P. 261 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 26001-26100   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	1cubr 26001	The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
⊢ 𝑅 = {1, ((-1 + (i · (√‘3))) / 2), ((-1 − (i · (√‘3))) / 2)}    ⇒   ⊢ (𝐴 ∈ 𝑅 ↔ (𝐴 ∈ ℂ ∧ (𝐴↑3) = 1))
Theorem	dcubic1lem 26002	Lemma for dcubic1 26004 and dcubic2 26003: simplify the cubic equation under the substitution 𝑋 = 𝑈 − 𝑀 / 𝑈. (Contributed by Mario Carneiro, 26-Apr-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   ⊢ (𝜑 → 𝑀 = (𝑃 / 3))    &   ⊢ (𝜑 → 𝑁 = (𝑄 / 2))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑈 ∈ ℂ)    &   ⊢ (𝜑 → 𝑈 ≠ 0)    &   ⊢ (𝜑 → 𝑋 = (𝑈 − (𝑀 / 𝑈)))    ⇒   ⊢ (𝜑 → (((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0 ↔ (((𝑈↑3)↑2) + ((𝑄 · (𝑈↑3)) − (𝑀↑3))) = 0))
Theorem	dcubic2 26003*	Reverse direction of dcubic 26005. 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.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   ⊢ (𝜑 → 𝑀 = (𝑃 / 3))    &   ⊢ (𝜑 → 𝑁 = (𝑄 / 2))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑈 ∈ ℂ)    &   ⊢ (𝜑 → 𝑈 ≠ 0)    &   ⊢ (𝜑 → 𝑋 = (𝑈 − (𝑀 / 𝑈)))    &   ⊢ (𝜑 → ((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0)    ⇒   ⊢ (𝜑 → ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = ((𝑟 · 𝑇) − (𝑀 / (𝑟 · 𝑇)))))
Theorem	dcubic1 26004	Forward direction of dcubic 26005: the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   ⊢ (𝜑 → 𝑀 = (𝑃 / 3))    &   ⊢ (𝜑 → 𝑁 = (𝑄 / 2))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑋 = (𝑇 − (𝑀 / 𝑇)))    ⇒   ⊢ (𝜑 → ((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0)
Theorem	dcubic 26005*	Solutions to the depressed cubic, a special case of cubic 26008. (The definitions of 𝑀, 𝑁, 𝐺, 𝑇 here differ from mcubic 26006 by scale factors of -9, 54, 54 and -27 respectively, to simplify the algebra and presentation.) (Contributed by Mario Carneiro, 26-Apr-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   ⊢ (𝜑 → 𝑀 = (𝑃 / 3))    &   ⊢ (𝜑 → 𝑁 = (𝑄 / 2))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    ⇒   ⊢ (𝜑 → (((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = ((𝑟 · 𝑇) − (𝑀 / (𝑟 · 𝑇))))))
Theorem	mcubic 26006*	Solutions to a monic cubic equation, a special case of cubic 26008. (Contributed by Mario Carneiro, 24-Apr-2015.)
⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = ((𝑁 + 𝐺) / 2))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) − (4 · (𝑀↑3))))    &   ⊢ (𝜑 → 𝑀 = ((𝐵↑2) − (3 · 𝐶)))    &   ⊢ (𝜑 → 𝑁 = (((2 · (𝐵↑3)) − (9 · (𝐵 · 𝐶))) + (;27 · 𝐷)))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    ⇒   ⊢ (𝜑 → ((((𝑋↑3) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / 3))))
Theorem	cubic2 26007*	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.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 0)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = ((𝑁 + 𝐺) / 2))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) − (4 · (𝑀↑3))))    &   ⊢ (𝜑 → 𝑀 = ((𝐵↑2) − (3 · (𝐴 · 𝐶))))    &   ⊢ (𝜑 → 𝑁 = (((2 · (𝐵↑3)) − ((9 · 𝐴) · (𝐵 · 𝐶))) + (;27 · ((𝐴↑2) · 𝐷))))    &   ⊢ (𝜑 → 𝑇 ≠ 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)))))
Theorem	cubic 26008*	The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4643 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)}    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 0)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 = (((𝑁 + (√‘𝐺)) / 2)↑𝑐(1 / 3)))    &   ⊢ (𝜑 → 𝐺 = ((𝑁↑2) − (4 · (𝑀↑3))))    &   ⊢ (𝜑 → 𝑀 = ((𝐵↑2) − (3 · (𝐴 · 𝐶))))    &   ⊢ (𝜑 → 𝑁 = (((2 · (𝐵↑3)) − ((9 · 𝐴) · (𝐵 · 𝐶))) + (;27 · ((𝐴↑2) · 𝐷))))    &   ⊢ (𝜑 → 𝑀 ≠ 0)    ⇒   ⊢ (𝜑 → ((((𝐴 · (𝑋↑3)) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ 𝑅 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / (3 · 𝐴))))
Theorem	binom4 26009	Work out a quartic binomial. (You would think that by this point it would be faster to use binom 15551, 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)))))
Theorem	dquartlem1 26010	Lemma for dquart 26012. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑆 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 = ((2 · 𝑆)↑2))    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 ∈ 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 = (-𝑆 − 𝐼))))
Theorem	dquartlem2 26011	Lemma for dquart 26012. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑆 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 = ((2 · 𝑆)↑2))    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 ∈ ℂ)    &   ⊢ (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))    &   ⊢ (𝜑 → 𝐷 ∈ 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)
Theorem	dquart 26012	Solve a depressed quartic equation. To eliminate 𝑆, which is the square root of a solution 𝑀 to the resolvent cubic equation, apply cubic 26008 or one of its variants. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑆 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 = ((2 · 𝑆)↑2))    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 ∈ ℂ)    &   ⊢ (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → (((𝑀↑3) + ((2 · 𝐵) · (𝑀↑2))) + ((((𝐵↑2) − (4 · 𝐷)) · 𝑀) + -(𝐶↑2))) = 0)    &   ⊢ (𝜑 → 𝐽 ∈ 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 = (-𝑆 − 𝐼)) ∨ (𝑋 = (𝑆 + 𝐽) ∨ 𝑋 = (𝑆 − 𝐽)))))
Theorem	quart1cl 26013	Closure lemmas for quart 26020. (Contributed by Mario Carneiro, 7-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    ⇒   ⊢ (𝜑 → (𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ ∧ 𝑅 ∈ ℂ))
Theorem	quart1lem 26014	Lemma for quart1 26015. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 = (𝑋 + (𝐴 / 4)))    ⇒   ⊢ (𝜑 → 𝐷 = ((((𝐴↑4) / ;;256) + (𝑃 · ((𝐴 / 4)↑2))) + ((𝑄 · (𝐴 / 4)) + 𝑅)))
Theorem	quart1 26015	Depress a quartic equation. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 = (𝑋 + (𝐴 / 4)))    ⇒   ⊢ (𝜑 → (((𝑋↑4) + (𝐴 · (𝑋↑3))) + ((𝐵 · (𝑋↑2)) + ((𝐶 · 𝑋) + 𝐷))) = (((𝑌↑4) + (𝑃 · (𝑌↑2))) + ((𝑄 · 𝑌) + 𝑅)))
Theorem	quartlem1 26016	Lemma for quart 26020. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑅 ∈ ℂ)    &   ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = ((-(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)))))
Theorem	quartlem2 26017	Closure lemmas for quart 26020. (Contributed by Mario Carneiro, 7-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 = -(𝐴 / 4))    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    &   ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    &   ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    ⇒   ⊢ (𝜑 → (𝑈 ∈ ℂ ∧ 𝑉 ∈ ℂ ∧ 𝑊 ∈ ℂ))
