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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))

$${psin}^{\mathrm{2}} {x}−{q}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)={p}−{q} \\ $$$${psin}^{\mathrm{2}} {x}−{q}+{qsin}^{\mathrm{2}} {x}={p}−{q} \\ $$$${sin}^{\mathrm{2}} {x}\left({p}+{q}\right)={p}\Rightarrow{sin}^{\mathrm{2}} {x}=\frac{{p}}{{p}+{q}}\Rightarrow{sin}^{\mathrm{4}} {x}=\frac{{p}^{\mathrm{2}} }{\left({p}+{q}\right)^{\mathrm{2}} } \\ $$$${cos}^{\mathrm{4}} {x}=\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} =\left(\mathrm{1}−\frac{{p}}{{p}+{q}}\right)^{\mathrm{2}} =\frac{{q}^{\mathrm{2}} }{\left({p}+{q}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{sin}^{\mathrm{4}} {x}}{{p}}+\frac{{cos}^{\mathrm{4}} {x}}{{q}}=\frac{{p}}{\left({p}+{q}\right)^{\mathrm{2}} }+\frac{{q}}{\left({p}+{q}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{p}+{q}} \\ $$

Commented by Spillover last updated on 07/Nov/24

great solution.thanks

$${great}\:{solution}.{thanks} \\ $$

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