Question

Evaluate $\sin (\frac{\pi }{6}+{\cos }^{-1}\frac{1}{4})$

Answer 1

$I=sin(\frac{\pi }{6}+co{s}^{-}1\frac{1}{4})$

$co{s}^{-1}\frac{1}{4}=P$

$cosp=\frac{1}{4}$ $sinp=\sqrt{1-(\frac{1}{4}{)}^2}=\frac{\sqrt{15}}{4}$

$tanp=\sqrt{15}$

$\Rightarrow p=ta{n}^{-1}\sqrt{15}$ and $\frac{\pi }{6}=ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$I=sin(\frac{\pi }{6}+co{s}^{-1}\frac{1}{4})=sin\left(ta{n}^{-1}\right(\frac{1}{\sqrt{3}})+ta{n}^{-1}\sqrt{15})$

$=sin\left(ta{n}^{-1}\left[\frac{\frac{1}{\sqrt{3}}+\sqrt{15}}{1-\sqrt{15}/3}\right]\right)$

$=sin\left[ta{n}^{-1}\left(\frac{1+\sqrt{45}}{(1-\sqrt{5})\sqrt{3}}\right)\right]$

$I=sin\left[ta{n}^{-1}\left(\frac{1+3\sqrt{5}}{\sqrt{3}-\sqrt{15}}\right)\right]$

$ta{n}^{-1}\left(\frac{1+3\sqrt{5}}{\sqrt{3}-\sqrt{15}}\right)=P$

$tanp=\frac{1+3\sqrt{5}}{\sqrt{3}-\sqrt{15}}\Rightarrow sinp=\frac{1+3\sqrt{5}}{(1+3\sqrt{5}{)}^2+(\sqrt{3}-\sqrt{15}{)}^2}$

$sinp=\frac{1+3\sqrt{5}}{(1+45+3+15+6\sqrt{5}-2\sqrt{45})}$

$sinp=\frac{43\sqrt{5}}{64+6\sqrt{5}-6\sqrt{5}}$

$sinp=\frac{1+3\sqrt{5}}{64}\Rightarrow p=si{n}^{-1}\left(\frac{1+3\sqrt{5}}{64}\right)$

$I=sin\left[si{n}^{-1}\left(\frac{1+3\sqrt{5}}{64}\right)\right]$

$\Rightarrow I=\frac{1+3\sqrt{5}}{64}$