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Question: How do you rewrite \(\cos 3\theta \) in terms of only \(\cos \theta \) and \(\sin \theta \)?...

How do you rewrite cos⁡3θ\cos 3\theta in terms of only cos⁡θ\cos \theta and sin⁡θ\sin \theta ?

Explanation

Solution

Hint : We first try to use the associative formula of cos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y for cos⁡3θ\cos 3\theta . We convert all the ratios to cos form by using cos⁡2θ=2cos⁡2θ−1\cos 2\theta =2{{\cos }^{2}}\theta -1 and sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta . Change of identity formula of sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta gives the solution.

Complete step-by-step answer :
We know the multiple angle formula of cos⁡2θ=2cos⁡2θ−1\cos 2\theta =2{{\cos }^{2}}\theta -1 and sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta .
We now break the given cos⁡3θ\cos 3\theta as cos⁡(2θ+θ)\cos \left( 2\theta +\theta \right).
We now use the associative angle formula of cos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y.
Placing the values x=2θ,y=θx=2\theta ,y=\theta we get
cos⁡(2θ+θ) =cos⁡(2θ)cos⁡θ−sin⁡(2θ)sin⁡θ \begin{aligned} & \cos \left( 2\theta +\theta \right) \\\ & =\cos \left( 2\theta \right)\cos \theta -\sin \left( 2\theta \right)\sin \theta \\\ \end{aligned}
Now we replace the values and get
cos⁡(2θ+θ) =(2cos⁡2θ−1)cos⁡θ−(2sin⁡θcos⁡θ)sin⁡θ =2cos⁡3θ−cos⁡θ−2sin⁡2θcos⁡θ \begin{aligned} & \cos \left( 2\theta +\theta \right) \\\ & =\left( 2{{\cos }^{2}}\theta -1 \right)\cos \theta -\left( 2\sin \theta \cos \theta \right)\sin \theta \\\ & =2{{\cos }^{3}}\theta -\cos \theta -2{{\sin }^{2}}\theta \cos \theta \\\ \end{aligned}
Now we try to convert the ratio sin into ratio of cos. We use sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta .
Therefore,
cos⁡(2θ+θ) =2cos⁡3θ−cos⁡θ−2sin⁡2θcos⁡θ =2cos⁡3θ−cos⁡θ−2(1−cos⁡2θ)cos⁡θ =2cos⁡3θ−cos⁡θ−2cos⁡θ+2cos⁡3θ =4cos⁡3θ−3cos⁡θ \begin{aligned} & \cos \left( 2\theta +\theta \right) \\\ & =2{{\cos }^{3}}\theta -\cos \theta -2{{\sin }^{2}}\theta \cos \theta \\\ & =2{{\cos }^{3}}\theta -\cos \theta -2\left( 1-{{\cos }^{2}}\theta \right)\cos \theta \\\ & =2{{\cos }^{3}}\theta -\cos \theta -2\cos \theta +2{{\cos }^{3}}\theta \\\ & =4{{\cos }^{3}}\theta -3\cos \theta \\\ \end{aligned}
Therefore, expressing cos⁡3θ\cos 3\theta in terms of only cos⁡θ\cos \theta , we get cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta .
Now we try to convert them into ratio sin.
cos⁡3θ =cos⁡θ(4cos⁡2θ−3) =cos⁡θ(4−4sin⁡2θ−3) =cos⁡θ(1−4sin⁡2θ) \begin{aligned} & \cos 3\theta \\\ & =\cos \theta \left( 4{{\cos }^{2}}\theta -3 \right) \\\ & =\cos \theta \left( 4-4{{\sin }^{2}}\theta -3 \right) \\\ & =\cos \theta \left( 1-4{{\sin }^{2}}\theta \right) \\\ \end{aligned}
So, the correct answer is “cos⁡θ(1−4sin⁡2θ)\cos \theta \left( 1-4{{\sin }^{2}}\theta \right)”.

Note : We cannot convert the whole expression into expression of sin⁡θ\sin \theta as that brings the root form of cos⁡θ=±1−sin⁡2θ\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }. This form is not a proper expression to get the exact value we need to know about the position of the angle to find the sign.