Prove the following-sin x-sin 3 x-cos x-cos 3 x-tan 2 x

We have L.H.S. = sin x + sin 3x/cos x + cos 3x = 2 sin (x+3x/2) cos ( x 3x/2)/2 cos (x+3x/2 )cos ( x 3x/2) [ ∵sin C + sin D = 2 sin ( C+D/2) cos ( C D/2); co ...

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Standard XI

Mathematics

Trigonometric Ratios of Multiples of an Angle

Prove the fol...

Question

Prove the following:
$\frac{sin x - sin 3x}{cos x + cos 3x} = tan 2x$

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Solution

We have

L.H.S. = $\frac{sin x + sin 3x}{cos x + cos 3x}$

= $\frac{2 sin (\frac{x + 3 x}{2}) cos (\frac{x - 3 x}{2})}{2 cos (\frac{x + 3 x}{2}) cos (\frac{x - 3 x}{2})}$

$⎡ ⎢ ⎣ \begin{matrix} ∵ sin C + sin D = 2 sin (\frac{C + D}{2}) cos (\frac{C - D}{2}) cos C+ cos D = 2 cos (\frac{C + D}{2}) cos (\frac{C - D}{2}) \end{matrix} ⎤ ⎥ ⎦$

= $\frac{2 sin 2x cos (-x)}{2 cos 2x cos(-x)}$

= tan 2x = R.H.S.

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Prove the following: sin x - sin 3xcos x + cos 3x=tan 2x

We have L.H.S. = sin x+sin 3xcos x+cos 3x = 2 sin(x+3x2)cos(x−3x2)2 cos(x+3x2)cos(x−3x2) ⎡⎢⎣∵sin C+sin D=2 sin(C+D2)cos(C−D2)cos C+ cos D = 2 cos(C+D2)cos(C−D2)⎤⎥⎦ = 2 sin 2x cos (-x)2 cos 2x cos(-x) = tan 2x = R.H.S.

Prove the following:
$\frac{sin x - sin 3x}{cos x + cos 3x} = tan 2x$