How do you simplify ${\cos ^4}x - {\sin ^4}x$ ?

Hint:
This problem can be solved with basic algebraic formulas. We will use the formula of \[{a^2} - {b^2}\] which will give us (a+b)(a-b). And after which we will use some basic trigonometric formulas to solve this problem.

Complete step by step solution:
We have to simplify ${\cos ^4}x - {\sin ^4}x$ for which we will use the formula \[{a^2} - {b^2}\]. Here we have ${\cos ^4}x$ and ${\sin ^4}x$ . So assume a = ${\cos ^2}x$and b = ${\sin ^2}x$then we get,
 $ \Rightarrow {({\cos ^2}x)^2} - {({\sin ^2}x)^2} = {\cos ^4}x - {\sin ^4}x$
By applying our algebraic formula that is (a+b)(a-b) we will get,
 $ = ({\cos ^2}x + {\sin ^2}x)({\cos ^2}x - {\sin ^2}x)$
We already know that ${\cos ^2}x + {\sin ^2}x = 1$ and ${\cos ^2}x - {\sin ^2}x = \cos 2x$. So our problem will get reduced by replacing the values to 1 and cos2x.
 $ = (1).(\cos 2x)$
After simplifying the above equation, we get,
= cos2x

Therefore, ${\cos ^4}x - {\sin ^4}x$ = cos2x.

Additional Information:
We have used the double-angle formula of cos2x. But Double-angle has 2 more formula that are, sin2x = 2 sin(x) cos(x) and tan(2x) = $\dfrac{{2\tan (x)}}{{1 - {{\tan }^2}(x)}}$ . In this problem we have also used the algebraic formula of ${a^2} - {b^2}$ which is (a+b) (a-b).
You need to note that trigonometric function comprises sin, cos, tan, cot, cosec and sec as well as their inverse. Whereas algebraic function doesn’t comprise any of these. They are simple arithmetic operations.

Note:
Here we have used the double angle formula of cos2x but cos2x can also be written as $1 - 2{\sin ^2}x$ and $2{\cos ^2}x - 1$ . So we can say that cos2x, $1 - 2{\sin ^2}x$ and $2{\cos ^2}x - 1$ all these 3 can be our answer. Moreover, we have to use the algebraic formula first because we have no formula for ${\cos ^4}x$ or ${\sin ^4}x$ . That’s why we have to use the algebraic formula first and then the trigonometric formula.