Question

Find the value of $sin\left(2ta{n}^{-1}\frac{1}{4}\right)+cos\left(ta{n}^{-1}2\sqrt{2}\right)$.

Answer 1

Let $y=sin\left(2ta{n}^{-1}\frac{1}{4}\right)+cos\left(ta{n}^{-1}2\sqrt{2}\right)=u+v$, where $u=sin2ta{n}^{-1}\frac{1}{4}$ and $v=costa{n}^{-1}2\sqrt{2}$
$\therefore u=sin\left(2ta{n}^{-1}\frac{1}{4}\right)=sinsi{n}^{-1}\left[\frac{2.\frac{1}{4}}{1+(\frac{1}{4}{)}^2}\right]=\frac{\frac{2}{4}}{\frac{17}{16}}=\frac{8}{17}$
Also, $v=costa{n}^{-1}2\sqrt{2}[Putta{n}^{-1}2\sqrt{2}=\theta \Rightarrow tan\theta =2\sqrt{2}\Rightarrow cos\theta =\frac{1}{3}\therefore \theta =co{s}^{-1}\frac{1}{3}$
$\therefore v=cos\left(co{s}^{-1}\frac{1}{3}\right)=\frac{1}{3}$
Now, $y=u+v=\frac{8}{17}+\frac{1}{3}=\frac{41}{51}.$