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

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.

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 $$\large \sin[\sin^{-1}x+\cos^{-1}x]=1$$

• I'm sorry, I don't understand how that helps. – LuminousNutria Nov 23 '18 at 19:28
• Then you can substitute $u=\sin^{-1}x$ and $u=\cos^{-1}x$ in the first and second equations respectively – Mostafa Ayaz Nov 23 '18 at 19:29
• So, instead of $\sin[\sin^{-1}(x) + \cos^{-1}(x)]$, I'll have $\sin[u + u]$? – LuminousNutria Nov 23 '18 at 19:32
• I clarified my answer a bit more. Hope it help now! – Mostafa Ayaz Nov 23 '18 at 19:36
• So, why does $\sqrt{1-\sin^2 (\sin^{-1}x)}\cdot \sqrt{1-\cos^2 (\cos^{-1}x)} = 1 - x^2$ ? – LuminousNutria Nov 23 '18 at 19:41

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.

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?

• You've got to be careful here $\sin y_1=\cos y_2$ doesn't necessarily imply that $y_2=\frac{\pi}{2}-y_1$. For example, $\sin (\frac{\pi}{2})=\cos (2\pi)$. – Anurag A Nov 23 '18 at 19:24
• But $2\pi = 0$ (in radians), so you get $\sin \frac{\pi}{2} = \cos 0$. – KM101 Nov 23 '18 at 19:27
• what do you mean by $2\pi=0$? !!! – Anurag A Nov 23 '18 at 19:30
• (Awful wording, I edited the comment.) I meant $0$ radians and $2\pi$ radians are the same. – KM101 Nov 23 '18 at 19:32
• No they are not. $0$ radian is $0^{\circ}$ and $2\pi$ radian is $360^{\circ}$. – Anurag A Nov 23 '18 at 19:34