Question

prove that sin ^6theta + cos^6 theta)= 1-3sin^2 Theta cos^2 theta​

Answer 1

Step-by-step explanation:

${ \sin( \alpha ) }^{6} + { \cos(\alpha ) }^{6} = {( {( \sin( \alpha ) )}^{2} )}^{3} + {( { \cos( \alpha ) }^{2}) }^{3} \\ \\ =( ({ \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2})( ({ { \sin( \alpha ) }^{2}) }^{2} + { {( \cos( \alpha ) }^{2}) }^{2} - { \sin( \alpha ) }^{2} { \cos( \alpha ) }^{2} ) \\ \\ = { \sin( \alpha ) }^{4} + { \cos( \alpha ) }^{4} - { \sin( \alpha ) }^{2} { \cos( \alpha ) }^{2} \\ \\ = 1 - 2 { \sin( \alpha ) }^{2} { \cos( \alpha ) }^{2} - { \sin( \alpha ) }^{2} { \cos( \alpha ) }^{2} \\ \\ = 1 - 3 { \sin( \alpha ) }^{2} { \cos( \alpha ) }^{2} \\ \\ hence \: proved$