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November 28, 2014

November 28, 2014

Posted by **Luis** on Sunday, May 13, 2012 at 5:04pm.

cosB= -8/17 in quad II and I have to find sin 7B.

I have gotten this far:

sin7B= sin(3B+4B)

sin3Bcos4B+sin4Bcos3B

my problem I can't figure out how to further break it down. After that I can do the rest.

- pre-calculus -
**Steve**, Sunday, May 13, 2012 at 5:31pmThis is just a make-work problem, providing no added insight. Just plug and chug.

You have a 8-15-17 triangle in QII or QIII. Let's go with QII for now.

sinB = 15/17

sin2B = 2sinVcosB = 2(-8/17)(15/17) = -240/17^2

cos2B = cos^2B - sin^2B = 64/289 - 225/289 = -161/17^2

sin4B = 2sin2Bcos2B = 2(-240/289)(-161/289) = 77280/17^4

cos4B = cos^2 2B - sin^2 2B = 64^2/17^4 - 240^2/17^4 = -53504/17^4

sin3B = sinBcos2B + cosBsin2B

sin7B = sin3Bcos4B + cos3Bsin4B

I'll let you do the arithmetic. *whew*

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