Tricky trig question from GRE

Question

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

Answer 1

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 $(\frac{\pi}{12})^2 = \frac{\pi^2}{144}$

Answer 2

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

Answer 3

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.

Answer 4

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

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