For $0 < x < 1$, express $\sin[\sin^{-1}(x) + \cos^{-1}(x)]$, in terms of $x$

Question

I need to express $\sin[\sin^{-1}(x) + \cos^{-1}(x)]$, in terms of $x$, if $0 < x < 1$.

I'm not sure how to solve this. If I knew the value of $x$, I would try and apply the identity, $\sin(A-B)=\sin(A)\cos(B)+ \cos(A)\sin(A)$, but since the answer is the real number $1$, I don't see how that would work.

For example: $$\sin[\sin^{-1}(x) + \cos^{-1}(x)]$$ $$\sin[\sin^{-1}(x)] \cdot \cos[\cos^{-1}(x)] + \cos[\sin^{-1}(x)] \cdot \sin[\cos^{-1}(x)]$$ $$x \cdot x + \cos[\sin^{-1}(x)] \cdot \sin[\cos^{-1}(x)]$$ $$x^2 + \cos[\sin^{-1}(x)] \cdot \sin[\cos^{-1}(x)]$$

That's where I'm stuck.

Answer 1

Using the fact $$\cos(u)=\sqrt{1-\sin^2 u}\\\sin(u)=\sqrt{1-\cos^2 u}$$we have $$\cos [\sin^{-1}x]\cdot \sin [\cos^{-1}x]=\sqrt{1-\sin^2 (\sin^{-1}x)}\cdot \sqrt{1-\cos^2 (\cos^{-1}x)}=1-x^2$$therefore $$\large \sin[\sin^{-1}x+\cos^{-1}x]=1$$

Answer 2

Write $\cos(\sin^{-1} x)$ as $\sin(\frac\pi2 - \sin^{-1}x)$ and expand using the identity.

Similarly, solve the other term by writing $\sin(\cos^{-1} x)$ as $\cos(\frac\pi2 - \cos^{-1} x)$ and expanding it using an identity.

Answer 3

Hint: Use

$$\arcsin x = y_1 \implies \sin y_1 = x$$

$$\arccos x = y_2 \implies \cos y_2 = x$$

and

$$\cos \theta = \sin \bigg(\frac{\pi}{2}-\theta\bigg)$$

to get

$$\implies y_2 = \frac{\pi}{2}-y_1$$

So what does $y_1+y_2$ become?