Theorem	quartlem3 26018	Closure lemmas for quart 26020. (Contributed by Mario Carneiro, 7-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 = -(𝐴 / 4))    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    &   ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    &   ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   ⊢ (𝜑 → 𝑆 = ((√‘𝑀) / 2))    &   ⊢ (𝜑 → 𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   ⊢ (𝜑 → 𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    ⇒   ⊢ (𝜑 → (𝑆 ∈ ℂ ∧ 𝑀 ∈ ℂ ∧ 𝑇 ∈ ℂ))
Theorem	quartlem4 26019	Closure lemmas for quart 26020. (Contributed by Mario Carneiro, 7-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 = -(𝐴 / 4))    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    &   ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    &   ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   ⊢ (𝜑 → 𝑆 = ((√‘𝑀) / 2))    &   ⊢ (𝜑 → 𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   ⊢ (𝜑 → 𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 = (√‘((-(𝑆↑2) − (𝑃 / 2)) + ((𝑄 / 4) / 𝑆))))    &   ⊢ (𝜑 → 𝐽 = (√‘((-(𝑆↑2) − (𝑃 / 2)) − ((𝑄 / 4) / 𝑆))))    ⇒   ⊢ (𝜑 → (𝑆 ≠ 0 ∧ 𝐼 ∈ ℂ ∧ 𝐽 ∈ ℂ))
Theorem	quart 26020	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 33136) if all the substitutions are performed. This is Metamath 100 proof #46. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 = -(𝐴 / 4))    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    &   ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    &   ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   ⊢ (𝜑 → 𝑆 = ((√‘𝑀) / 2))    &   ⊢ (𝜑 → 𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   ⊢ (𝜑 → 𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 = (√‘((-(𝑆↑2) − (𝑃 / 2)) + ((𝑄 / 4) / 𝑆))))    &   ⊢ (𝜑 → 𝐽 = (√‘((-(𝑆↑2) − (𝑃 / 2)) − ((𝑄 / 4) / 𝑆))))    ⇒   ⊢ (𝜑 → ((((𝑋↑4) + (𝐴 · (𝑋↑3))) + ((𝐵 · (𝑋↑2)) + ((𝐶 · 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
Syntax	casin 26021	The arcsine function.
class arcsin
Syntax	cacos 26022	The arccosine function.
class arccos
Syntax	catan 26023	The arctangent function.
class arctan
Definition	df-asin 26024	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)))))))
Definition	df-acos 26025	Define the arccosine function. See also remarks for df-asin 26024. 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)))
Definition	df-atan 26026	Define the arctangent function. See also remarks for df-asin 26024. 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))))))
Theorem	asinlem 26027	The argument to the logarithm in df-asin 26024 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → ((i · 𝐴) + (√‘(1 − (𝐴↑2)))) ≠ 0)
Theorem	asinlem2 26028	The argument to the logarithm in df-asin 26024 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)
Theorem	asinlem3a 26029	Lemma for asinlem3 26030. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (ℑ‘𝐴) ≤ 0) → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))
Theorem	asinlem3 26030	The argument to the logarithm in df-asin 26024 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))
Theorem	asinf 26031	Domain and range of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ arcsin:ℂ⟶ℂ
Theorem	asincl 26032	Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (arcsin‘𝐴) ∈ ℂ)
Theorem	acosf 26033	Domain and range of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ arccos:ℂ⟶ℂ
Theorem	acoscl 26034	Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (arccos‘𝐴) ∈ ℂ)
Theorem	atandm 26035	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))
Theorem	atandm2 26036	This form of atandm 26035 is a bit more useful for showing that the logarithms in df-atan 26026 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 − (i · 𝐴)) ≠ 0 ∧ (1 + (i · 𝐴)) ≠ 0))
Theorem	atandm3 26037	A compact form of atandm 26035. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (𝐴↑2) ≠ -1))
Theorem	atandm4 26038	A compact form of atandm 26035. (Contributed by Mario Carneiro, 3-Apr-2015.)
