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OpenStudy (jdoe0001):
\(\bf sin\left(-\cfrac{11\pi}{12}\right) \ \ ?\)
OpenStudy (anonymous):
yes
OpenStudy (jdoe0001):
so how did you get that answer?
OpenStudy (anonymous):
i solved for regular 11pi/12 and then switched the signs
OpenStudy (jdoe0001):
hmm, dunno, I'd have to do it, but yes
sin( -x ) = - sin(x)
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OpenStudy (anonymous):
so i was right?
OpenStudy (jdoe0001):
well, lemme do it
OpenStudy (anonymous):
ok
OpenStudy (jdoe0001):
$$\bf sin\left(-\cfrac{11\pi}{12}\right) \implies -sin\left(\cfrac{11\pi}{12}\right)\\
sin\left(\cfrac{\pi}{4}+\cfrac{2\pi}{3}\right)\\
\cfrac{\sqrt{2}}{2}\cdot\cfrac{-1}{2}+\cfrac{\sqrt{2}}{2}\cdot\cfrac{\sqrt{3}}{2} \implies -\cfrac{\sqrt{2}}{4}+\cfrac{\sqrt{6}}{4}\\
\cfrac{\sqrt{6}-\sqrt{2}}{4}\\
\text{now recall we have a -1 in front of the sine}\\
\left(\cfrac{\sqrt{6}-\sqrt{2}}{4}\right) \implies \cfrac{\sqrt{2}-\sqrt{6}}{4}
$$
OpenStudy (anonymous):
ohhhhh so i had it inverted by doing what i did
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