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Question: What is the value of \({{\cos }^{2}}\theta \) in terms of \(\sin \theta \)?...

What is the value of cos⁡2θ{{\cos }^{2}}\theta in terms of sin⁡θ\sin \theta ?

Explanation

Solution

We should know that this problem is from trigonometric ratios. Here we have been asked to deduce a relation and more specifically, write the value of cos⁡2θ{{\cos }^{2}}\theta in terms of sin⁡θ\sin \theta or maybe as a function of sin⁡θ\sin \theta . For that we have to first deduce a relation with only sin⁡θ\sin \theta and cos⁡θ\cos \theta in it. We can also use some trigonometric identities such as sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 to get the desired answer.

Complete step-by-step solution:
In the given question, we have been asked the relation of cos⁡2θ{{\cos }^{2}}\theta in terms of sin⁡θ\sin \theta . Now, following the rules of trigonometry, we have an identity which has both cos⁡2θ{{\cos }^{2}}\theta and sin⁡θ\sin \theta in it, which is given as,
sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
Rearranging the terms in this, we can say that,
cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta
We know the identity given by (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} . Here, if we check the RHS, we can see that it is in the form if a2−b2{{a}^{2}}-{{b}^{2}} , where a is 1 and b is sin⁡θ\sin \theta .
Applying the identity, we get
cos⁡2θ=(1+sin⁡θ)(1−sin⁡θ){{\cos }^{2}}\theta =\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)
Hence, we have derived cos⁡2θ{{\cos }^{2}}\theta in terms of sin⁡θ\sin \theta .
We can also take another approach and check if it satisfies the given condition or not. We know that,
tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta }
Now, this can also be written as,
cos⁡θ=sin⁡θtan⁡θ\cos \theta =\dfrac{\sin \theta }{\tan \theta }
Now when we square both sides, we can write it as
cos⁡2θ=sin⁡2θtan⁡2θ{{\cos }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta }{{{\tan }^{2}}\theta }
We can see that this is also another relation which has cos⁡2θ{{\cos }^{2}}\theta and sin⁡θ\sin \theta , but since it also contains tan⁡θ\tan \theta , we will not consider it. We must follow the earlier approach to satisfy the condition given in the question.

Note: While solving this question, we should remember that trigonometry involves a lot of ratios and identities. But the main point that we have to consider is to choose an identity which includes sin⁡θ\sin \theta and cos⁡θ\cos \theta only and has no other trigonometric ratios.