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 + (𝐴↑2)) ≠ 0))
Theorem	atanf 26039	Domain and range of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ arctan:(ℂ ∖ {-i, i})⟶ℂ
Theorem	atancl 26040	Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ dom arctan → (arctan‘𝐴) ∈ ℂ)
Theorem	asinval 26041	Value of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (arcsin‘𝐴) = (-i · (log‘((i · 𝐴) + (√‘(1 − (𝐴↑2)))))))
Theorem	acosval 26042	Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (arccos‘𝐴) = ((π / 2) − (arcsin‘𝐴)))
Theorem	atanval 26043	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))))))
Theorem	atanre 26044	A real number is in the domain of the arctangent function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℝ → 𝐴 ∈ dom arctan)
Theorem	asinneg 26045	The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (arcsin‘-𝐴) = -(arcsin‘𝐴))
Theorem	acosneg 26046	The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (arccos‘-𝐴) = (π − (arccos‘𝐴)))
Theorem	efiasin 26047	The exponential of the arcsine function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (exp‘(i · (arcsin‘𝐴))) = ((i · 𝐴) + (√‘(1 − (𝐴↑2)))))
Theorem	sinasin 26048	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 26051 is only true under limited circumstances. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (sin‘(arcsin‘𝐴)) = 𝐴)
Theorem	cosacos 26049	The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (cos‘(arccos‘𝐴)) = 𝐴)
Theorem	asinsinlem 26050	Lemma for asinsin 26051. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → 0 < (ℜ‘(exp‘(i · 𝐴))))
Theorem	asinsin 26051	The arcsine function composed with sin is equal to the identity. This plus sinasin 26048 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 26055). (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → (arcsin‘(sin‘𝐴)) = 𝐴)
Theorem	acoscos 26052	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)
Theorem	asin1 26053	The arcsine of 1 is π / 2. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (arcsin‘1) = (π / 2)
Theorem	acos1 26054	The arccosine of 1 is 0. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (arccos‘1) = 0
Theorem	reasinsin 26055	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)
Theorem	asinsinb 26056	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))
Theorem	acoscosb 26057	Relationship between cosine and arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (ℜ‘𝐵) ∈ (0(,)π)) → ((arccos‘𝐴) = 𝐵 ↔ (cos‘𝐵) = 𝐴))
Theorem	asinbnd 26058	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)))
Theorem	acosbnd 26059	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))
Theorem	asinrebnd 26060	Bounds on the arcsine function. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ (-(π / 2)[,](π / 2)))
Theorem	asinrecl 26061	The arcsine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ ℝ)
Theorem	acosrecl 26062	The arccosine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (𝐴 ∈ (-1[,]1) → (arccos‘𝐴) ∈ ℝ)
Theorem	cosasin 26063	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))))
Theorem	sinacos 26064	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))))
Theorem	atandmneg 26065	The domain of the arctangent function is closed under negatives. (Contributed by Mario Carneiro, 3-Apr-2015.)
⊢ (𝐴 ∈ dom arctan → -𝐴 ∈ dom arctan)
Theorem	atanneg 26066	The arctangent function is odd. (Contributed by Mario Carneiro, 3-Apr-2015.)
⊢ (𝐴 ∈ dom arctan → (arctan‘-𝐴) = -(arctan‘𝐴))
Theorem	atan0 26067	The arctangent of zero is zero. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (arctan‘0) = 0
Theorem	atandmcj 26068	The arctangent function distributes under conjugation. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ dom arctan → (∗‘𝐴) ∈ dom arctan)
Theorem	atancj 26069	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 26066 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))))
Theorem	atanrecl 26070	The arctangent function is real for all real inputs. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℝ → (arctan‘𝐴) ∈ ℝ)
Theorem	efiatan 26071	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)))))
Theorem	atanlogaddlem 26072	Lemma for atanlogadd 26073. (Contributed by Mario Carneiro, 3-Apr-2015.)
⊢ ((𝐴 ∈ dom arctan ∧ 0 ≤ (ℜ‘𝐴)) → ((log‘(1 + (i · 𝐴))) + (log‘(1 − (i · 𝐴)))) ∈ ran log)
Theorem	atanlogadd 26073	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)
Theorem	atanlogsublem 26074	Lemma for atanlogsub 26075. (Contributed by Mario Carneiro, 4-Apr-2015.)
