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April 18, 2014

April 18, 2014

Posted by **Kimmie** on Thursday, January 28, 2010 at 10:14pm.

cos B = -12/13 when Pi<B<3Pi/2, then what is the value of sin(a+b)?

I don't get the question... Help appreciated.

- pre-calculus -
**Reiny**, Thursday, January 28, 2010 at 10:27pmmake two separate diagrams,

one will be the 3,4,5 right-angled triangle,

the other will be the 5,12,13 right-angled triangle

given sin a = 3/5 and a is in the II quadrant, so cos a = -4/5

given cos B = -12/13 and B is in III, so

sin B = - 5/13

now you have to know the expansion for

sin(a+b) = sina cosb + cosa sinb

=(3/5)(-12/13) + (-4/5)(-5/13) = -16/65

- pre-calculus -
**Kimmie**, Thursday, January 28, 2010 at 10:57pmThanks!

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