Question

cos 6x =32 cos ^6X - 48 cos ^ 4 x +18 cos ^2x-1​ (need to prove )​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{{\bf \: cos2x = 2 {cos}^{2} x - 1}}$

$\boxed{{\bf \: cos3x = 4 {cos}^{3} x - 3cosx}}$

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\:cos6x$

$\sf \: \: \: \: = \: \: cos2(3x)$

$\sf \: \: \: \: = \: \: 2 {cos}^{2} 3x - 1$

$\sf \: \: \: \: = \: \: 2 {(cos3x)}^{2} - 1$

$\sf \: \: \: \: = \: \: 2 {\bigg( {4cos}^{3}x - 3cosx \bigg) }^{2} - 1$

$\sf \: \: \: \: = \: \: 2\bigg(16 {cos}^{6}x + {9cos}^{2}x - 24 {cos}^{4} x \bigg) - 1$

$\sf \: \: \: \: = \: \: {32cos}^{6}x - {48cos}^{4}x + {18cos}^{2}x - 1$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\boxed{{\bf \: sin2x = 2sinxcosx}}$

$\boxed{{\bf \: cos2x = {cos}^{2} x - {sin}^{2} x = 1 - {2sin}^{2} x = {2cos}^{2} x - 1}}$

$\boxed{{\bf \: tan2x = \dfrac{2tanx}{1 - {tan}^{2} x} }}$

$\boxed{{\bf \: sin3x = 3sinx - 4 {sin}^{3}x}}$

$\boxed{{\bf \: tan3x = \dfrac{3tanx - {tan}^{3} x}{1 - 3 {tan}^{2}x }}}$

$\boxed{{\bf \: 1 - cos2x = {2sin}^{2} x}}$

$\boxed{{\bf \: 1 + cos2x = {2cos}^{2} x}}$