Question

How do you find the exact value of sin (x) when cos (x) = 3/5 and the terminal side of x is in quadrant 4?

Answer 1

See below

Explanation:

Remember that `sin^2x+cos^2x=1` for all x-value

If `cosx=3/5`, then `cos^x=1-sin^2x=1-9/25=16/25`

`cosx=+-sqrt(16/25)=+-sqrt16/sqrt25=+-4/5`

However `x` is in 4th quadrant where `cosx` is positive, we conclude that `cosx=4/5`