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Question: The value of \[{\cos ^2}30^\circ - {\sin ^2}30^\circ \]is: a. \[\cos 60^\circ \] b. \[\sin 60...

The value of cos⁡230∘−sin⁡230∘{\cos ^2}30^\circ - {\sin ^2}30^\circ is:

a. cos⁡60∘\cos 60^\circ
b. sin⁡60∘\sin 60^\circ
c. 0
d. 1

Explanation

Solution

In this problem we are to find the value of cos⁡230∘−sin⁡230∘{\cos ^2}30^\circ - {\sin ^2}30^\circ . And to solve this problem we are using the given formula, cos⁡2x−sin⁡2x=cos⁡2x{\cos ^2}x - {\sin ^2}x = \cos 2x. Then we will analyze the given options and find which one is the right option.

Complete step-by-step answer:
In this problem we are to find, the value of cos⁡230∘−sin⁡230∘{\cos ^2}30^\circ - {\sin ^2}30^\circ ,
Now, this is known to us that, cos⁡2x−sin⁡2x=cos⁡2x{\cos ^2}x - {\sin ^2}x = \cos 2x
So, we are going to use this formula here,
We are given here, x=30∘x = 30^\circ ,
So, for, cos⁡230∘−sin⁡230∘{\cos ^2}30^\circ - {\sin ^2}30^\circ , we will get
{\cos ^2}30^\circ - {\sin ^2}30^\circ = \cos (2 \times 30^\circ )$$$$ = \cos (60^\circ )
Hence, option (a) is correct.

Note: We prove, cos⁡2x−sin⁡2x=cos⁡2x{\cos ^2}x - {\sin ^2}x = \cos 2x,
We are given, cos⁡2x\cos 2x in this problem,
Now, Applying the angle-sum identity for cosine to cos(x+x).cos(x + x).
We will get,
The identity needed is the angle-sum identity for cosine.
cos(α+β)=cos(α)cos(β)−sin(α)sin(β)cos(\alpha + \beta ) = cos(\alpha )cos(\beta ) - sin(\alpha )sin(\beta )
So, thus, With that, we have
cos(2x)=cos(x+x)cos(2x) = cos(x + x)
=cos(x)cos(x)−sin(x)sin(x)= cos(x)cos(x) - sin(x)sin(x)
=cos2(x)−sin2(x)= co{s^2}(x) - si{n^2}(x)
So, we get, cos⁡2x−sin⁡2x=cos⁡2x{\cos ^2}x - {\sin ^2}x = \cos 2x.