For angles u and v in quadrant II, if sin u = 3/5 and cos v = –5/13, how do you find the exact value of sin (v – u)?

We use the sum/difference angle formula for sin:

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

We know #sin u = 3/5#, and #u# is in quadrant II, so that means #cos u = -4/5.# (Draw the reference triangle for #u# and use the Pythagorean theorem to see why.)

Similarly, since #cos v= -5/13# and #v# is in quadrant II, we know #sin v = 12/13.#

We now plug these 4 values into the sum angle formula above:

#sin(v-u) = sin v cos u - cos v sin u#
#color(white)(sin(v-u)) = (12/13)(-4/5) - (-5/13)(3/5)#
#color(white)(sin(v-u)) = -48/65 - (-15/65)#
#color(white)(sin(v-u)) = -33/65#

Use the identity:

#sin(v-u) = sin(v)cos(u)- cos(v)sin(u)" [1]"#

We are given #sin(u) = 3/5" [2]"#:

We are given #cos(v) = -5/13" [3]"#:

Please observe that equation [1] requires two values that we do not know, #sin(v)#, and #cos(u)#; the next part shows you who to compute those two values.

We need to use the identity:

#sin(v) = +-sqrt(1-cos^2(v))#

Substitute #cos^2(v) = (-5/13)^2#:

#sin(v) = +-sqrt(1-(-5/13)^2)#

#sin(v) = +-sqrt(169/169-25/169)#

#sin(v) = +-sqrt(144/169)#

#sin(v) = +-12/13#

We know that the sine function is positive in the second quadrant, therefore, we choose the positive value.

#sin(v) = 12/13" [4]"#

We need to use the identity:

#cos(u) = +-sqrt(1-sin^2(u))#

Substitute #sin^2(u) = (3/5)^2#:

#cos(u) = +-sqrt(1-(3/5)^2)#

#cos(u) = +-sqrt(25/25-9/25)#

#cos(u) = +-sqrt(16/25)#

#cos(u) = +-4/5#

We know that the cosine function is negative in the second quadrant, therefore, we choose the negative value:

#cos(u) = -4/5" [5]"#

Please observe that we have all 4 values that equation [1] requires expressed by equations [2], [3], [4], and [5].

We shall substitute those 4 values into equation [1]:

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

Simplify:

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