𝑧) ∈ (-∞, -1) ∪ (1, +∞)}. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan = (𝑥 ∈ (ℂ ∖ {-i, i}) ↦ ((i / 2) · ((log‘(1 − (i · 𝑥))) − (log‘(1 + (i · 𝑥))))))

Theoremasinlem 25450 The argument to the logarithm in df-asin 25447 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → ((i · 𝐴) + (√‘(1 − (𝐴↑2)))) ≠ 0)

Theoremasinlem2 25451 The argument to the logarithm in df-asin 25447 has the property that replacing 𝐴 with -𝐴 in the expression gives the reciprocal. (Contributed by Mario Carneiro, 1-Apr-2015.)
(𝐴 ∈ ℂ → (((i · 𝐴) + (√‘(1 − (𝐴↑2)))) · ((i · -𝐴) + (√‘(1 − (-𝐴↑2))))) = 1)

Theoremasinlem3a 25452 Lemma for asinlem3 25453. (Contributed by Mario Carneiro, 1-Apr-2015.)
((𝐴 ∈ ℂ ∧ (ℑ‘𝐴) ≤ 0) → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))

Theoremasinlem3 25453 The argument to the logarithm in df-asin 25447 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.)
(𝐴 ∈ ℂ → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))

Theoremasinf 25454 Domain and range of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin:ℂ⟶ℂ

Theoremasincl 25455 Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (arcsin‘𝐴) ∈ ℂ)

Theoremacosf 25456 Domain and range of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos:ℂ⟶ℂ

Theoremacoscl 25457 Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (arccos‘𝐴) ∈ ℂ)

Theorematandm 25458 Since the property is a little lengthy, we abbreviate 𝐴 ∈ ℂ ∧ 𝐴 ≠ -i ∧ 𝐴 ≠ i as 𝐴 ∈ dom arctan. This is the necessary precondition for the definition of arctan to make sense. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ 𝐴 ≠ -i ∧ 𝐴 ≠ i))

Theorematandm2 25459 This form of atandm 25458 is a bit more useful for showing that the logarithms in df-atan 25449 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 − (i · 𝐴)) ≠ 0 ∧ (1 + (i · 𝐴)) ≠ 0))

Theorematandm3 25460 A compact form of atandm 25458. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (𝐴↑2) ≠ -1))

Theorematandm4 25461 A compact form of atandm 25458. (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 + (𝐴↑2)) ≠ 0))

Theorematanf 25462 Domain and range of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan:(ℂ ∖ {-i, i})⟶ℂ

Theorematancl 25463 Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan → (arctan‘𝐴) ∈ ℂ)

Theoremasinval 25464 Value of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (arcsin‘𝐴) = (-i · (log‘((i · 𝐴) + (√‘(1 − (𝐴↑2)))))))

Theoremacosval 25465 Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (arccos‘𝐴) = ((π / 2) − (arcsin‘𝐴)))

Theorematanval 25466 Value of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan → (arctan‘𝐴) = ((i / 2) · ((log‘(1 − (i · 𝐴))) − (log‘(1 + (i · 𝐴))))))

Theorematanre 25467 A real number is in the domain of the arctangent function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℝ → 𝐴 ∈ dom arctan)

Theoremasinneg 25468 The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.)
(𝐴 ∈ ℂ → (arcsin‘-𝐴) = -(arcsin‘𝐴))

Theoremacosneg 25469 The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ ℂ → (arccos‘-𝐴) = (π − (arccos‘𝐴)))

Theoremefiasin 25470 The exponential of the arcsine function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (exp‘(i · (arcsin‘𝐴))) = ((i · 𝐴) + (√‘(1 − (𝐴↑2)))))

Theoremsinasin 25471 The arcsine function is an inverse to sin. This is the main property that justifies the notation arcsin or sin↑-1. Because sin is not an injection, the other converse identity asinsin 25474 is only true under limited circumstances. (Contributed by Mario Carneiro, 1-Apr-2015.)
(𝐴 ∈ ℂ → (sin‘(arcsin‘𝐴)) = 𝐴)

Theoremcosacos 25472 The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 1-Apr-2015.)
(𝐴 ∈ ℂ → (cos‘(arccos‘𝐴)) = 𝐴)

