Question

Prove that: $\cos 4x = 1 - 8si{n^2}xco{s^2}x$

Answer 1

Here in this problem we are going to use the formula of $cos2x = 2co{s^2}x - 1$ and use it for $\cos 4x = \cos (2(2x))$, and again we use $cos2x = 2co{s^2}x - 1$ for further simplification, on solving the problem we will reach to our desired result.

Complete step by step Answer:

Given, $cos4x = 1 - 8si{n^2}xco{s^2}x$
Taking LHS, $cos4x$,
We know, $cos2x = 2co{s^2}x - 1$
Replacing x by $2x$ , we get,
$cos2\left( {2x} \right) = 2co{s^2}\left( {2x} \right) - 1$
$\Rightarrow cos4x = 2co{s^2}2x - 1$
Using $cos2x = 2co{s^2}x - 1$ again, we get,
$= 2{\left( {2co{s^2}x - 1} \right)^2} - 1\;$
Using ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, we get,
$= 2\left[ {{{\left( {2co{s^2}x} \right)}^2} + {{\left( 1 \right)}^2} - 2\left( {2co{s^2}x} \right) \times 1} \right] - 1$
On Simplification we get,
$= 2\left[ {4{{\cos }^4}x + 1 - 4{{\cos }^2}x} \right] - 1$
On opening the bracket we get,
$= 8{\cos ^4}x + 2 - 8{\cos ^2}x - 1$
On further simplification we get,
$= 8{\cos ^4}x + 1 - 8{\cos ^2}x$
On taking terms common we get,
$= 8co{s^2}x\left( {co{s^2}x - 1} \right) + 1$
On taking -1 common we get,
$= 8co{s^2}x\left[ { - \left( {1 - co{s^2}x} \right)} \right] + 1$
$= - 8co{s^2}x\left[ {\left( {1 - co{s^2}x} \right)} \right] + 1$
Using, $si{n^2}x = 1 - co{s^2}x$, we get,
$= - 8co{s^2}xsi{n^2}x + 1\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;$
On rearranging we get,
$= 1 - 8si{n^2}xco{s^2}x$
=RHS
Hence, $\cos 4x = 1 - 8si{n^2}xco{s^2}x$

Note: We can prove the result, $cos2x = 2co{s^2}x - 1$ by,
Using the identity: $cos{\text{ }}\left( {a{\text{ }} + {\text{ }}b} \right){\text{ }} = {\text{ }}cos{\text{ }}a.cos{\text{ }}b{\text{ }} - {\text{ }}sin{\text{ }}a.sin{\text{ }}b$
cos2x = cos(x + x) = cosx.cosx - sinx.sinx = co{s^2}x - si{n^2}x$$$$ = co{s^2}x - (1 - co{s^2}x) = 2co{s^2}x - 1