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I need to prove the following trigonometric identity: $$ \frac{\sin^2(\frac{5\pi}{6} - \alpha )}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2(-\alpha - \frac{13\pi}{2}) =\sin^2(\alpha)$$

I cannot express $\sin(\frac{5\pi}{6}-\alpha)$ as a function of $\alpha$. Could it be a textbook error?

  • 0
    Have you tried using $\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$?2012-03-12
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    I should only use the properties of the trigonometric functions(even, odd, periodic). Sum and difference identities are not allowed.2012-03-12
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    I don't think what you have is correct. $\alpha = 0$ gives the lhs $\neq$ rhs2012-03-12
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    If you can't use sum/difference identities, I don't think you'll be able to do anything with that term with the $5\pi/6$.2012-07-02

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