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Question Number 213504 by Spillover last updated on 06/Nov/24

Answered by A5T last updated on 06/Nov/24

psin^2 x−q(1−sin^2 x)=p−q psin^2 x−q+qsin^2 x=p−q sin^2 x(p+q)=p⇒sin^2 x=(p/(p+q))⇒sin^4 x=(p^2 /((p+q)^2 )) cos^4 x=(1−sin^2 x)^2 =(1−(p/(p+q)))^2 =(q^2 /((p+q)^2 )) ⇒((sin^4 x)/p)+((cos^4 x)/q)=(p/((p+q)^2 ))+(q/((p+q)^2 ))=(1/(p+q))

psin2x−q(1−sin2x)=p−qpsin2x−q+qsin2x=p−qsin2x(p+q)=p⇒sin2x=pp+q⇒sin4x=p2(p+q)2cos4x=(1−sin2x)2=(1−pp+q)2=q2(p+q)2⇒sin4xp+cos4xq=p(p+q)2+q(p+q)2=1p+q

Commented by Spillover last updated on 07/Nov/24

great solution.thanks

greatsolution.thanks

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