Posted by Luis on Sunday, May 13, 2012 at 5:04pm.
Currently in school we're learning about proving identities and for extra credit I got this problem.
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.

precalculus  Steve, Sunday, May 13, 2012 at 5:31pm
This is just a makework problem, providing no added insight. Just plug and chug.
You have a 81517 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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