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July 10th, 2016, 01:13 PM   #1
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Sin(2*theta) into 2*theta

Hey,

I've gone a long way in this question, but I can't seem to get to the final answer. I've gotten up to sin(2*theta) = -1+(1/sqrt(2)). Doesn't seem like the double of a common angle and the question doesn't give any odd sines values.

Any help would be really appreciated!

Last edited by Itika; July 10th, 2016 at 01:58 PM.
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July 10th, 2016, 01:16 PM   #2
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Oh, my bad! I was careless and promptly posted this thread in the wrong forums, I'm sorry. Is there any way to lock it or erase it?
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July 10th, 2016, 02:13 PM   #3
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show your work.
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July 10th, 2016, 02:23 PM   #4
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$\sin(2\theta) = 2\sin{\theta}\cos{\theta} = -2\sqrt{\dfrac{1}{4} - \dfrac{1}{8}} = -\dfrac{1}{\sqrt{2}}$

$2\theta = \arcsin\left(-\dfrac{1}{\sqrt{2}}\right) = -\dfrac{\pi}{4}$


don't worry ... a mod will move it if necessary
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July 10th, 2016, 02:29 PM   #5
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Quote:
Originally Posted by skeeter View Post


$\sin(2\theta) = 2\sin{\theta}\cos{\theta} = -2\sqrt{\dfrac{1}{4} - \dfrac{1}{8}} = -\dfrac{1}{\sqrt{2}}$

$2\theta = \arcsin\left(-\dfrac{1}{\sqrt{2}}\right) = -\dfrac{\pi}{4}$


don't worry ... a mod will move it if necessary
The problem is that the answer sheet shows that 2theta=15pi/4. The use of calculators is also forbidden in the exam, so using arcsin is impossible.
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July 10th, 2016, 02:46 PM   #6
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Quote:
using arcsin is impossible
I did not use a calculator to find $-\dfrac{\pi}{4}$ ... it's a unit circle value.


The principle value for $2\theta = -\dfrac{\pi}{4} \implies \theta = -\dfrac{\pi}{8}$.

Since $0 \le \theta < 2\pi$, that means $\theta = \dfrac{15\pi}{8}$ is the coterminal angle in the given interval ... which means $2\theta = ?$
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July 10th, 2016, 03:11 PM   #7
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Originally Posted by Itika View Post
Oh, my bad! I was careless and promptly posted this thread in the wrong forums, I'm sorry. Is there any way to lock it or erase it?
I've merged the two threads.
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