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

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

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

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


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

to get

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

So what does $y_1+y_2$ become?

  • $\begingroup$ 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, $\sin (\frac{\pi}{2})=\cos (2\pi)$. $\endgroup$ – Anurag A Nov 23 '18 at 19:24
  • $\begingroup$ But $2\pi = 0 $ (in radians), so you get $\sin \frac{\pi}{2} = \cos 0$. $\endgroup$ – KM101 Nov 23 '18 at 19:27
  • $\begingroup$ what do you mean by $2\pi=0$? !!! $\endgroup$ – Anurag A Nov 23 '18 at 19:30
  • $\begingroup$ (Awful wording, I edited the comment.) I meant $0$ radians and $2\pi$ radians are the same. $\endgroup$ – KM101 Nov 23 '18 at 19:32
  • $\begingroup$ No they are not. $0$ radian is $0^{\circ}$ and $2\pi$ radian is $360^{\circ}$. $\endgroup$ – Anurag A Nov 23 '18 at 19:34

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