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Question

Mathematics Question on Trigonometric Functions

sin⁡x−sin⁡3xsin⁡2x−cos⁡2x\frac{\sin x - \sin 3x}{\sin^{2} x -\cos^{2} x} is equal to

A

−2 sin x- 2\, sin\, x

B

2sin⁡ x\frac{2}{\sin\, x }

C

1sin⁡ x\frac{1}{\sin\, x }

D

2 sin x2 \,sin\, x

Answer

2 sin x2 \,sin\, x

Explanation

Solution

sin⁡x−sin⁡3xsin⁡2x−cos⁡2x=2 cos⁡ 2x sin⁡(−x)−cos⁡ 2x\frac{\sin x - \sin 3x}{\sin^{2} x -\cos^{2} x} = \frac{2 \, \cos \, 2x \, \sin (-x)}{ -\cos\, 2 x}
=2 sin⁡ x= 2 \, \sin \, x