Theoremasinsinlem 25473 Lemma for asinsin 25474. (Contributed by Mario Carneiro, 2-Apr-2015.)
((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → 0 < (ℜ‘(exp‘(i · 𝐴))))

Theoremasinsin 25474 The arcsine function composed with sin is equal to the identity. This plus sinasin 25471 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 = (π / 2) − i𝑦 for nonnegative real 𝑦 and also symmetrically at 𝐴 = i𝑦 − (π / 2). In particular, when restricted to reals this identity extends to the closed interval [-(π / 2), (π / 2)], not just the open interval (see reasinsin 25478). (Contributed by Mario Carneiro, 2-Apr-2015.)
((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → (arcsin‘(sin‘𝐴)) = 𝐴)

Theoremacoscos 25475 The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 2-Apr-2015.)
((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (0(,)π)) → (arccos‘(cos‘𝐴)) = 𝐴)

Theoremasin1 25476 The arcsine of 1 is π / 2. (Contributed by Mario Carneiro, 2-Apr-2015.)
(arcsin‘1) = (π / 2)

Theoremacos1 25477 The arcsine of 1 is π / 2. (Contributed by Mario Carneiro, 2-Apr-2015.)
(arccos‘1) = 0

Theoremreasinsin 25478 The arcsine function composed with sin is equal to the identity. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ (-(π / 2)[,](π / 2)) → (arcsin‘(sin‘𝐴)) = 𝐴)

Theoremasinsinb 25479 Relationship between sine and arcsine. (Contributed by Mario Carneiro, 2-Apr-2015.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (ℜ‘𝐵) ∈ (-(π / 2)(,)(π / 2))) → ((arcsin‘𝐴) = 𝐵 ↔ (sin‘𝐵) = 𝐴))

Theoremacoscosb 25480 Relationship between sine and arcsine. (Contributed by Mario Carneiro, 2-Apr-2015.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (ℜ‘𝐵) ∈ (0(,)π)) → ((arccos‘𝐴) = 𝐵 ↔ (cos‘𝐵) = 𝐴))

Theoremasinbnd 25481 The arcsine function has range within a vertical strip of the complex plane with real part between -π / 2 and π / 2. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ ℂ → (ℜ‘(arcsin‘𝐴)) ∈ (-(π / 2)[,](π / 2)))

Theoremacosbnd 25482 The arccosine function has range within a vertical strip of the complex plane with real part between 0 and π. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ ℂ → (ℜ‘(arccos‘𝐴)) ∈ (0[,]π))

Theoremasinrebnd 25483 Bounds on the arcsine function. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ (-(π / 2)[,](π / 2)))

Theoremasinrecl 25484 The arcsine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ ℝ)

Theoremacosrecl 25485 The arccosine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ (-1[,]1) → (arccos‘𝐴) ∈ ℝ)

Theoremcosasin 25486 The cosine of the arcsine of 𝐴 is √(1 − 𝐴↑2). (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ ℂ → (cos‘(arcsin‘𝐴)) = (√‘(1 − (𝐴↑2))))

Theoremsinacos 25487 The sine of the arccosine of 𝐴 is √(1 − 𝐴↑2). (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ ℂ → (sin‘(arccos‘𝐴)) = (√‘(1 − (𝐴↑2))))

Theorematandmneg 25488 The domain of the arctangent function is closed under negatives. (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan → -𝐴 ∈ dom arctan)

Theorematanneg 25489 The arctangent function is odd. (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan → (arctan‘-𝐴) = -(arctan‘𝐴))

Theorematan0 25490 The arctangent of zero is zero. (Contributed by Mario Carneiro, 31-Mar-2015.)
(arctan‘0) = 0

Theorematandmcj 25491 The arctangent function distributes under conjugation. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan → (∗‘𝐴) ∈ dom arctan)

Theorematancj 25492 The arctangent function distributes under conjugation. (The condition that ℜ(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 25489 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 -1 and 1, though.) (Contributed by Mario Carneiro, 31-Mar-2015.)
((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ≠ 0) → (𝐴 ∈ dom arctan ∧ (∗‘(arctan‘𝐴)) = (arctan‘(∗‘𝐴))))

