How do you find the exact value of #sin(v-u)# given that #sinu=5/13# and #cosv=-3/5#?

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How do you find the exact value of #sin(v-u)# given that #sinu=5/13# and #cosv=-3/5#?

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Answer 1 · 2017-11-28T13:43:03

#sin(v-u) = -33/65#

Explanation:

Given #sin u = 5/13#
#13^2 = 5^2 + 12^2#
In effect it’s a right angle triangle with sides 5, 12, 13.
#:. cos u = 12/13#

Given #cos v = -3/5#
#5^2 = 3^2 + 4^2#
In effect it’s a right angle triangle with sides 3, 4, 5.
#:. sin v = -4/5# in the third quadrant where cos and sin are negative.

#sin (v-u) = (sin v * cos u )- (cos v * sin u)#

#sin (v-u) = ((-4/5)(12/13)) - ((-3/5)(5/13))#

#sin(v-u) = (-48/65 )+ (15/65) = -33/65#

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How do you find the exact value of #sin(v-u)# given that #sinu=5/13# and #cosv=-3/5#?

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