Question

If $\sin \theta =\frac{1}{2}$ and $\sin \varphi =\frac{1}{3},$ find the value of $\sin (\theta +\varphi )$ and $\sin (2\theta +2\varphi )$.

Answer 1

Given, $\sin \theta =\frac{1}{2}$ and $\sin \varphi =\frac{1}{3}$

$\cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-\frac{1}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$

$\cos \varphi =\sqrt{1-{\sin }^2\varphi }=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$

$\Rightarrow$ $\sin (\theta +\varphi )=\sin \theta \cos \varphi +\cos \theta \sin \varphi$

$=\frac{1}{2}\times \frac{2\sqrt{2}}{3}+\frac{\sqrt{3}}{2}\times \frac{1}{3}$

$=\frac{2\sqrt{2}+\sqrt{3}}{6}$

We know,

${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$

$\therefore$ $\cos 2\theta =2{\cos }^2\theta -1$

$=2\times \frac{3}{4}-1$

$=\frac{1}{2}$

Similarly, $\cos 2\varphi =2{\cos }^2\varphi -1$

$=2\times \frac{8}{9}-1$

$=\frac{7}{9}$

$\Rightarrow$ $\sin 2\theta =\sqrt{1-{\cos }^{2}2\theta }=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

$\Rightarrow$ $\sin 2\varphi =\sqrt{1-{\cos }^{2}2\varphi }=\sqrt{1-\frac{49}{81}}=\frac{\sqrt{32}}{9}=\frac{4\sqrt{2}}{9}$

$\Rightarrow$ $\sin (2\theta +2\varphi )=\sin 2\theta \cos 2\varphi +\cos 2\theta \sin 2\varphi$

$=\frac{\sqrt{3}}{2}\times \frac{7}{9}+\frac{1}{2}\times \frac{4\sqrt{2}}{9}$

$=\frac{7\sqrt{3}+4\sqrt{2}}{18}$