If #Sin u= -12/37# and #cos v=-99/101#, with u being in QIII and v being in QII, how do you find the exact value of sin(u-v)?

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Answer 1 · 2015-06-04T03:00:23

Use the trig identity: sin (u - v) = sin u.cos v - sin v.cos u

#sin u = -12/37# -> #cos^2 u = 1 - sin^2 u = 1 - 144/1369 = 1225/1369 =

0.89 -> cos u = -0.94# (Q. III)

#cos v = -99/101 -> sin^2 v = 1 - cos^2 v = 1 - 9801/10201 = #

#2400/10201 = 0.24 -> sin v = 0.49.# (Q. II)

#sin (u - v) = (-12/37)(-99/101) - (0.94)(0.49) = #

#= 1188/3737 - 0.46 = 0.32 - 0.46 = -0.14#