Question

$prove \: that \: \frac{sin \: x \: + \: sin \: 3x}{cos \: x \: + \: cos \: 3x} = \: tan \: 2x$
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Answer 1

Topic :-

Trigonometry

To Prove :-

$\dfrac{\sin x + \sin 3x}{\cos x + \cos 3x}=\tan 2x$

Formula to be Used :-

$\sin A + \sin B =2\sin \left ( \dfrac{A+B}{2}\right )\cos \left ( \dfrac{A-B}{2}\right )$

$\cos A+ \cos B =2\cos \left ( \dfrac{A+B}{2}\right )\cos \left ( \dfrac{A-B}{2}\right )$

Proof :-

Solving LHS,

$\dfrac{\sin x + \sin 3x}{\cos x + \cos 3x}$

Applying formula,

$\dfrac{2\sin \left ( \dfrac{x+3x}{2}\right )\cos \left ( \dfrac{x-3x}{2}\right )}{2\cos \left ( \dfrac{x+3x}{2}\right )\cos \left ( \dfrac{x-3x}{2}\right )}$

$\dfrac{2\sin \left ( \dfrac{4x}{2}\right )\cos \left ( \dfrac{-2x}{2}\right )}{2\cos \left ( \dfrac{4x}{2}\right )\cos \left ( \dfrac{-2x}{2}\right )}$

$\dfrac{2\sin \left ( 2x\right )\cos \left ( -x\right )}{2\cos \left ( 2x\right )\cos \left ( -x\right )}$

Canceling 2cos(-x),

$\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x$

$\left ( \because \dfrac{ \sin \theta}{\cos \theta}=\tan \theta\right )$

RHS,

$\tan 2x$

We observe that LHS = RHS.

Hence, Proved !!

Additional Formulae :-

$\sin A - \sin B =2\cos \left ( \dfrac{A+B}{2}\right )\sin \left ( \dfrac{A-B}{2}\right )$

$\cos A - \cos B =2\sin \left ( \dfrac{A+B}{2}\right )\sin \left ( \dfrac{B-A}{2}\right )$