Theorematanrecl 25493 The arctangent function is real for all real inputs. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℝ → (arctan‘𝐴) ∈ ℝ)

Theoremefiatan 25494 Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ dom arctan → (exp‘(i · (arctan‘𝐴))) = ((√‘(1 + (i · 𝐴))) / (√‘(1 − (i · 𝐴)))))

((𝐴 ∈ dom arctan ∧ 0 ≤ (ℜ‘𝐴)) → ((log‘(1 + (i · 𝐴))) + (log‘(1 − (i · 𝐴)))) ∈ ran log)

Theorematanlogadd 25496 The rule √(𝑧𝑤) = (√𝑧)(√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 (-π, π], so there are some cases such as this one with 𝑧 = 1 + i𝐴 and 𝑤 = 1 − i𝐴 which are true unconditionally. This result can also be stated as "√(1 + 𝑧) + √(1 − 𝑧) is analytic". (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan → ((log‘(1 + (i · 𝐴))) + (log‘(1 − (i · 𝐴)))) ∈ ran log)

Theorematanlogsublem 25497 Lemma for atanlogsub 25498. (Contributed by Mario Carneiro, 4-Apr-2015.)
((𝐴 ∈ dom arctan ∧ 0 < (ℜ‘𝐴)) → (ℑ‘((log‘(1 + (i · 𝐴))) − (log‘(1 − (i · 𝐴))))) ∈ (-π(,)π))

Theorematanlogsub 25498 A variation on atanlogadd 25496, to show that √(1 + i𝑧) / √(1 − i𝑧) = √((1 + i𝑧) / (1 − i𝑧)) under more limited conditions. (Contributed by Mario Carneiro, 4-Apr-2015.)
((𝐴 ∈ dom arctan ∧ (ℜ‘𝐴) ≠ 0) → ((log‘(1 + (i · 𝐴))) − (log‘(1 − (i · 𝐴)))) ∈ ran log)

Theoremefiatan2 25499 Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan → (exp‘(i · (arctan‘𝐴))) = ((1 + (i · 𝐴)) / (√‘(1 + (𝐴↑2)))))

Theorem2efiatan 25500 Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ dom arctan → (exp‘(2 · (i · (arctan‘𝐴)))) = (((2 · i) / (𝐴 + i)) − 1))

