Question

If `sinx-cosx=1/3`, find `cos(4x)`?

Answer 1

cos 4x = - 0.58
`x = 58^@63`; and `x = 211^@37`

Explanation:

Use trig identity:
`sin x - cos x = sqrt2sin (x - pi/4)`
In this case
`sqrt2sin (x - pi/4) = 1/3`
`sin (x - pi/4) = 1/(3sqrt2) = sqrt2/6`
Calculator and unit circle give -->

a. x - 45 = 13.63 --> `x = 45 + 13.63 = 58^@63`
4x = 234^@52 --> `cos 4x = cos 234^@52 = - 0.58`

b. x - 45 = 180 - 13.63 = 166.37
`x = 166.37 + 45 = 211^@37`
`4x = 845.48 = 845.48 - 720 = 125^@48`
`cos 4x = cos 125^@48 = - 0.58`
Check by calculator.
x = 58.63 --> sin x = 0.85 --> cos x = 0.52.
sin x - cos x = 0.33 = 1/3. Proved
x = 211.37 --> sin x = - 0.52 --> cos x = - 0.85
sin x - cos x = - 0.52 - (-0.85) = 0.33 = 1/3. Proved

Answer 2

`cos(4x) = - 47/81`

Explanation:

We have `sinx-cosx=1/3`

`sin^2x + cos^2x - 2sinxcosx = 1/9`

`1 - 1/9 = sin(2x)`

`8/9 = sin(2x)`

We know that

`sin^2(2x) + cos^2(2x) = 1`

`(8/9)^2 + cos^2(2x) = 1`

`cos^2(2x) =1 - 64/81`

`cos^2(2x) = 17/81`

We can rewrite

`cos(4x)` as `2cos^2(2x) - 1`

`cos(4x) = 2xx17/81 - 1`

`cos(4x) = - 47/81`