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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?

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  • $\begingroup$ Have you tried using $\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$? $\endgroup$ – Alex Becker Mar 12 '12 at 23:14
  • $\begingroup$ I should only use the properties of the trigonometric functions(even, odd, periodic). Sum and difference identities are not allowed. $\endgroup$ – Adam Mar 12 '12 at 23:18
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    $\begingroup$ I don't think what you have is correct. $\alpha = 0$ gives the lhs $\neq$ rhs $\endgroup$ – user17762 Mar 12 '12 at 23:20
  • $\begingroup$ If you can't use sum/difference identities, I don't think you'll be able to do anything with that term with the $5\pi/6$. $\endgroup$ – Gerry Myerson Jul 2 '12 at 5:24
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Since all the trig values are squared, it seems as though the exercise is simply playing with shifts by odd or even multiples of $\pi/2$.

Loosely,

  • Shifting by "$\frac{\pi}{2} \cdot \text{odd}$" switches "sin" and "cos" (and possibly affects the sign)
  • Shifting by "$\frac{\pi}{2} \cdot \text{even}$" ($=$ "$\pi \cdot \text{any}$") preserves "sin" and "cos" (and possibly affects sign)
  • Negating the argument preserves "sin" and "cos" (and possibly affects sign)

Since squaring eliminates sign considerations, we have, simply:

$$\begin{align} \mathrm{trig}^2\left( \pm \; \theta \pm \frac{\pi}{2} \text{odd} \right) &= \mathrm{cotrig}^2\theta \\ \mathrm{trig}^2\left( \pm \; \theta \pm \frac{\pi}{2} \text{even} \right) &= \mathrm{trig}^2\left( \pm \; \theta \pm \pi \cdot \text{any} \right) = \mathrm{trig}^2\theta \end{align}$$

where each "$\pm$" is independent, "any" means (of course) "any integer", and "trig" can in fact be any of the six trig functions.

This makes pretty quick work of the simplification process ... $$\begin{align} \frac{\sin^2\left(\frac{5\pi}{6}-\alpha\right)}{\cos^2\left(\alpha-4\pi\right)} - \cot^2\left(\alpha-11\pi\right) \; \sin^2\left(-\alpha-\frac{13\pi}{2}\right) &\stackrel{?}{=} \sin^2\alpha \\[1em] \frac{\sin^2\left(\frac{5\pi}{6}-\alpha\right)}{\cos^2\alpha} - \cot^2\alpha \; \cos^2\alpha &\stackrel{?}{=} \sin^2\alpha \end{align}$$ ... right up to the point at which the process shudders to a halt.

Given the nature of all the other terms (and @Adam's comment that sum and difference identities are not allowed), I suspect that "$\frac{5\pi}{6}$" is a typo of "$\frac{5\pi}{2}$", which would get us a little further ...

$$\frac{\cos^2\alpha}{\cos^2\alpha} - \cot^2\alpha \; \cos^2\alpha = 1 - \cot^2\alpha \;\cos^2\alpha \stackrel{?}{=} \sin^2\alpha$$

... but we hit another snag. Could it be that "$\sin^2\left(-\alpha-\frac{13\pi}{2}\right)$" is a typo of "$\cos^2(...)$"? If so, then that factor should've simplified to "$\sin^2\alpha$", and we'd have

$$1 - \cot^2\alpha \;\sin^2\alpha = 1 - \cos^2\alpha = \sin^2\alpha$$

as desired.

(It's also possible that, instead of a sin-cos typo, "$\cot$" is a typo for "$\tan$", but it seems like that would be an easier one for the OP to notice.)

