Trigonometric Function
Prove the identity $\frac{\cos 2x}{\sin^{2} x + \frac{1}{2} \sin 2x } $ = $\cot x - 1.$
1 Answer
Prove: \frac{\cos 2x}{\sin^{2} x + \frac{1}{2} \sin 2x} = \cot x - 1
sin
2
x+
2
1
sin2x
cos2x
=cotx−1
Start with LHS: \frac{\cos 2x}{\sin^{2} x + \frac{1}{2} \sin 2x}
sin
2
x+
2
1
sin2x
cos2x
Use double angle identity for cosine: \cos 2x = \cos^2 x - \sin^2 xcos2x=cos
2
x−sin
2
x.
Factor the numerator: \frac{(\cos x + \sin x)(\cos x - \sin x)}{\sin^{2} x + \frac{1}{2} \sin 2x}
sin
2
x+
2
1
sin2x
(cosx+sinx)(cosx−sinx)

16
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