Question

If $\frac{{\sin }^{4}x}{a}+\frac{{\cos }^{4}x}{b}=\frac{1}{a+b}$, prove that
(i) $\frac{{\sin }^{8}x}{a^3}+\frac{{\cos }^{8}x}{b^3}=\frac{1}{{(a+b)}^3}$
(ii) $\frac{{\sin }^{4n}x}{a^{2n-1}}+\frac{{\cos }^{4n}x}{b^{2n-1}}=\frac{1}{{(a+b)}^{2n-1}},n\in N$

Answer 1

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

$\therefore \frac{({\sin }^{2}x{)}^2}{a}+\frac{(1-{\sin }^{2}x{)}^2}{b}=\frac{1}{a+b}$

$\Rightarrow (b{l}^2+a(1-l{)}^2\left)\right(a+b)=ab$ ; let ${\sin }^{2}x=l$

$\Rightarrow (ab+b^2)+(a^2+ab)(1+t^2-2t)=ab$

$\Rightarrow t^2(a+b{)}^2-2(a^2+ab)t+a^2=0$

$t=\frac{2(a^2+ab)\pm \sqrt{4(a^4+a^2{b}^2+2a^{3}b)-4a^2(a^2+b^2+2ab)}}{2(a+b{)}^2}$

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

$\Rightarrow t=\frac{a}{a+b}$ $\therefore {\sin }^{2}x=\frac{a}{a+b}$

$\therefore {\cos }^{2}x=\frac{b}{a+b}$

$\therefore \frac{{\sin }^{4x}x}{a^{2n-1}}+\frac{{\cos }^{4x}x}{b^{2n-1}}=\frac{({\sin }^{2}x{)}^{2n}}{a^{2n-1}}+\frac{({\cos }^{2}x{)}^{2n}}{b^{2n-1}}$

$=\frac{a^{2n}}{(a+b{)}^{2n}{a}^{2n-1}}+\frac{b^{2n}}{(a+b{)}^{2n}{b}^{2n-1}}$

$=\frac{a+b}{(a+b{)}^{2n}}=\frac{1}{(a+b{)}^{2n-1}}$

$\therefore \frac{{\sin }^{4x}x}{a^{2n-1}}+\frac{{\cos }^{4x}x}{b^{2n-1}}=\frac{1}{(a+b{)}^{2n-1}}$ (proved)

put $n=2$

$\frac{{\sin }^{8}x}{a^3}+\frac{{\cos }^{8}x}{b^3}=\frac{1}{(a+b{)}^3}$ (proved)