Question

prove that cos 8theta=128 cos^8 theta-256 cos^6 theta 2+160 cos^4theta-32cos^theta+1​

Answer 1

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

Consider

$\rm \: {cos}8\theta$

can be rewritten as

$\rm \: = \: cos2(4\theta )$

We know,

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

So, using this identity, we get

$\rm \: = \: {2cos}^{2}4\theta - 1$

$\rm \: = \: 2 {(cos4\theta )}^{2} - 1$

$\rm \: = \: 2 {[(cos2(2\theta )]}^{2} - 1$

$\rm \: = \: 2 {[ {2cos}^{2}2\theta - 1]}^{2} - 1$

$\rm \: = \: 2 {[ {2(cos2\theta )}^{2} - 1]}^{2} - 1$

$\rm \: = \: 2 {[ {2( {2cos}^{2}\theta - 1)}^{2} - 1]}^{2} - 1$

$\rm \: = \: 2 {[ {2( {4cos}^{4}\theta + 1 - {4cos}^{2}\theta)} - 1]}^{2} - 1$

$\rm \: = \: 2 {[ {{8cos}^{4}\theta + 2 - {8cos}^{2}\theta} - 1]}^{2} - 1$

$\rm \: = \: 2 {[ {{8cos}^{4}\theta + 1 - {8cos}^{2}\theta}]}^{2} - 1$

$\rm \: = 2( {64cos}^{8}\theta + 1 + {64cos}^{4}\theta + 16 {cos}^{4}\theta - {16cos}^{2}\theta - 128 {cos}^{6}\theta) - 1$

$\rm \: = 2( {64cos}^{8}\theta + 1 + {80cos}^{4}\theta - {16cos}^{2}\theta - 128 {cos}^{6}\theta) - 1$

$\rm \: = {128cos}^{8}\theta + 2 + {160cos}^{4}\theta - {32cos}^{2}\theta - 256 {cos}^{6}\theta - 1$

$\rm \: = {128cos}^{8}\theta + {160cos}^{4}\theta - {32cos}^{2}\theta - 256 {cos}^{6}\theta + 1$

can be re-arranged as

$\rm \: = {128cos}^{8}\theta - 256 {cos}^{6}\theta+{160cos}^{4}\theta-{32cos}^{2}\theta+1$

Hence,

$\boxed{\rm cos8\theta = {128cos}^{8}\theta - 256 {cos}^{6}\theta+{160cos}^{4}\theta-{32cos}^{2}\theta+1}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered} { \boxed{ \begin{array}{c} \underline{\underline{ \color{orange} \text{Additional \: lnformation}}} \\& \rm \: sin2x \: = 2 \: sinx \: cosx\:\\ & \rm \: cos2x = 1 - {2sin}^{2}x \\ & \rm \: cos2x = {2cos}^{2}x - 1 \\ & \rm \: cos2x = {cos}^{2}x - {sin}^{2}x \\ & \rm \:tan2x = \dfrac{2tanx}{1 - {tan}^{2} x} \\ & \rm \: sin2x = \dfrac{2tanx}{1 + {tan}^{2}x } \\ & \rm \:sin3x = 3sinx - {4sin}^{3}x \\ & \rm \: cos3x = {4cos}^{3}x - 3cosx \\ & \rm \: tan3x = \dfrac{3tanx - {tan}^{3} x}{1 - {3tan}^{2}x} \end{array}}}\end{gathered}$

Answer 2

Step-by-step explanation:

prove that cos 8theta=128 cos^8 theta-256 cos^6 theta 2+160 cos^4theta-32cos^theta+1