Question

sove above. question

Answer 1

SOLUTION -:

➡️TAKING L.H.S

\cos6x = \cos2(3x)cos6x=cos2(3x)
➡️USING THE Formula⤵️

\cos(2x) = 2 \cos^{2}x - 1cos(2x)=2cos
2
x−1
================

\begin{lgathered}\cos2(3x) \\ = 2 \cos {}^{2} (3x) - 1 \\ = 2( \cos3x) {}^{2} - 1 \\ Using \: the \: formula \: of \: \cos(3x ) = 4 \cos^{3}x - 3cosx \\ = 2(4 \cos {}^{3}x - 3cosx) {}^{2} - 1 \\ = 2(16cos {}^{6}x + 9cos {}^{2} x - 24cos {}^{4}x) - 1 \\ = 32cos {}^{6} + 18cos {}^{2} x - 48cos {}^{4}x - 1\end{lgathered}
cos2(3x)
=2cos
2
(3x)−1
=2(cos3x)
2
−1
Usingtheformulaofcos(3x)=4cos
3
x−3cosx
=2(4cos
3
x−3cosx)
2
−1
=2(16cos
6
x+9cos
2
x−24cos
4
x)−1
=32cos
6
+18cos
2
x−48cos
4
x−1
​