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

Jun 5, 2018

$\frac{7}{9}$

#### Explanation:

$\cos \left(2 \theta\right)$

= ${\cos}^{2} \theta - {\sin}^{2} \theta$

=$1 - {\sin}^{2} \theta - {\sin}^{2} \theta$

=$1 - 2 {\sin}^{2} \theta$

=$1 - 2 {\left(\frac{1}{3}\right)}^{2}$

=$1 - \frac{2}{9}$

=$\frac{7}{9}$

Jun 5, 2018

$\cos 2 \theta = \frac{7}{9}$

#### Explanation:

$\text{using the "color(blue)"trigonometric identities}$

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

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

$\Rightarrow \cos \theta = \pm \sqrt{1 - {\sin}^{2} \theta}$

$\theta \text{ is in first quadrant as is } 2 \theta$

$\cos \theta = \sqrt{1 - {\left(\frac{1}{3}\right)}^{2}} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}}$

$\Rightarrow \cos \theta = \frac{2 \sqrt{2}}{3}$

$\cos 2 \theta = {\left(\frac{2 \sqrt{2}}{3}\right)}^{2} - {\left(\frac{1}{3}\right)}^{2}$

$\textcolor{w h i t e}{\cos 2 \theta} = \frac{8}{9} - \frac{1}{9} = \frac{7}{9}$