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

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Answer 1 · 2017-02-22T10:05:42

# 1-((p^2-1)/2)^2#

#(sintheta+costheta)^2= 1+2sinthetacostheta = p^2#

so

#sinthetacostheta = (p^2-1)/2#

now

#sin^2theta+cos^4theta=sin^2theta+(1-sin^2theta)cos^2theta=1-sin^2thetacos^2theta#

and putting all together

#sin^2theta+cos^4theta = 1-((p^2-1)/2)^2#