Question

If $\frac{{\sin }^4\theta }{a}+\frac{{\cos }^4\theta }{b}=\frac{1}{(a+b)}$ then prove that $\frac{{\sin }^8\theta }{a^3}+\frac{{\cos }^8\theta }{b^3}=\frac{2}{(a+b{)}^3}$.

Answer 1

$\frac{{\sin }^4\theta }{a}+\frac{{\cos }^4\theta }{b}=\frac{1}{a+b}$

$b{\sin }^4\theta +a{\cos }^4\theta =\frac{ab}{a+b}$

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

$b{\sin }^2\theta -b{\sin }^2\theta {\cos }^2\theta +a{\cos }^4\theta =\frac{ab}{a+b}$

$b(1-{\cos }^2\theta )-b(1-{\cos }^2\theta ){\cos }^2\theta +a{\cos }^4\theta =\frac{ab}{a+b}$

$b-b{\cos }^2\theta -b{\cos }^2\theta +b{\cos }^4\theta +a{\cos }^4\theta =\frac{ab}{a+b}$

$(a+b){\cos }^4\theta -2b{\cos }^2\theta +b-\frac{ab}{a+b}=0$

$(a+b){\cos }^4\theta -2b{\cos }^2\theta +\frac{ab+b^2-ab}{a+b}=0$

$(a+b){\cos }^4\theta -2b{\cos }^2\theta +\frac{b^2}{a+b}=0$

${(a+b)}^2{\cos }^4\theta -2b(a+b){\cos }^2\theta +b^2=0$

Discriminant$={\left(2b(a+b)\right)}^2-4\times {(a+b)}^2{b}^2=0$

$\therefore$ roots are equal Hence ${\cos }^2\theta =\frac{b}{a+b}$

${\sin }^8\theta ={(1-{\cos }^2\theta )}^3={(1-\frac{b}{a+b})}^3=\frac{a^3}{{(a+b)}^3}$

${\cos }^8\theta ={\left(\frac{b}{a+b}\right)}^3=\frac{b^3}{{(a+b)}^3}$

$\therefore \frac{{\sin }^8\theta }{a^3}+\frac{{\sin }^8\theta }{a^3}$

$=\frac{a^3}{a^3{(a+b)}^3}+\frac{b^3}{b^3{(a+b)}^3}$

$=\frac{2}{{(a+b)}^3}$

Hence proved.