# If sin theta + cos theta =p, what is sin^2 theta + cos^4theta in terms of p ?

Feb 22, 2017

$1 - {\left(\frac{{p}^{2} - 1}{2}\right)}^{2}$

#### Explanation:

${\left(\sin \theta + \cos \theta\right)}^{2} = 1 + 2 \sin \theta \cos \theta = {p}^{2}$

so

$\sin \theta \cos \theta = \frac{{p}^{2} - 1}{2}$

now

${\sin}^{2} \theta + {\cos}^{4} \theta = {\sin}^{2} \theta + \left(1 - {\sin}^{2} \theta\right) {\cos}^{2} \theta = 1 - {\sin}^{2} \theta {\cos}^{2} \theta$

and putting all together

${\sin}^{2} \theta + {\cos}^{4} \theta = 1 - {\left(\frac{{p}^{2} - 1}{2}\right)}^{2}$