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April 21, 2014

April 21, 2014

Posted by **stacey graham** on Friday, March 26, 2010 at 12:24am.

sin(A-B)?

- trig -
**Reiny**, Friday, March 26, 2010 at 9:04amif cosA = 1/3 in I, then sinA = √8/3 using Pythagoras

if sinB= -1/2 in IV, the cosB = √3/2

sin (A-B) = sinAcosB - cosAsinB

= (√8/3)(√3/2) - (1/3)(-1/2)

= (√24 + 1)/6

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