# trig proof

• May 5th 2011, 02:00 AM
andtom
trig proof
$\frac{sin^{4} \alpha}{a}+ \frac{cos ^{4}\alpha }{b}= \frac{1}{a+b}$

Prove that:

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

• May 5th 2011, 02:38 AM
abhishekkgp
Quote:

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?
• May 5th 2011, 03:35 AM
andtom
Nope, I still cannot prove it :/
• May 5th 2011, 03:45 AM
FernandoRevilla
Quote:

Originally Posted by andtom
Nope, I still cannot prove it

Show what roots do you obtain for

$\dfrac{x^2}{a}+\dfrac{(1-x)^2}{b}=\dfrac{1}{a+b}$
• May 5th 2011, 04:00 AM
andtom
so there's only 1 root
since $\Delta = 0$,
$x=sin^{2}\alpha=\frac{2a}{2(a+b)}=\frac{a}{a+b}$
• May 5th 2011, 04:46 AM
abhishekkgp
Quote:

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!
• May 5th 2011, 06:44 AM
andtom
OK, i got it :)
Thank you guys