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  4. If 8cos2θ + 8sec2θ = 65, 0 < θ < π /2, then the value of 4cos4θ is equal to

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Q. If $8cos2\theta + 8sec2\theta = 65$, $0 < \theta < \pi /2$, then the value of $4cos4\theta$ is equal to

Trigonometric Functions

Solution:

$8cos2\theta + 8sec2\theta = 65$, $0 < \theta < \pi/2$
$\Rightarrow 8cos^{2}2\theta + 8 = 65\, cos2\theta$
$\Rightarrow 8cos^{2}2\theta - 65\, cos2\theta + 8 = 0$
$\Rightarrow \left(cos2\theta - 8\right)\left(8cos2\theta - 1\right) = 0$
$\Rightarrow cos2\theta = 1/8$,
$cos2\theta = 8$,
$\left(cos2\theta \ne 8 \, {\text{ as}} \,cos2\theta\,\in \left[- 1, 1\right]\right)$
So,$ cos2\theta = 1/8$. Now, $4 \,cos\, 4\theta = 4(2cos^22\theta - 1)$
$= 4(2\cdot (1/64) - 1) = - 31/8$