# trig proof

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

Prove that:

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

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

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

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

Prove that:

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

Thanks in advance

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

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

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

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