Tricky trig question from GRE

Please evaluate \begin{align} 1-\sin^2\left(\arccos \frac{\pi}{12}\right) \end{align}

What I've tried so far is to use Pythagorean identity and I got \begin{align} \cos^2\left(\arccos \frac{\pi}{12}\right) \end{align}

If \begin{align} \arccos \frac{\pi}{12}=y \end{align} then \begin{align} \cos y =\frac{\pi}{12} \end{align}

and here I can't continue because the answers are in such form:

• (A) $\sqrt{\frac{1-\cos\frac\pi{24}}{2}}$
• (B) $\sqrt{\frac{1-\cos\frac\pi{6}}{2}}$
• (C) $\sqrt{\frac{1+\cos\frac\pi{24}}{2}}$
• (D) $\frac\pi6$

and one more is missing, but that's pdf's issue

• Please add information about what you have attempted so far. Otherwise you are unlikely to get any helpful responses. – MathInferno Sep 27 '15 at 12:03
• What's your effort to solve this problem? – akusaja Sep 27 '15 at 12:11
• @marty Hey, you may consider add a few words to indicate that your tries came after some answers had appeared; otherwise people cannot tell why there are some seemingly iterative repeats below... – Gary Moore Sep 27 '15 at 12:48
• I can tell you that none of those options you've posted is the right one. What's the missing one? Even without calculation I can exclude the first three for being algebraic numbers. The fourth option is more plausible but also wrong. – Deepak Sep 27 '15 at 12:52
• @Marty Because I've worked out the right answer (see mine, and the other answers as well). I can immediately exclude the fourth one because it's clearly not equal to what I've found. You can exclude the first three by tedious simplification or calculation, but if you know some slightly advanced theory (Chebyshev polynomials, transcendental and algebraic number theory), you can immediately "see" none of those can be correct. – Deepak Sep 27 '15 at 13:01

Once you get to this stage:

$\cos^2 (\arccos \frac{\pi}{12})$,

Use the facts that $\cos(\arccos x) = x$ and $\cos^2 y = (\cos y)^2$ to get to the final answer.

In this case, that's simply $(\frac{\pi}{12})^2 = \frac{\pi^2}{144}$

• Heh, it is almost as if the testmaker made $\cos^2(\arccos(\pi/12)) \color{Red}{=} \cos(\pi/12)=\sqrt{\frac{1+\cos(\pi/6)}2}$... – Ian Mateus Sep 28 '15 at 0:21
• @IanMateus Yes, that is a distinct possibility. :) But the last option is not given, and since the third option is plausible, as I said, we don't know if it's a correct question with (ironically) the only correct option not being mentioned by the asker. – Deepak Sep 28 '15 at 0:50

Hint: For all $x \in \mathbb{R}$ we have $$1 - \sin^{2}(\arccos x) = \cos^{2}(\arccos x).$$

• can you explain why $\cos^{2}(\cos^{-1}x) = \cos x$? – Dimitri C Sep 27 '15 at 12:05
• $\cos^2(\cos^{-1}x) = x^2$ – Deepak Sep 27 '15 at 12:05
• @Deepak can you explain why you simplified this way in your comment? – Dimitri C Sep 27 '15 at 12:10
• @DimitriC Because $\cos(\cos^{-1}x) = x$ and $\cos^2 y = (\cos y)^2$. (I know, the notation is fairly appalling, but that's not my abuse, it's convention!). – Deepak Sep 27 '15 at 12:14
• Use $\arccos$ instead of $\cos^{-1}$. – Surb Sep 27 '15 at 12:42

Hint: Try drawing a right triangle with one angle $\theta$ where $\cos(\theta)=\frac{\pi}{12}$. A good choice would be a right triangle where the hypotenuse is length $1$ and the adjacent side is length $\frac{\pi}{12}$.

Now, use the Pythagorean theorem to get the length of the opposite side and then compute the sine as opposite over hypotenuse, square your result and you're almost done.

You have that $\sin(\arccos(x))=\sqrt{1-x^2}$, therefore

$$\sin^2\left(\arccos\left(\frac{\pi}{12}\right)\right)=1-\frac{\pi^2}{144}\implies 1-\sin^2\left(\arccos\left(\frac{\pi}{12}\right)\right)=\frac{\pi^2}{144}$$