Question

Find the value of sin$\left(2ta{n}^{-1}\frac{1}{3}\right)+cos\left(ta{n}^{-1}2\sqrt{2}\right).$

Answer 1

We have, sin$\left(2ta{n}^{-1}\frac{1}{3}\right)+cos\left(ta{n}^{-1}2\sqrt{2}\right).$

$=sin\left[si{n}^{-1}\left\{\frac{2\times \frac{1}{3}}{1+{\left(\frac{1}{3}\right)}^2}\right\}\right]+cos\left(co{s}^{-1}\frac{1}{3}\right)[\because ta{n}^{-1}x=co{s}^{-1}\frac{1}{\sqrt{1+x^2}}][\because 2ta{n}^{-1}x=si{n}^{-1}\frac{2x}{1+x^2}.-1\le 1andta{n}^{-1}(2\sqrt{2})=co{s}^{-1}\frac{1}{3}]$

$=sin\left[si{n}^{-1}\left(\frac{\frac{2}{3}}{1+\frac{1}{9}}\right)\right]+\frac{1}{3}\{\because cos(co{s}^{-1}x)=x,x\in [-1,1\left]\right\}=sin\left[si{n}^{-1}\left(\frac{2\times 9}{3\times 10}\right)\right]+\frac{1}{3}=sin\left[si{n}^{-1}\left(\frac{3}{5}\right)\right]+\frac{1}{3}[\because sin(si{n}^{-1}x)=x]=\frac{3}{5}+\frac{1}{3}=\frac{9+5}{15}=\frac{14}{15}$