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I am stuck with the simple expression $$ \frac{\cos^2(\theta + \alpha)}{1 - \cos^2(\theta - \alpha)} = \text{const.} $$ where $\theta$ is a variable and $\alpha$ is the number satisfying $$ \alpha = \tan^{-1} (\frac{4}{3})\,. $$ I cannot see it to be immediate, somehow I am missing a particular trigonometric identity. Or does this require a more detailed calculuation? Thanks for any hints, I'll fill in the details myself!

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  • $\begingroup$ Is this an equation to solve for $\theta$? If so, here is what the lhs function looks like, approximating $\alpha$ by $0.93$ radians. Infinitely many (not surprising, given the periodicity) solutions (not surprising, given the limiting values) for any nonnegative number on the rhs. None if the number is negative, of course. If you simplify all this, you'll end up with an equation in $\tan\theta$. $\endgroup$
    – Julien
    May 4, 2013 at 20:54

4 Answers 4

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Note that $1 - \cos^2(\theta - \alpha) = \sin^2(\theta - \alpha)$

That gives you:

$$\frac{\cos^2(\theta + \alpha)}{1 - \cos^2(\theta - \alpha)} = \text{const.} = \frac{\cos^2(\theta + \alpha)}{\sin^2(\theta - \alpha)}$$

For the numerator: $\cos(\theta+\alpha) = \cos\theta\cos\alpha-\sin\theta\sin\alpha.\tag{1}$

For the denominator: $\sin(\theta - \alpha) = \sin \theta \cos \alpha - \cos \theta \sin \alpha \tag{2}$

Note that since $\tan \alpha = \dfrac 43$, we have a $3:4:5$ triangle, using the Pythagorean Theorem, and noting that the leg opposite $\alpha$ must be of length $4$, and the leg adjacent to $\alpha$ is length $3$. This gives us a hypotenuse of length $5$. Calculating $\cos \alpha = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac 35$. Likewise $\sin \alpha = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac 45.$

So $(1)$ becomes $\cos^2(\theta + \alpha) = \left(\dfrac 35 \cos\theta - \dfrac 45 \sin\theta\right)^2$.

And $(2)$ becomes $\sin^2(\theta - \alpha) = \left(\dfrac 35\sin\theta - \dfrac45 \cos\theta\right)^2$

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    $\begingroup$ A problem: $\dfrac\cos\sin=\cot$ if you're taking the sine and cosine of the same thing. But you're taking the cosine of one thing and the sine of another thing. $\endgroup$ May 4, 2013 at 20:24
  • $\begingroup$ This $C$ in your second equation seems to be the remains of a former equation. $\endgroup$
    – Julien
    May 4, 2013 at 21:03
  • $\begingroup$ Thanks, @julien. I lost track of the fact that I equated the last term to the constant, already. Extra "C" deleted. $\endgroup$
    – amWhy
    May 4, 2013 at 21:06
  • $\begingroup$ @amWhy: Even a nice example +1 $\endgroup$
    – Amzoti
    May 5, 2013 at 0:45
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$$ \cos(\theta+\alpha) = \cos\theta\cos\alpha-\sin\theta\sin\alpha. $$

We have $\tan\alpha=\dfrac43$ so if $\mathrm{opp}=4$ and $\mathrm{adj}=3$ then $\mathrm{hyp}=5$ by the Pythagorean theorem, so $\sin\alpha=\dfrac{\mathrm{opp}}{\mathrm{hyp}}= \dfrac45$ and $\cos\alpha=\dfrac35$.

Consequently $\cos^2(\theta+\alpha)=\left(\dfrac35\cos\theta+\dfrac45\sin\theta\right)^2$.

Do something similar in the denominator. At some point you should probably multiply the top and bottom both by $5^2$ to clear out the fractions-within-fractions.

Next: $$ 1-\cos^2(\theta-\alpha) = \sin^2(\theta-\alpha), $$ so we recall that $$ \sin(\theta-\alpha) = \sin\theta\cos\alpha-\cos\theta\sin\alpha= \frac35\sin\theta - \frac45\cos\theta. $$ Then we have $$ \frac{\left(\dfrac35\cos\theta+\dfrac45\sin\theta\right)^2}{\left(\dfrac35\sin\theta - \dfrac45\cos\theta\right)^2} = \left(\frac{3\cos\theta+4\sin\theta}{3\sin\theta-4\cos\theta}\right)^2 = \left(\frac{3+4\tan\theta}{3\tan\theta-4}\right)^2. $$

This is not constant as a function of $\theta$, but I surmise that what you mean is that one is to solve for $\theta$. By a bit of simple algebra one goes from what you see above to $\tan\theta=\text{something}$, and from there to $\theta=\arctan\text{something}$.

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Hint:

Draw a right angled triangle. You know the sides(How?)

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$\tan\alpha=\frac{4}{3}$,then one solution of $\sin\alpha=\frac{4}{5},\cos\alpha=\frac{3}{5}$,the other is $\sin\alpha=-\frac{4}{5},\cos\alpha=-\frac{3}{5}$.

If your conclusion is right. The value of left hand does not depend on $\theta$, so we choose free, for easy calculation. I prefer $\theta=0$,the result is $\frac{9}{16}$. But when choosing $\theta=\pi/2$, the result is $\frac{16}{9}$.

Could you check your problem again?

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