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144 14301-14400 145 14401-14500 146 14501-14600 147 14601-14700 148 14701-14800 149 14801-14900 150 14901-15000 151 15001-15100 152 15101-15200 153 15201-15300 154 15301-15400 155 15401-15500 156 15501-15600 157 15601-15700 158 15701-15800 159 15801-15900 160 15901-16000 161 16001-16100 162 16101-16200 163 16201-16300 164 16301-16400 165 16401-16500 166 16501-16600 167 16601-16700 168 16701-16800 169 16801-16900 170 16901-17000 171 17001-17100 172 17101-17200 173 17201-17300 174 17301-17400 175 17401-17500 176 17501-17600 177 17601-17700 178 17701-17800 179 17801-17900 180 17901-18000 181 18001-18100 182 18101-18200 183 18201-18300 184 18301-18400 185 18401-18500 186 18501-18600 187 18601-18700 188 18701-18800 189 18801-18900 190 18901-19000 191 19001-19100 192 19101-19200 193 19201-19300 194 19301-19400 195 19401-19500 196 19501-19600 197 19601-19700 198 19701-19800 199 19801-19900 200 19901-20000 201 20001-20100 202 20101-20200 203 20201-20300 204 20301-20400 205 20401-20500 206 20501-20600 207 20601-20700 208 20701-20800 209 20801-20900 210 20901-21000 211 21001-21100 212 21101-21200 213 21201-21300 214 21301-21400 215 21401-21500 216 21501-21600 217 21601-21700 218 21701-21800 219 21801-21900 220 21901-22000 221 22001-22100 222 22101-22200 223 22201-22300 224 22301-22400 225 22401-22500 226 22501-22600 227 22601-22700 228 22701-22800 229 22801-22900 230 22901-23000 231 23001-23100 232 23101-23200 233 23201-23300 234 23301-23400 235 23401-23500 236 23501-23600 237 23601-23700 238 23701-23800 239 23801-23900 240 23901-24000 241 24001-24100 242 24101-24200 243 24201-24300 244 24301-24400 245 24401-24500 246 24501-24600 247 24601-24700 248 24701-24800 249 24801-24900 250 24901-25000 251 25001-25100 252 25101-25200 253 25201-25300 254 25301-25400 255 25401-25500 256 25501-25600 257 25601-25700 258 25701-25800 259 25801-25900 260 25901-26000 261 26001-26100 262 26101-26200 263 26201-26300 264 26301-26400 265 26401-26500 266 26501-26600 267 26601-26700 268 26701-26800 269 26801-26900 270 26901-27000 271 27001-27100 272 27101-27200 273 27201-27300 274 27301-27400 275 27401-27500 276 27501-27600 277 27601-27700 278 27701-27800 279 27801-27900 280 27901-28000 281 28001-28100 282 28101-28200 283 28201-28300 284 28301-28400 285 28401-28500 286 28501-28600 287 28601-28700 288 28701-28800 289 28801-28900 290 28901-29000 291 29001-29100 292 29101-29200 293 29201-29300 294 29301-29400 295 29401-29500 296 29501-29600 297 29601-29700 298 29701-29800 299 29801-29900 300 29901-30000 301 30001-30100 302 30101-30200 303 30201-30300 304 30301-30400 305 30401-30500 306 30501-30600 307 30601-30700 308 30701-30800 309 30801-30900 310 30901-31000 311 31001-31100 312 31101-31200 313 31201-31300 314 31301-31400 315 31401-31500 316 31501-31600 317 31601-31700 318 31701-31800 319 31801-31900 320 31901-32000 321 32001-32100 322 32101-32200 323 32201-32300 324 32301-32400 325 32401-32500 326 32501-32600 327 32601-32700 328 32701-32800 329 32801-32900 330 32901-33000 331 33001-33100 332 33101-33200 333 33201-33300 334 33301-33400 335 33401-33500 336 33501-33600 337 33601-33700 338 33701-33800 339 33801-33900 340 33901-34000 341 34001-34100 342 34101-34200 343 34201-34300 344 34301-34400 345 34401-34500 346 34501-34600 347 34601-34700 348 34701-34800 349 34801-34900 350 34901-35000 351 35001-35100 352 35101-35200 353 35201-35300 354 35301-35400 355 35401-35500 356 35501-35600 357 35601-35700 358 35701-35800 359 35801-35900 360 35901-36000 361 36001-36100 362 36101-36200 363 36201-36300 364 36301-36400 365 36401-36500 366 36501-36600 367 36601-36700 368 36701-36800 369 36801-36900 370 36901-37000 371 37001-37100 372 37101-37200 373 37201-37300 374 37301-37400 375 37401-37500 376 37501-37600 377 37601-37700 378 37701-37800 379 37801-37900 380 37901-38000 381 38001-38100 382 38101-38200 383 38201-38300 384 38301-38400 385 38401-38500 386 38501-38600 387 38601-38700 388 38701-38800 389 38801-38900 390 38901-39000 391 39001-39100 392 39101-39200 393 39201-39300 394 39301-39400 395 39401-39500 396 39501-39600 397 39601-39700 398 39701-39800 399 39801-39900 400 39901-40000 401 40001-40100 402 40101-40200 403 40201-40300 404 40301-40400 405 40401-40500 406 40501-40600 407 40601-40700 408 40701-40800 409 40801-40900 410 40901-41000 411 41001-41100 412 41101-41200 413 41201-41300 414 41301-41400 415 41401-41500 416 41501-41600 417 41601-41700 418 41701-41800 419 41801-41900 420 41901-42000 421 42001-42100 422 42101-42200 423 42201-42300 424 42301-42400 425 42401-42500 426 42501-42600 427 42601-42700 428 42701-42800 429 42801-42900 430 42901-43000 431 43001-43100 432 43101-43200 433 43201-43300 434 43301-43400 435 43401-43500 436 43501-43600 437 43601-43700 438 43701-43800 439 43801-43900 440 43901-44000 441 44001-44100 442 44101-44200 443 44201-44300 444 44301-44400 445 44401-44500 446 44501-44600 447 44601-44700 448 44701-44800 449 44801-44900 450 44901-45000 451 45001-45100 452 45101-45184
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