Question

ex.1. $\sin(\tan^{-1}(\frac{\sqrt{3}}{2}))$ ex.2. $\tan(\cos^{-1}(-\frac{1}{3}))$

Explanation:

Step1: Let $\theta=\tan^{-1}(\frac{\sqrt{3}}{2})$

By the definition of inverse tangent, $\tan\theta=\frac{\sqrt{3}}{2}$, where $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$. We can consider a right - triangle where the opposite side $y = \sqrt{3}$ and the adjacent side $x = 2$. Then, by the Pythagorean theorem, the hypotenuse $r=\sqrt{x^{2}+y^{2}}=\sqrt{4 + 3}=\sqrt{7}$.

Step2: Find $\sin\theta$

Since $\sin\theta=\frac{y}{r}$ (for a right - triangle with angle $\theta$, opposite side $y$, adjacent side $x$, and hypotenuse $r$), and $y=\sqrt{3}$, $r = \sqrt{7}$, we have $\sin\theta=\frac{\sqrt{3}}{\sqrt{7}}=\frac{\sqrt{21}}{7}$.

Answer:

$\frac{\sqrt{21}}{7}$