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Question: If we have an equation as \(8\cos 2\theta +8\sec 2\theta =65,0 < \theta < \dfrac{\pi }{2}\) then the...

If we have an equation as 8cos⁡2θ+8sec⁡2θ=65,0<θ<π28\cos 2\theta +8\sec 2\theta =65,0 < \theta < \dfrac{\pi }{2} then the value of 4cos⁡4θ4\cos 4\theta is equal to:
(1) −338\dfrac{-33}{8}
(2) −318\dfrac{-31}{8}
(3) −3132\dfrac{-31}{32}
(4) −3332\dfrac{-33}{32}

Explanation

Solution

Here in this question we have been asked to find the value of 4cos⁡4θ4\cos 4\theta given that 8cos⁡2θ+8sec⁡2θ=65,0<θ<π28\cos 2\theta +8\sec 2\theta =65,0 < \theta < \dfrac{\pi }{2} . To answer this question we will convert sec⁡2θ\sec 2\theta into cos⁡2θ\cos 2\theta and then simplify the quadratic expression that we get and find the value of cos⁡2θ\cos 2\theta and use it to derive the required result.

Complete step-by-step solution:
Now considering from the question we have been asked to find the value of 4cos⁡4θ4\cos 4\theta given that 8cos⁡2θ+8sec⁡2θ=65,0<θ<π28\cos 2\theta +8\sec 2\theta =65,0 < \theta < \dfrac{\pi }{2} .
Now by using sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x}in the given expression, we will have 8cos⁡2θ+8(1cos⁡2θ)=65 8\cos 2\theta +8\left( \dfrac{1}{\cos 2\theta } \right)=65 .
Now by further simplifying this expression we will have 8cos⁡22θ+8=65cos⁡2θ8{{\cos }^{2}}2\theta +8=65\cos 2\theta.
By further simplifying this expression we will have
⇒8cos⁡22θ−64cos⁡2θ−cos⁡2θ+8=0 ⇒(8cos⁡2θ−1)(cos⁡2θ−8)=0 \begin{aligned} & \Rightarrow 8{{\cos }^{2}}2\theta -64\cos 2\theta -\cos 2\theta +8=0 \\\ & \Rightarrow \left( 8\cos 2\theta -1 \right)\left( \cos 2\theta -8 \right)=0 \\\ \end{aligned} .
Now we know that the value of cos⁡x\cos x always lies between -1 and 1.
Hence now we will have cos⁡2θ=18\cos 2\theta =\dfrac{1}{8} .
From the basic concepts of trigonometry, we know that cos⁡2x=2cos⁡2x−1\cos 2x=2{{\cos }^{2}}x-1 .
Hence now we can say that
cos⁡4θ=2cos⁡22θ−1 ⇒2(18)2−1=132−1 ⇒−3132 \begin{aligned} & \cos 4\theta =2{{\cos }^{2}}2\theta -1 \\\ & \Rightarrow 2{{\left( \dfrac{1}{8} \right)}^{2}}-1=\dfrac{1}{32}-1 \\\ & \Rightarrow \dfrac{-31}{32} \\\ \end{aligned} .
Therefore we can conclude that when it is given that 8cos⁡2θ+8sec⁡2θ=65,0<θ<π28\cos 2\theta +8\sec 2\theta =65,0 < \theta < \dfrac{\pi }{2} then we will have the value of 4cos⁡4θ4\cos 4\theta as −318\dfrac{-31}{8} .
Hence we will mark the option “2” as correct.

Note: While answering questions of this type we should be sure with the formula that we are going to use in the process of simplification in between the steps. This question seems to be a complex one but when we start applying the appropriate it gets solved within a short span of time. The formula for cos⁡2x\cos 2x can also be given as 1−2sin⁡2x=cos⁡2x−sin⁡2x1-2{{\sin }^{2}}x={{\cos }^{2}}x-{{\sin }^{2}}x .