Question

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)?

Im lost

Answer 1

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

Explanation:

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`

Answer 2

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`