Question

. Simplify sin cube theta + cos square theta upon sin theta + cos theta

Answer 1

Hi ,

Here I used A instead of theta.


( sin³ A + cos³ A )/( sinA + cosA )

= (sinA+cosA)[ sin² A + sinAcosA+cos²A]/(sinA+ cosA )

= ( sin²A + cos² A + sinAcosA )

= 1 + SinA cosA

I hope this helps you.

: )

Answer 2

$\frac{ \sin ^{3} ( \alpha ) + \cos ^{3} ( \alpha ) }{ \sin( \alpha ) + \cos ( \alpha ) } \\ \\ \frac{( \sin( \alpha ) + \cos( \alpha ) )( { \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} + \sin( \alpha ) \cos( \alpha ) )}{ \sin( \alpha ) + \cos( \alpha ) } \\ \\ \sin^{2} ( \alpha ) + { \cos^{2} ( \alpha ) } + \sin( \alpha ) \cos( \alpha ) \\ \\ 1 + \sin( \alpha ) \cos( \alpha )$