Question

if sin theta +cos theta = a, then sin^4 theta, then sin^4 + cos^4 theta = ?

Answer 1

Answer:

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

Step-by-step explanation:

${ \sin( \alpha ) }^{} + { \cos( \alpha ) }^{} = a \\ \\ on \: squaring \: both \: the \: sides \\ \\ {( \sin( \alpha ) + \cos( \alpha ) ) }^{2} = {a}^{2} \\ \\ { \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} + 2 \sin( \alpha ) \cos( \alpha ) = {a}^{2} \\ \\ \\ \\ therefore \: \: \: \: \: 1 + 2 \sin( \alpha ) \cos( \alpha ) = {a}^{2} \\ \\ then \: \: \: \: \sin( \alpha ) \cos( \alpha ) = \frac{ {a}^{2} - 1 }{2} \\ \\ { \sin( \alpha ) }^{4} + { \cos( \alpha ) }^{4} = 1 - 2 { \sin( \alpha ) }^{2} { \cos( \alpha ) }^{2} \\ \\ = 1 - 2 \frac{ {( {a}^{2} - 1)}^{2} }{ {2}^{2} } \\ \\ = 1 - \frac{ {{( {a}^{2} - 1) }^{2} }^{} }{2} \\ \\ = \frac{2 - {a}^{4 } - 1 + 2 {a}^{2} }{2} \\ \\ = \frac{ 1 - {a}^{4} + 2 {a}^{2} }{2} \\ \\ = \frac{{(1 - {a}^{2} )}^{2} }{2}$