TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-03

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For $-1<y<1$ , $(\arcsin y)^{\prime}=1/\sqrt{1-y^2}$ and $(\arccos y)^{\prime}=-1/\sqrt{1-y^2}$

Statement

For $-1<y<1$ ,

$(\arcsin y)'=\frac{1}{\sqrt{1-y^2}},\qquad(\arccos y)'=-\frac{1}{\sqrt{1-y^2}}.$

Facts & Assumptions

Given: A real number $y$ with $-1<y<1$ .

[L1]

Principal inverse sine and cosine are the inverses of the indicated restricted functions (Principal inverse sine and inverse cosine).

[L2]

Sine and cosine are differentiable, hence continuous, with derivatives $\cos$ and $-\sin$ (The derivatives of sine and cosine are cosine and minus sine, A function differentiable at $c$ is continuous at $c$ ).

[L3]

Sine is strictly increasing on $[-\pi/2,\pi/2]$ and strictly decreasing on $[\pi/2,3\pi/2]$ , while cosine is strictly decreasing on $[0,\pi]$ ; the special values are $\sin0=\sin\pi=0$ and $\cos(\pi/2)=0$ , and cosine is even (Signs, monotonicity intervals, and ranges of sine and cosine, Quarter-turn values and shifts by pi/2 and pi, Parity and the Pythagorean identity for sine and cosine).

[L4]

$\sin^2t+\cos^2t=1$ for every $t$ (Parity and the Pythagorean identity for sine and cosine).

[L5]

Every nonnegative real has a unique nonnegative square root (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ ).

[L6]

The inverse of a continuous injective function on a nondegenerate interval has derivative the reciprocal of the original derivative wherever that derivative is nonzero (Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f'(c) \ne 0$ , then the inverse $g$ is differentiable at $f(c)$ with $g'(f(c)) = 1/f'(c)$ ; and if $f'(c) = 0$ then $g$ is not differentiable at $f(c)$ ).

Proof

technique · direct

1.1

Put $a:=\arcsin y$ and $b:=\arccos y$ . Then $\sin a=y$ , $\cos b=y$ , and $a,b$ lie in the interiors of their respective principal intervals.

L1given

2.1

The interval placement of step 1.1, the monotonicity and special values in [L3], and evenness of cosine give $\cos a>0$ and $\sin b>0$ . The Pythagorean identity then gives $\cos a=\sqrt{1-y^2}$ and $\sin b=\sqrt{1-y^2}$ .

step 1.1L3L4L5

3.1

Apply [L6] to sine on $[-\pi/2,\pi/2]$ at $a$ : [L1] supplies injectivity and [L2] supplies continuity. Since its derivative there is $\cos a\ne0$ , the inverse is differentiable at $y$ with $(\arcsin y)'=1/\cos a=1/\sqrt{1-y^2}$ .

step 1.1step 2.1L1L2L6

3.2

Apply [L6] to cosine on $[0,\pi]$ at $b$ : [L1] supplies injectivity and [L2] supplies continuity. Its derivative is $-\sin b\ne0$ , so $(\arccos y)'=-1/\sin b=-1/\sqrt{1-y^2}$ .

step 1.1step 2.1L1L2L6

4.1

Steps 3.1 and 3.2 prove the two derivative formulas.

step 3.1step 3.2∎

Depends on

Used by

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Sources

NIST Digital Library of Mathematical Functions, Inverse Trigonometric Functions (standard reference, not scraped)

For −1<y<1-1<y<1−1<y<1, (arcsin⁡y)′=1/1−y2(\arcsin y)^{\prime}=1/\sqrt{1-y^2}(arcsiny)′=1/1−y2​ and (arccos⁡y)′=−1/1−y2(\arccos y)^{\prime}=-1/\sqrt{1-y^2}(arccosy)′=−1/1−y2​