⊢ ((𝐴 ∈ dom arctan ∧ 0 < (ℜ‘𝐴)) → (ℑ‘((log‘(1 + (i · 𝐴))) − (log‘(1 − (i · 𝐴))))) ∈ (-π(,)π))
Theorem	atanlogsub 26075	A variation on atanlogadd 26073, 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)
Theorem	efiatan2 26076	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)))))
Theorem	2efiatan 26077	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))
Theorem	tanatan 26078	The arctangent function is an inverse to tan. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (𝐴 ∈ dom arctan → (tan‘(arctan‘𝐴)) = 𝐴)
Theorem	atandmtan 26079	The tangent function has range contained in the domain of the arctangent. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (cos‘𝐴) ≠ 0) → (tan‘𝐴) ∈ dom arctan)
Theorem	cosatan 26080	The cosine of an arctangent. (Contributed by Mario Carneiro, 3-Apr-2015.)
⊢ (𝐴 ∈ dom arctan → (cos‘(arctan‘𝐴)) = (1 / (√‘(1 + (𝐴↑2)))))
Theorem	cosatanne0 26081	The arctangent function has range contained in the domain of the tangent. (Contributed by Mario Carneiro, 3-Apr-2015.)
⊢ (𝐴 ∈ dom arctan → (cos‘(arctan‘𝐴)) ≠ 0)
Theorem	atantan 26082	The arctangent function is an inverse to tan. (Contributed by Mario Carneiro, 5-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → (arctan‘(tan‘𝐴)) = 𝐴)
Theorem	atantanb 26083	Relationship between tangent and arctangent. (Contributed by Mario Carneiro, 5-Apr-2015.)
⊢ ((𝐴 ∈ dom arctan ∧ 𝐵 ∈ ℂ ∧ (ℜ‘𝐵) ∈ (-(π / 2)(,)(π / 2))) → ((arctan‘𝐴) = 𝐵 ↔ (tan‘𝐵) = 𝐴))
Theorem	atanbndlem 26084	Lemma for atanbnd 26085. (Contributed by Mario Carneiro, 5-Apr-2015.)
⊢ (𝐴 ∈ ℝ+ → (arctan‘𝐴) ∈ (-(π / 2)(,)(π / 2)))
Theorem	atanbnd 26085	The arctangent function is bounded by π / 2 on the reals. (Contributed by Mario Carneiro, 5-Apr-2015.)
⊢ (𝐴 ∈ ℝ → (arctan‘𝐴) ∈ (-(π / 2)(,)(π / 2)))
Theorem	atanord 26086	The arctangent function is strictly increasing. (Contributed by Mario Carneiro, 5-Apr-2015.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (𝐴 < 𝐵 ↔ (arctan‘𝐴) < (arctan‘𝐵)))
Theorem	atan1 26087	The arctangent of 1 is π / 4. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (arctan‘1) = (π / 4)
Theorem	bndatandm 26088	A point in the open unit disk is in the domain of the arctangent. (Contributed by Mario Carneiro, 5-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) < 1) → 𝐴 ∈ dom arctan)
Theorem	atans 26089*	The "domain of continuity" of the arctangent. (Contributed by Mario Carneiro, 7-Apr-2015.)
⊢ 𝐷 = (ℂ ∖ (-∞(,]0))    &   ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷}    ⇒   ⊢ (𝐴 ∈ 𝑆 ↔ (𝐴 ∈ ℂ ∧ (1 + (𝐴↑2)) ∈ 𝐷))
Theorem	atans2 26090*	It suffices to show that 1 − i𝐴 and 1 + i𝐴 are in the continuity domain of log to show that 𝐴 is in the continuity domain of arctangent. (Contributed by Mario Carneiro, 7-Apr-2015.)
⊢ 𝐷 = (ℂ ∖ (-∞(,]0))    &   ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷}    ⇒   ⊢ (𝐴 ∈ 𝑆 ↔ (𝐴 ∈ ℂ ∧ (1 − (i · 𝐴)) ∈ 𝐷 ∧ (1 + (i · 𝐴)) ∈ 𝐷))
Theorem	atansopn 26091*	The domain of continuity of the arctangent is an open set. (Contributed by Mario Carneiro, 7-Apr-2015.)
