If the inverse trigonometric functions take principal values then ^-1(310(^-143)+25(^-143)…

Question

If the inverse trigonometric functions take principal values then $\cos^{-1}\left(\dfrac{3}{10}\cos\left(	an^{-1}\dfrac43ight)+\dfrac25\sin\left(	an^{-1}\dfrac43ight)ight)$ is equal to:

Accepted Answer

For $\tan^{-1}\dfrac43$: $\sin=\dfrac45,\ \cos=\dfrac35$. So

$$\frac{3}{10}\cdot\frac35+\frac25\cdot\frac45=\frac{9}{50}+\frac{16}{50}=\frac{25}{50}=\frac12,\qquad \cos^{-1}\left(\frac12\right)=\frac{\pi}{3}.$$

Solutions · 1