1. ## trig proof

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

Prove that:

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

2. Originally Posted by andtom
$\frac{sin^{4} \alpha}{a}+ \frac{cos ^{4}\alpha }{b}= \frac{1}{a+b}$
take $sin^{2} \alpha=x$. then $cos^{2} \alpha=1-x$.

Now substitute in the given equation. SOLVE the quadratic. the roots are nice! you will get $sin^{2} \alpha=\frac{a}{a+b}$ . Substitute this below

Prove that:

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

did this help?

3. Nope, I still cannot prove it :/

4. Originally Posted by andtom
Nope, I still cannot prove it

Show what roots do you obtain for

$\dfrac{x^2}{a}+\dfrac{(1-x)^2}{b}=\dfrac{1}{a+b}$

5. so there's only 1 root
since $\Delta = 0$,
$x=sin^{2}\alpha=\frac{2a}{2(a+b)}=\frac{a}{a+b}$

6. Originally Posted by andtom
so there's only 1 root
since $\Delta = 0$,
$x=sin^{2}\alpha=\frac{2a}{2(a+b)}=\frac{a}{a+b}$
correct!

7. OK, i got it
Thank you guys