Question

If `sin s = -12/13` and `sin t = 4/5`, what is `cos(s+t)` and `cos (s-t)`?

`s` is in quadrant IV and `t` in quadrant II.

Answer 1

`cos(s+t) = 33/65`

`cos(s-t) = -63/65`

Explanation:

We will need the values of `cos(s) and cos(t)`, therefore, we shall use the identity:

`cos(x) = +-sqrt(1-sin^2(x))" [1]"`

Substitute `sin(s)= -12/13` into equation [1]:

`cos(s) = +-sqrt(1-(-12/13)^2)`

`cos(s) = +-sqrt(169/169-144/169)`

`cos(s) = +-sqrt(25/169)`

`cos(s) = +-5/13" [2]"`

We are told that `s` is in the fourth quadrant, therefore, we shall choose the positive value:

`cos(s) = 5/13`

Substitute `sin(t)= 4/5` into equation [1]:

`cos(t) = +-sqrt(1-(4/5)^2)`

`cos(t) = +-sqrt(25/25-16/25)`

`cos(t) = +-sqrt(9/25)`

`cos(t) = +-3/5`

We are told that `t` is in the second quadrant, therefore, we shall choose the negative value:

`cos(t) = -3/5" [3]"`

Using the identity,

`cos(s+t) = cos(s)cos(t) - sin(s)sin(t)`

, we substitute `cos(s) = 5/13, cos(t) = -3/5, sin(s) = -12/13, and sin(t) = 4/5`:

`cos(s+t) = (5/13)(-3/5) - (-12/13)(4/5)`

`cos(s+t) = 33/65`

Using the identity,

`cos(s-t) = cos(s)cos(t) + sin(s)sin(t)`

, we substitute `cos(s) = 5/13, cos(t) = -3/5, sin(s) = -12/13, and sin(t) = 4/5`:

`cos(s-t) = (5/13)(-3/5) + (-12/13)(4/5)`

`cos(s-t) = -63/65`