Question

The value of \[{\cos ^2}30^\circ - {\sin ^2}30^\circ \]is:

a. \[\cos 60^\circ \]
b. \[\sin 60^\circ \]
c. 0
d. 1

Answer 1

Hint: In this problem we are to find the value of \[{\cos ^2}30^\circ - {\sin ^2}30^\circ \]. And to solve this problem we are using the given formula, \[{\cos ^2}x - {\sin ^2}x = \cos 2x\]. 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 \[{\cos ^2}30^\circ - {\sin ^2}30^\circ \],
Now, this is known to us that, \[{\cos ^2}x - {\sin ^2}x = \cos 2x\]
So, we are going to use this formula here,
We are given here, \[x = 30^\circ \],
So, for, \[{\cos ^2}30^\circ - {\sin ^2}30^\circ \], 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, \[{\cos ^2}x - {\sin ^2}x = \cos 2x\],
We are given, \[\cos 2x\] in this problem,
Now, Applying the angle-sum identity for cosine to \[cos(x + x).\]
We will get,
The identity needed is the angle-sum identity for cosine.
\[cos(\alpha + \beta ) = cos(\alpha )cos(\beta ) - sin(\alpha )sin(\beta )\]
So, thus, With that, we have
\[cos(2x) = cos(x + x)\]
\[ = cos(x)cos(x) - sin(x)sin(x)\]
\[ = co{s^2}(x) - si{n^2}(x)\]
So, we get, \[{\cos ^2}x - {\sin ^2}x = \cos 2x\].