# Given $\cos (\theta)$ and $\sin (\theta)$, find $2\theta$

I am working on my scholarship exam. I worked through almost final step but got my answer wrong. Could you please have a look?

If $$\cos (\theta) = \sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}}$$ and $$\sin (\theta) = -\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}$$ with $$0\leq\theta<2\pi$$, it follows that $$2\theta = ..... \pi$$

What I have got is below:

$$\sin(2\theta)=2\sin\theta\cos\theta$$

Then, $$\sin(2\theta)=-\frac{1}{\sqrt{2}}$$

Hence, $$2\theta = \frac{5\pi}{4}$$ or $$\frac{7\pi}{4}$$ (quadrant 3 or 4)

$$\theta=\frac{5\pi}{8}$$ or $$\frac{7\pi}{8}$$

Since $$\cos\theta$$ is positive and $$\sin\theta$$ is negative, $$\theta$$ should be in quadrant 4 but my $$\theta$$'s are not. So I cannot use my $$2\theta$$ as a final answer.

However, the answer key provided is $$\theta=\frac{15\pi}{4}$$, Why do you think that is the case? How did they get to this answer? Please help.

If $$0\le \theta<2\pi$$, then $$0\le 2\theta<4\pi$$. You should have $$4$$ possible values of $$\theta$$.

$$\cos (\theta) = \sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}}$$ and $$\sin (\theta) = -\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}$$ imply that $$\sin(2\theta)=-\dfrac1{\sqrt2}$$, but not the other way round.

Note that $$\displaystyle \tan\theta=\frac{-\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}}{\sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}}}=-\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}=1-\sqrt{2}$$.

So $$\theta=n\pi-\dfrac{\pi}{8}$$, ($$n\in\mathbb{Z}$$).

[Note: $$\frac{\tan(-\frac\pi8)}{1-\tan^2(-\frac\pi8)}=\tan2(-\frac\pi8)=-1$$ implies that $$\tan(-\frac\pi8)=1-\sqrt2$$.]

As $$\theta\in[0,2\pi)$$, $$\cos\theta>0$$ and $$\sin\theta<0$$, we have $$\theta=2\pi-\dfrac{\pi}{8}$$ and hence $$2\theta=\dfrac{15\pi}4$$.

• Oh yes, that makes sense! Now we have $2\theta = \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{13\pi}{4} and \frac{15\pi}{4}$. And $\frac{15\pi}{4}$ is the best candidate since its $\theta$ is the only one that is in quadrant 4, thank you so much! – Trey Anupong Jun 16 at 8:18
• @Trey, not quite. Two of those candidates put $\theta$ in quadrant 4. – Peter Taylor Jun 16 at 15:05
• @PeterTaylor Oh right! $\theta$ would be $\frac{2.5\pi}{4}$,$\frac{3.5\pi}{4}$, $\frac{6.5\pi}{4}$ and $\frac{7.5\pi}{4}$ and the last two are in quadrant 4 (since $\frac{6\pi}{4}\leq\theta\leq2\pi$). Does that mean we have two answers although the answer key provided one? – Trey Anupong Jun 16 at 15:42
• Both $\theta=\frac{6.5\pi}{4}$ and $\frac{7.5\pi}{4}$ satisfy $\sin2\theta=-\frac1{\sqrt2}$. But you still have to check whether both of them satisfy the original system of equations. If they do, then both of them are solutions. – CY Aries Jun 16 at 15:48
• @TreyAnupong Actually, the system implies that $\sin2\theta=-\frac1{\sqrt2}$. But the opposite direction does not hold. My answer is just to give a hint that why we have more than two possibilities. I will edit it to give a full solution. – CY Aries Jun 16 at 15:55

Remember that if the range of $$\theta$$ is $$0 \leq \theta \lt 2 \pi$$ then the range od $$2\theta$$ will be $$0\leq\theta \lt 4\pi$$. So $$2\theta = \ldots$$

Your mistake is at this step: $$\sin(2\theta)=-\frac{1}{\sqrt{2}}$$, hence $$2\theta = \frac{5\pi}{4}$$ or $$\frac{7\pi}{4}$$. The correct step is: $$\sin(2\theta)=-\frac{1}{\sqrt{2}}$$, hence $$2\theta = \frac{5\pi}{4}+2k\pi$$ or $$\frac{7\pi}{4}+2k\pi$$, where $$k$$ is any integer. So dividing by 2 we get $$\theta = \frac{5\pi}{8}+k\pi$$ or $$\frac{7\pi}{8}+k\pi$$, where $$k$$ is any integer. If you take $$k=0$$, the answer doesn't work. So you need to use a different $$k$$.

Hence, $$2\theta = \frac{5\pi}{4}$$ or $$\frac{7\pi}{4}$$

Don't forget $$+2\pi n$$.

Since the signs of $$\cos\theta$$ and $$\sin\theta$$ place $$\theta$$ in the fourth quadrant, only two candidates hold up. You can choose between them by testing whether $$\cos\theta > \cos\frac{7\pi}{4}$$