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Some important translations:

$$\tag 1\sin(x\pm 2 \pi) = \sin x $$ $$\tag {1'}\cos(x\pm 2 \pi) = \cos x $$ $$\tag 2\cot(x\pm \pi)= \cot x$$ $$\tag {2'}\tan(x\pm \pi)= \tan x$$

$$\tag 3 \sin \left(\frac \pi 2 -x \right)=\cos x$$ $$\tag 4 \cos \left(\frac \pi 2 -x \right)=\sin x$$ $$\tag 5 \sin(\pi-x)=\sin x$$ and $$\tag 6\cos(\pi-x)=-\cos x$$

and $$\tag 7 \sin(-x)=-\sin x$$ $$\tag 8 \cos (-x) = \cos x$$

Although taking $\alpha =0$ reveals that the equality doesn't hold, assume that there is no typo, then, we could move on as follows.

You have that

$$ \frac{\sin^2 \left(\frac{5\pi}{6} - \alpha \right)}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2 \left(-\alpha - \frac{13\pi}{2}\right) =\sin^2(\alpha)$$

Using the above, we can write

$$\eqalign{ & {\sin ^2}\left( {\frac{{5\pi }}{6} - \alpha } \right) = {\left[ { - \sin \left( {\alpha - \frac{{5\pi }}{6}} \right)} \right]^2} = {\sin ^2}\left( {\alpha - \frac{{5\pi }}{6}} \right) \cr & {\cot ^2}(\alpha - 11\pi ) = {\cot ^2}\left( {\alpha - 10\pi } \right) = \cdots = {\cot ^2}\alpha \cr & {\sin ^2}\left( { - \alpha - \frac{{13\pi }}{2}} \right) = {\left[ { - \sin \left( {\alpha + \frac{{13\pi }}{2}} \right)} \right]^2} = \sin {\left( {\alpha + \frac{{13\pi }}{2}} \right)^2} \cr & {\cos ^2}(\alpha - 4\pi ) = {\cos ^2}(\alpha - 2\pi ) = {\cos ^2}\alpha \cr} $$

so that we have

$$\frac{{{{\sin }^2}\left( {\alpha - \frac{{5\pi }}{6}} \right)}}{{{{\cos }^2}\alpha }} - {\cot ^2}\alpha {\sin ^2}\left( {\alpha + \frac{{13\pi }}{2}} \right) = {\sin ^2}\alpha $$

Now

$${\sin ^2}\left( {\alpha - \frac{{5\pi }}{6}} \right) = {\sin ^2}\left( {\alpha - \frac{{3\pi }}{6} - \frac{{2\pi }}{6}} \right) = {\sin ^2}\left( {\alpha - \frac{\pi }{3} - \frac{\pi }{2}} \right) = {\left( { - 1} \right)^2}{\sin ^2}\left( {\frac{\pi }{2} - \left( {\alpha - \frac{\pi }{3}} \right)} \right) = {\cos ^2}\left( {\alpha - \frac{\pi }{3}} \right)$$

and

$$\eqalign{ & {\sin ^2}\left( {\alpha + \frac{{13\pi }}{2}} \right) = {\sin ^2}\left( {\alpha + \frac{{12\pi }}{2} + \frac{\pi }{2}} \right) = {\sin ^2}\left( {\alpha + 6\pi + \frac{\pi }{2}} \right) = {\sin ^2}\left( {\alpha + \frac{\pi }{2}} \right) \cr & = {\sin ^2}\left( {\frac{\pi }{2} - \left( { - \alpha } \right)} \right) = {\cos ^2}\left( { - \alpha } \right) = {\cos ^2}\alpha \cr} $$

so that you have

$$\frac{{{{\cos }^2}\left( {\alpha - \frac{\pi }{3}} \right)}}{{{{\cos }^2}\alpha }} - {\cot ^2}\alpha {\cos ^2}\alpha = {\sin ^2}\alpha $$

Now, solving for $${{{\cos }^2}\left( {\alpha - \frac{\pi }{3}} \right)}$$

gives

$${\cos ^2}\left( {\alpha - \frac{\pi }{3}} \right) = {\cos ^2}\alpha {\sin ^2}\alpha + {\cos ^6}\alpha \frac{1}{{{{\sin }^2}\alpha }}$$

There is some typo in your excercise, since letting $\alpha =0$ gives $1/4$ on the LHS and is not defined for the RHS. When you discover what the typo is, move on with the listed translations.

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