# How do you find the exact values of costheta and sintheta when tantheta=1?

Nov 15, 2016

$\sin \left(\theta\right) = \cos \left(\theta\right) = \frac{\sqrt{2}}{2}$

OR

$\sin \left(\theta\right) = \cos \left(\theta\right) = - \frac{\sqrt{2}}{2}$

#### Explanation:

Given: $\tan \left(\theta\right) = 1$

Use the identity:

${\tan}^{2} \left(\theta\right) + 1 = {\sec}^{2} \left(\theta\right)$

Substitute ${1}^{2}$ for ${\tan}^{2} \left(\theta\right)$

${1}^{2} + 1 = {\sec}^{2} \left(\theta\right)$

$2 = {\sec}^{2} \left(\theta\right)$

Because $\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right)$ we can change the above equation to:

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

$\cos \left(\theta\right) = \pm \frac{1}{\sqrt{2}}$

$\cos \left(\theta\right) = \pm \frac{\sqrt{2}}{2}$

$\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right) = 1$

$\sin \left(\theta\right) = \cos \left(\theta\right)$

$\sin \left(\theta\right) = \pm \frac{\sqrt{2}}{2}$

Because we are given nothing to determine whether $\theta$ is in the first of the third quadrant:

$\sin \left(\theta\right) = \cos \left(\theta\right) = \frac{\sqrt{2}}{2}$

OR

$\sin \left(\theta\right) = \cos \left(\theta\right) = - \frac{\sqrt{2}}{2}$

Nov 15, 2016

 cosθ=sinθ=+-sqrt2/2

#### Explanation:

Draw a right isosceles triangle. The angles are 45, 45,90
Tan 45 =1,
Using Pythagoras the sides are in the ratio $1 : 1 : \sqrt{2}$
cos 45=sin45=$\frac{1}{\sqrt{2}}$=$\frac{\sqrt{2}}{2}$
But then we need to look at the angle in the third quadrant to get the negative result.