S sean b New member Apr 8, 2010 #1 ?/12 = 9?/12-8?/12 = 3?/4-2?/3 sin(3?/4-2?/3) = sin(3?/4)cos(2?/3) - sin(2?/3)cos(3?/4) =?2/2(-1/2) - (?3/2)(-?2/2) =-?2/4 - (-?6/4) =-?2+?6/4 =[?6-?2]/4
?/12 = 9?/12-8?/12 = 3?/4-2?/3 sin(3?/4-2?/3) = sin(3?/4)cos(2?/3) - sin(2?/3)cos(3?/4) =?2/2(-1/2) - (?3/2)(-?2/2) =-?2/4 - (-?6/4) =-?2+?6/4 =[?6-?2]/4
H heyeveryone New member Apr 8, 2010 #2 the sum and difference formula = sin(x1+x2) = sinx1cos2+cos1sin2, sun(x1-x2) = sinx1cosx2 - cosx1sinx2. how would i use that formula on sin ?/12 ??? show would so i know how you did it thanks!!! show work*
the sum and difference formula = sin(x1+x2) = sinx1cos2+cos1sin2, sun(x1-x2) = sinx1cosx2 - cosx1sinx2. how would i use that formula on sin ?/12 ??? show would so i know how you did it thanks!!! show work*
V vitran New member Apr 8, 2010 #3 Im guessing you have to evaluate pi/12? pi/12 isnt on the unit circle, so you have to split it up into pieces that are so you can do 3pi/4 = 9pi/12, and -2pi/3= -8pi/12 so you get sin(3pi/4 - 2pi/3) sincos+cossin sin(3pi/4)cos(2pi/3) - sin(2pi/3)cos(3pi/4) solve it!
Im guessing you have to evaluate pi/12? pi/12 isnt on the unit circle, so you have to split it up into pieces that are so you can do 3pi/4 = 9pi/12, and -2pi/3= -8pi/12 so you get sin(3pi/4 - 2pi/3) sincos+cossin sin(3pi/4)cos(2pi/3) - sin(2pi/3)cos(3pi/4) solve it!