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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.

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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 $2\theta=\dfrac{15\pi}4$.

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  • $\begingroup$ 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! $\endgroup$ – Trey Anupong Jun 16 at 8:18
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    $\begingroup$ @Trey, not quite. Two of those candidates put $\theta$ in quadrant 4. $\endgroup$ – Peter Taylor Jun 16 at 15:05
  • $\begingroup$ @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? $\endgroup$ – Trey Anupong Jun 16 at 15:42
  • $\begingroup$ 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. $\endgroup$ – CY Aries Jun 16 at 15:48
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    $\begingroup$ @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. $\endgroup$ – CY Aries Jun 16 at 15:55
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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 $2\theta = \ldots$

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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 $k$.

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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 $\cos\theta > \cos\frac{7\pi}{4}$

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