Question

How do you evaluate the expression `cos(u-v)` given `cosu=4/7` with `0<u<pi/2` and `sinv=-9/10` with `pi<v<(3pi)/2`?

Answer 1

`- (4sqrt19 + 9sqrt33)/70`

Explanation:

Use the trig identity:
cos (u - v) = cos u.cos v + sin u.sin v
In this case, we have the values of cos u, and sin v.
We must find sin u, and cos v.
`sin^2 u = 1 - cos^2 u = 1 - 16/49 = 33/49` --> `sin u = +- sqrt33/7`
Because u is in Quadrant I, then sin u is positive.
`cos^2 v = 1 - sin^2 v = 1 - 81/100 = 19/100` --> `cos v = +- sqrt19/10`
Because v is in Quadrant III, then cos v is negative. We get:
`cos (u - v) = (4/7)(- sqrt19/10) + (-9/10)(sqrt33/7)`
`= - (4sqrt19 + 9sqrt33)/70`