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NCERT Class XI Mathematics - Trigonometric Functions - Solutions

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Question : 37 of 61
Marks: +1, -0
sin⁡x−sin⁡3xsin⁡2x−cos⁡2x\frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2 sin x
Solution:  
We have, L.H.S.= sin⁡x−sin⁡3xsin⁡2x−cos⁡2x\frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2cos⁡(x+3x2)sin⁡(x−3x2)−(cos⁡2x−sin⁡2x)\frac{2\cos\left(\frac{x+3x}{2}\right)\sin\left(\frac{x-3x}{2}\right)}{-\left(\cos^2 x - \sin^2 x\right)}
= 2cos⁡2xsin⁡(−x)−cos⁡2x\frac{2\cos 2x \sin(-x)}{-\cos 2x} [Since cos⁡2x−sin⁡2\cos^2 x - \sin^2 x = cos 2x]
= −2sin⁡x−1\frac{-2\sin x}{-1} [Since sin (θ) = - isn θ] = sin x = R.H.S.
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