Question

Prove the following identities.
$2(si{n}^6\alpha +co{s}^6\alpha )-3(si{n}^4\alpha +co{s}^4\alpha )+1=0$

Answer 1

${\sin }^6\alpha +{\cos }^6\alpha$

$=({\sin }^2\alpha {)}^3+({\cos }^2\alpha {)}^3$

$=({\sin }^2\alpha +{\cos }^2\alpha )({\sin }^4\alpha +{\cos }^4\alpha -{\sin }^2\alpha {\cos }^2\alpha )$

$=({\sin }^4\alpha +{\cos }^4\alpha -{\sin }^2\alpha {\cos }^2\alpha )$

Now-

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

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

$=1-2{\sin }^2\alpha {\cos }^2\alpha -({\sin }^4\alpha +{\cos }^4\alpha )$

$=1-2{\sin }^2\alpha {\cos }^2\alpha -\left\{\right({\sin }^2\alpha +{\cos }^2\alpha {)}^2-2{\sin }^2\alpha {\cos }^2\alpha \}$

$=1-2{\sin }^2\alpha {\cos }^2\alpha -\{1-2{\sin }^2\alpha {\cos }^2\alpha \}$

$=0$