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Question: If \[\sin \theta + {\sin ^2}\theta = 1\], then what is the value of \({\cos ^2}\theta + {\cos ^4}\th...

If sin⁡θ+sin⁡2θ=1\sin \theta + {\sin ^2}\theta = 1, then what is the value of cos⁡2θ+cos⁡4θ?{\cos ^2}\theta + {\cos ^4}\theta ?
A. 00
B. 2\sqrt 2
C. 11
D. 2

Explanation

Solution

First of all this is a very simple and a very easy problem. In order to solve this problem we need to have some basic knowledge of trigonometry, which includes basic trigonometric identities and basic trigonometric formulas. Along with this we also need to understand and should be able to solve simple mathematical equations.
Here the trigonometric identity which is used here is as given below:
⇒sin⁡2θ+cos⁡2θ=1\Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1
Hence by rearranging the terms, the above expression becomes, as given below:
⇒sin⁡2θ=1−cos⁡2θ\Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta

Complete step-by-step solution:
Given that sin⁡θ+sin⁡2θ=1\sin \theta + {\sin ^2}\theta = 1, we have to find the value of cos⁡2θ+cos⁡4θ{\cos ^2}\theta + {\cos ^4}\theta .
Consider sin⁡θ+sin⁡2θ=1\sin \theta + {\sin ^2}\theta = 1, as given below:
⇒sin⁡θ+sin⁡2θ=1\Rightarrow \sin \theta + {\sin ^2}\theta = 1
⇒sin⁡θ=1−sin⁡2θ\Rightarrow \sin \theta = 1 - {\sin ^2}\theta
As we know that from the most important trigonometric identity that sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
∴1−sin⁡2θ=cos⁡2θ\therefore 1 - {\sin ^2}\theta = {\cos ^2}\theta
Substituting the above expression in the equation sin⁡θ=1−sin⁡2θ\sin \theta = 1 - {\sin ^2}\theta , replacing 1−sin⁡2θ1 - {\sin ^2}\theta with cos⁡θ\cos \theta , as given below:
⇒sin⁡θ=cos⁡2θ\Rightarrow \sin \theta = {\cos ^2}\theta
Now we have found out that cos⁡2θ{\cos ^2}\theta is equal to sin⁡θ\sin \theta .
∴cos⁡2θ=sin⁡θ\therefore {\cos ^2}\theta = \sin \theta
Now squaring the above equation on both sides, as given below:
⇒(cos⁡2θ)2=(sin⁡θ)2\Rightarrow {\left( {{{\cos }^2}\theta } \right)^2} = {\left( {\sin \theta } \right)^2}
⇒cos⁡4θ=sin⁡2θ\Rightarrow {\cos ^4}\theta = {\sin ^2}\theta
∴cos⁡4θ=sin⁡2θ\therefore {\cos ^4}\theta = {\sin ^2}\theta
We obtained the expressions for both cos⁡4θ{\cos ^4}\theta and cos⁡2θ{\cos ^2}\theta , which are as given below:
⇒cos⁡2θ=sin⁡θ\Rightarrow {\cos ^2}\theta = \sin \theta and
⇒cos⁡4θ=sin⁡2θ\Rightarrow {\cos ^4}\theta = {\sin ^2}\theta
Substituting these above obtained expressions in the expression as given below:
⇒cos⁡2θ+cos⁡4θ\Rightarrow {\cos ^2}\theta + {\cos ^4}\theta
⇒cos⁡2θ+cos⁡4θ=sin⁡θ+sin⁡2θ\Rightarrow {\cos ^2}\theta + {\cos ^4}\theta = \sin \theta + {\sin ^2}\theta
But already given that, sin⁡θ+sin⁡2θ=1\sin \theta + {\sin ^2}\theta = 1
Hence the expression will also be equal to, as given below
⇒cos⁡2θ+cos⁡4θ=1\Rightarrow {\cos ^2}\theta + {\cos ^4}\theta = 1
Hence the value of cos⁡2θ+cos⁡4θ=1{\cos ^2}\theta + {\cos ^4}\theta = 1

The value of cos⁡2θ+cos⁡4θ{\cos ^2}\theta + {\cos ^4}\theta is 1.

Note: While solving this problem we should understand that we are substituting in this equation sin⁡θ=1−sin⁡2θ\sin \theta = 1 - {\sin ^2}\theta , in place of 1−sin⁡2θ1 - {\sin ^2}\theta , replacing 1−sin⁡2θ1 - {\sin ^2}\theta with cos⁡θ\cos \theta . There is a chance that we might be able to confuse while substituting this. One should take care. We have to remember all the trigonometric identities such as: sin⁡2θ+cos⁡2θ=1,sec⁡2θ−tan⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1,{\sec ^2}\theta - {\tan ^2}\theta = 1 and cos⁡ec2θ−cot⁡2θ=1\cos e{c^2}\theta - {\cot ^2}\theta = 1.