Question

What is the value of `cos(2pi+x)` if `sinx=0.3`?

I'm having trouble solving this problem:

Find the value of `cos(2pi+x)` if `sinx=0.3`

Answer 1

`+- 0.95`

Explanation:

`cos (x + 2pi) = cos x`.
`sin x = 0.3`
`cos^2 x = 1 - sin^2 x = 1 - 0.09 = 0.91`
`cos x = +- sqrt0.91 = +- 0.95`

Note. There are 2 values of cos x, because if sin x = 0.3,
x could either be in Quadrant 1 or Quadrant 3.

Answer 2

`+-sqrt91/10`

Explanation:

`"expand using "color(blue)"addition formula for cosine"`

`•color(white)(x)cos(A+-B)=cosAcosB∓sinAsinB`

`rArrcos(2pi+x)`

`=cos(2pi)cosx-sin(2pi)sinx`

`=cosx-0=cosx`

`•color(white)(x)cosx=+-sqrt(1-sin^2x)`

`rArrcosx=+-sqrt(1-(3/10)^2)`

`color(white)(rArrcosx)=+-sqrt(91/100)`

`color(white)(rArrcosx)=+-sqrt91/10larrcolor(red)" exact values"`