⊢ 𝐷 = (ℂ ∖ (-∞(,]0))    &   ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷}    ⇒   ⊢ 𝑆 ∈ (TopOpen‘ℂfld)
Theorem	atansssdm 26092*	The domain of continuity of the arctangent is a subset of the actual domain of the arctangent. (Contributed by Mario Carneiro, 7-Apr-2015.)
⊢ 𝐷 = (ℂ ∖ (-∞(,]0))    &   ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷}    ⇒   ⊢ 𝑆 ⊆ dom arctan
Theorem	ressatans 26093*	The real number line is a subset of the domain of continuity of the arctangent. (Contributed by Mario Carneiro, 7-Apr-2015.)
⊢ 𝐷 = (ℂ ∖ (-∞(,]0))    &   ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷}    ⇒   ⊢ ℝ ⊆ 𝑆
Theorem	dvatan 26094*	The derivative of the arctangent. (Contributed by Mario Carneiro, 7-Apr-2015.)
⊢ 𝐷 = (ℂ ∖ (-∞(,]0))    &   ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷}    ⇒   ⊢ (ℂ D (arctan ↾ 𝑆)) = (𝑥 ∈ 𝑆 ↦ (1 / (1 + (𝑥↑2))))
Theorem	atancn 26095*	The arctangent is a continuous function. (Contributed by Mario Carneiro, 7-Apr-2015.)
⊢ 𝐷 = (ℂ ∖ (-∞(,]0))    &   ⊢ 𝑆 = {𝑦 ∈ ℂ ∣ (1 + (𝑦↑2)) ∈ 𝐷}    ⇒   ⊢ (arctan ↾ 𝑆) ∈ (𝑆–cn→ℂ)
Theorem	atantayl 26096*	The Taylor series for arctan(𝐴). (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ (((i · ((-i↑𝑛) − (i↑𝑛))) / 2) · ((𝐴↑𝑛) / 𝑛)))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) < 1) → seq1( + , 𝐹) ⇝ (arctan‘𝐴))
Theorem	atantayl2 26097*	The Taylor series for arctan(𝐴). (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(2 ∥ 𝑛, 0, ((-1↑((𝑛 − 1) / 2)) · ((𝐴↑𝑛) / 𝑛))))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) < 1) → seq1( + , 𝐹) ⇝ (arctan‘𝐴))
Theorem	atantayl3 26098*	The Taylor series for arctan(𝐴). (Contributed by Mario Carneiro, 7-Apr-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ0 ↦ ((-1↑𝑛) · ((𝐴↑((2 · 𝑛) + 1)) / ((2 · 𝑛) + 1))))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) < 1) → seq0( + , 𝐹) ⇝ (arctan‘𝐴))
Theorem	leibpilem1 26099	Lemma for leibpi 26101. (Contributed by Mario Carneiro, 7-Apr-2015.) (Proof shortened by Steven Nguyen, 23-Mar-2023.)
⊢ ((𝑁 ∈ ℕ0 ∧ (¬ 𝑁 = 0 ∧ ¬ 2 ∥ 𝑁)) → (𝑁 ∈ ℕ ∧ ((𝑁 − 1) / 2) ∈ ℕ0))
Theorem	leibpilem2 26100*	The Leibniz formula for π. (Contributed by Mario Carneiro, 7-Apr-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ0 ↦ ((-1↑𝑛) / ((2 · 𝑛) + 1)))    &   ⊢ 𝐺 = (𝑘 ∈ ℕ0 ↦ if((𝑘 = 0 ∨ 2 ∥ 𝑘), 0, ((-1↑((𝑘 − 1) / 2)) / 𝑘)))    &   ⊢ 𝐴 ∈ V    ⇒   ⊢ (seq0( + , 𝐹) ⇝ 𝐴 ↔ seq0( + , 𝐺) ⇝ 𝐴)