Question

Identity transformations of Trigonometric Expressions.
prove the following identities.
$si{n}^6\alpha +co{s}^6\alpha +3si{n}^2\alpha co{s}^2\alpha =1.$

Answer 1

$si{n}^6\alpha +co{s}^6\alpha +3si{n}^2\alpha co{s}^2\alpha =1$

$=(si{n}^2\alpha +co{s}^2\alpha {)}^3-3si{n}^2\alpha co{s}^2\alpha +3si{n}^2\alpha co{s}^2\alpha$

$=(si{n}^2\alpha +co{s}^2\alpha {)}^3$

$=1^3$

$=1$