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Thread: trig proof

  1. #1
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    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
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by andtom View Post
    $\displaystyle \frac{sin^{4} \alpha}{a}+ \frac{cos ^{4}\alpha }{b}= \frac{1}{a+b}$
    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?
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  3. #3
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    Nope, I still cannot prove it :/
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by andtom View Post
    Nope, I still cannot prove it

    Show what roots do you obtain for

    $\displaystyle \dfrac{x^2}{a}+\dfrac{(1-x)^2}{b}=\dfrac{1}{a+b}$
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  5. #5
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    so there's only 1 root
    since $\displaystyle \Delta = 0$,
    $\displaystyle x=sin^{2}\alpha=\frac{2a}{2(a+b)}=\frac{a}{a+b}$
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  6. #6
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by andtom View Post
    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!
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  7. #7
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    OK, i got it
    Thank you guys
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