2-[^-145+^-1513+^-11665] is equal to:

Question

$2\pi-\left[\sin^{-1}\dfrac45+\sin^{-1}\dfrac{5}{13}+\sin^{-1}\dfrac{16}{65}\right]$ is equal to:

Accepted Answer

Convert to tangents: $\sin^{-1}\dfrac45=\tan^{-1}\dfrac43$, $\sin^{-1}\dfrac{5}{13}=\tan^{-1}\dfrac{5}{12}$, $\sin^{-1}\dfrac{16}{65}=\tan^{-1}\dfrac{16}{63}$.

$$\tan^{-1}\frac43+\tan^{-1}\frac{5}{12}=\tan^{-1}\left(\frac{\frac43+\frac{5}{12}}{1-\frac43\cdot\frac{5}{12}}\right)=\tan^{-1}\frac{63}{16}.$$

$$\tan^{-1}\frac{63}{16}+\tan^{-1}\frac{16}{63}=\tan^{-1}\frac{63}{16}+\cot^{-1}\frac{63}{16}=\frac{\pi}{2}.$$

So the bracket $=\dfrac{\pi}{2}$, and $2\pi-\dfrac{\pi}{2}=\dfrac{3\pi}{2}$.

Solutions · 1