Question

If `sintheta=1/3` and `theta` is in quadrant I, how do you evaluate `cos2theta`?

Answer 1

`7/9`

Explanation:

`cos(2theta)`

= `cos^2theta-sin^2theta`

=`1-sin^2theta-sin^2theta`

=`1-2sin^2theta`

=`1-2(1/3)^2`

=`1-2/9`

=`7/9`

Answer 2

`cos2theta=7/9`

Explanation:

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)cos2theta=cos^2theta-sin^2theta`

`•color(white)(x)sin^2theta+cos^2theta=1`

`rArrcostheta=+-sqrt(1-sin^2theta)`

`theta" is in first quadrant as is "2theta`

`costheta=sqrt(1-(1/3)^2)=sqrt(1-1/9)=sqrt(8/9)`

`rArrcostheta=(2sqrt2)/3`

`cos2theta=((2sqrt2)/3)^2-(1/3)^2`

`color(white)(cos2theta)=8/9-1/9=7/9`