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Mathematics 13 Online
OpenStudy (anonymous):

sin -11 pi /12... is the answer (-sqrt6-sqrt2)/4

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)

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

OpenStudy (anonymous):

thank you!

OpenStudy (jdoe0001):

yw

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