Question

If $\cos \theta =\frac{-5}{13}$ and $\frac{\pi }{2}<\theta <\pi$ then find the values of

(i) $\sin 3\theta +\sin 5\theta$

(ii) $\tan 3\theta$

Answer 1

Given;

$\cos \theta =\frac{-5}{13}$ and $\frac{\pi }{2}<\theta <\pi$

$\sin \theta =\sqrt{1-{\cos }^2\theta }$ [Since, $\sin \theta$ is positive in second quadrat]

$\Rightarrow \sin \theta =\sqrt{1-(\frac{-5}{13}{)}^2}$

$\Rightarrow \sin \theta =\sqrt{\frac{169-24}{169}}$

$\Rightarrow \sin \theta =\sqrt{\frac{144}{169}}$

$\Rightarrow \sin \theta =\frac{12}{13}$ ------(1)

Now, $\tan \theta =\frac{\sin \theta }{\cos \theta }$

$\tan \theta =\frac{\frac{12}{13}}{\frac{-5}{13}}$

$\tan \theta =\frac{-12}{5}$-------(2)

Now,

(i) $\sin 3\theta +\sin 5\theta$

$=2\sin \left(\frac{3\theta +5\theta }{2}\right)\cos \left(\frac{3\theta -5\theta }{2}\right)$

$=2\sin 4\theta \cos \theta$

$=2\left(2\sin 2\theta \cos 2\theta \right)\left(\cos \theta \right)$

$=4\left(2\sin \theta \cos \theta \right)(1-2{\sin }^2\theta )\cos \theta$

$=8\left(\frac{12}{13}\right)\left(\frac{-5}{13}\right)(1-2(\frac{12}{13}{)}^2\left)\right(\frac{-5}{13})$

$=\frac{-480}{169}\left(\frac{169-288}{169}\right)\left(\frac{-5}{13}\right)$

$=\frac{-480}{169}\left(\frac{-119}{169}\right)\left(\frac{-5}{13}\right)$

$=\frac{-285600}{371293}$

(ii) $\tan 3\theta$

$\tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

$=\frac{3\left(\frac{-12}{5}\right)-(\frac{-12}{5}{)}^3}{1-3\left(\frac{-12}{5}\right)}$

$=\frac{3\left(\frac{-12}{5}\right)-\left(\frac{-1728}{125}\right)}{1-3(\frac{-12}{5}{)}^2}$

$=\frac{\frac{-36\left(25\right)+1728}{125}}{\frac{25-432}{25}}$

$=\frac{\frac{-828}{125}}{\frac{-407}{25}}$

$=\frac{-828}{125}\times \frac{25}{-407}$

$=\frac{828}{814}$

$=\frac{414}{407}$