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  4. For any angle θ, the expression ( 2 cos 8 θ + 1 / 2 cos θ + 1 ) =

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Q. For any angle $\theta$, the expression $\frac{ 2 \cos 8 \theta + 1 }{ 2 \cos \theta + 1 } = $

Trigonometric Functions

Solution:

$ \frac{2 \cos 8\theta + 1 }{2\cos \theta +1} = \frac{2\left(2\cos^{2} 4 \theta - 1\right) - 1}{2 \cos \theta + 1 }$
$ = \frac{\left(2 \cos 4 \theta -1\right)\left(2\cos 4 \theta +1 \right)}{2 \cos \theta +1}$
$ = \frac{\left(2\cos4 \theta - 1 \right)\left(4 \cos^{2} 2\theta -1\right)}{2\cos \theta +1}$
$ =\frac{\left(2 \cos 4 \theta -1\right)\left(2\cos 2\theta -1\right)\left(2\cos 2 \theta +1\right)}{2 \cos \theta +1}$
$ =\frac{\left(2\cos 4\theta -1\right)\left(2 \cos 2 \theta-1\right) \left(4\cos^{2} \theta -1\right) }{2 \cos \theta +1} $