Dear Uncle Colin,

In my non-calculator paper, I’m told $\cos(\theta) = \sqrt{\frac{1}{2}+ \frac{1}{2\sqrt{2}}}$ and that $\sin(\theta) = -\left(\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}\right)$. Given that $0 \le \theta \lt 2\pi$, find $\theta$. I’ve no idea how to approach it!

- Trigonometric Headaches Evaluating This Angle

Hi, THETA, and thanks for your message!

My third thought - after ‘yuk’ and ‘that’s in the fourth quadrant’ - is to wonder whether finding $\sin(2\theta)$, which is $2\sin(\theta)\cos(\theta)$ helps.

The product of $\cos(\theta)$ and $\sin(\theta)$ is $-\sqrt{\left(\frac{1}{2}+\frac{1}{2\sqrt{2}}\right) \left(\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}$, and under the square root, I spy a difference of two squares!

You end up with $-\sqrt{\frac{1}{4} - \frac{1}{8}}$, or $-\sqrt{\frac{1}{8}}$.

That’s $\sin(\theta)\cos(\theta)$, so $\sin(2\theta) = -\frac{1}{\sqrt{2}}$.

Since $\theta$ is in the fourth quadrant, $\frac{3}{2}\pi \lt \theta \lt 2\pi$, so $3\pi \lt 2\theta \lt 4\pi$. This gives $2\theta = \frac{13}{4}\pi$ or $\frac{15}{4}\pi$ and $\theta = \frac{13}{8}\pi$ or $\frac{15}{8}\pi$.

Come on now, it can’t be both!

To figure out which of the two you want, you’re probably meant to work the same sort of trick with $\cos(2\theta)$. You can do that on your own time if you want to.

Personally, I’m going to note that $|\sin(\theta)| < |\cos(\theta)|$, so $|\tan(\theta)|<1$ and $\theta$ must be within $\piby 4$ of a multiple of $2\pi$. The only answer that satisfies that is $\theta = \frac{15}{8}\pi$.

Oh, hello sensei, didn’t see you there

“… that’s part of the shtick.”

“I suppose.”

“$\tan(\theta) = -\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$, which - using a conjugate trick - is $1-\sqrt{2}$”.

“… I suppose that, too.”

“And since $\tan\left(\frac{\pi}{8}\right) = \sqrt{2}-1$, $\theta = -\frac{\pi}{8} + 2n\pi$.”

“… For integer $n$.”

“I said that. You just didn’t hear it.”

Hope that helps!

- Uncle Colin