If ^-1(17)+^-1(45)-^-1(7736)=0, 0

Question

If $\sin^{-1}\left(\dfrac{\alpha}{17}ight)+\cos^{-1}\left(\dfrac45ight)-	an^{-1}\left(\dfrac{77}{36}ight)=0$, $0<\alpha<13$, then $\sin^{-1}(\sin\alpha)+\cos^{-1}(\cos\alpha)$ is equal to

Accepted Answer

$\cos^{-1}\dfrac45=\tan^{-1}\dfrac34$, so

$$\sin^{-1}\frac{\alpha}{17}=\tan^{-1}\frac{77}{36}-\tan^{-1}\frac34=\tan^{-1}\left(\frac{\frac{77}{36}-\frac34}{1+\frac{77}{36}\cdot\frac34}\right)=\tan^{-1}\frac{8}{15}=\sin^{-1}\frac{8}{17}.$$

So $\dfrac{\alpha}{17}=\dfrac{8}{17}\Rightarrow\alpha=8$.

$$\sin^{-1}(\sin8)+\cos^{-1}(\cos8)=(3\pi-8)+(8-2\pi)=\pi.$$

Solutions · 1