Question

How do you express $\cos 4\theta $ in terms of $\cos 2\theta ?$

Answer 1

Hint:Split $4\theta $ into two $2\theta $ and then apply the cosine compound angle formula and further simplify, use some trigonometric identities in order to simplify it.
The compound formula for cosine function which will be used is
$\cos (a + b) = \cos a\cos b - \sin a\sin b$ , where $a\;{\text{and}}\;b$ are respective arguments of the cosine function.

Complete step by step solution:
In order to express $\cos 4\theta $ in terms of $\cos 2\theta $, we can clearly see that we need $2\theta $ as the argument, so splitting $4\theta $ into $2\theta $
$ \Rightarrow 4\theta = 2\theta + 2\theta $
Now taking the cosine function both sides, we will get
$ \Rightarrow \cos 4\theta = \cos (2\theta + 2\theta )$
Do we have seen this type of expression that is $\cos (2\theta + 2\theta )$ somewhere else?
Yes this is a trigonometric identity and belongs to the compound angle formula for cosine function which is given as
$\cos (a + b) = \cos a\cos b - \sin a\sin b$
With the help of this formula simplifying the equation further,
$
\Rightarrow \cos 4\theta = \cos (2\theta + 2\theta ) \\
\Rightarrow \cos 4\theta = \cos 2\theta \cos 2\theta - \sin 2\theta \sin 2\theta \\
\Rightarrow \cos 4\theta = {\cos ^2}2\theta - {\sin ^2}2\theta \\
$
Now from the identity ${\sin ^2}x + {\cos ^2}x = 1$ we know that
${\sin ^2}x = 1 - {\cos ^2}x$
Substituting this in above equation we will get,
$
\Rightarrow \cos 4\theta = {\cos ^2}2\theta - {\sin ^2}2\theta \\
\Rightarrow \cos 4\theta = {\cos ^2}2\theta - (1 - {\cos ^2}2\theta ) \\

\Rightarrow \cos 4\theta = {\cos ^2}2\theta - 1 + {\cos ^2}2\theta \\
\Rightarrow \cos 4\theta = 2{\cos ^2}2\theta - 1 \\
$
So, $\cos 4\theta $ is expressed in terms of $\cos 2\theta $ as $2{\cos ^2}2\theta - 1$

Note: There is one more way to express $\cos 4\theta $ in terms of $\cos 2\theta $, let us see that from the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$, we can write that
$
\Rightarrow {\cos ^2}x = 1 - {\sin ^2}x \\
\Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x} \\
$
With help of this we can write,
$ \Rightarrow \cos 4\theta = \sqrt {1 - {{\sin }^2}4\theta } $
Using here the algebraic identity ${a^2} - {b^2} = (a + b)(a - b)$ to simplify further
$
\Rightarrow \cos 4\theta = \sqrt {{1^2} - {{\sin }^2}4\theta } \\
\Rightarrow \cos 4\theta = \sqrt {(1 + \sin 4\theta )(1 - \sin 4\theta )} \\
$
Now applying the compound angle formula of sine angle which is given as
$\sin 2x = 2\sin x\cos x$
With help of this simplifying the equation further, we will get
$
\Rightarrow \cos 4\theta = \sqrt {(1 + \sin 2(2\theta ))(1 - \sin 2(2\theta ))} \\
\Rightarrow \cos 4\theta = \sqrt {(1 + 2\sin 2\theta \cos 2\theta )(1 - 2\sin 2\theta \cos 2\theta )}
\\
$
Now from ${\sin ^2}x + {\cos ^2}x = 1$, we will get
$
\Rightarrow \cos 4\theta = \sqrt {(1 + 2\sin 2\theta \cos 2\theta )(1 - 2\sin 2\theta \cos 2\theta )}
\\
\Rightarrow \cos 4\theta = \sqrt {({{\sin }^2}2\theta + {{\cos }^2}2\theta + 2\sin 2\theta \cos
2\theta )({{\sin }^2}2\theta + {{\cos }^2}2\theta - 2\sin 2\theta \cos 2\theta )} \\
$
From the algebraic identity ${(a + b)^2} = ({a^2} + {b^2} + 2ab)\;{\text{and}}\;{(a - b)^2} = ({a^2} +
{b^2} - 2ab)$ we can write
\[
\Rightarrow \cos 4\theta = \sqrt {({{\cos }^2}2\theta + {{\sin }^2}2\theta + 2\sin 2\theta \cos
2\theta )({{\cos }^2}2\theta + {{\sin }^2}2\theta - 2\sin 2\theta \cos 2\theta )} \\
\Rightarrow \cos 4\theta = \sqrt {{{(\cos 2\theta + \sin 2\theta )}^2}{{(\cos 2\theta - \sin 2\theta
)}^2}} \\
\]
We can further write it as
\[
\Rightarrow \cos 4\theta = \sqrt {{{(\cos 2\theta + \sin 2\theta )}^2}{{(\cos 2\theta - \sin 2\theta
)}^2}} \\
\Rightarrow \cos 4\theta = (\cos 2\theta + \sin 2\theta )(\cos 2\theta - \sin 2\theta ) \\
\]
Again from ${a^2} - {b^2} = (a + b)(a - b)$ we can write
\[
\Rightarrow \cos 4\theta = (\cos 2\theta + \sin 2\theta )(\cos 2\theta - \sin 2\theta ) \\
\Rightarrow \cos 4\theta = {\cos ^2}2\theta - {\sin ^2}2\theta \\
\]
Now from ${\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$
\[
\Rightarrow \cos 4\theta = {\cos ^2}2\theta - {\sin ^2}2\theta \\
\Rightarrow \cos 4\theta = {\cos ^2}2\theta - (1 - {\cos ^2}2\theta ) \\
\Rightarrow \cos 4\theta = {\cos ^2}2\theta - 1 + {\cos ^2}2\theta \\
\Rightarrow \cos 4\theta = 2{\cos ^2}2\theta - 1 \\
\]
So we again expressed $\cos 4\theta $ in terms of $\cos 2\theta $, but this method is a bit lengthy and complex, go for the upper one.