# If sinθ =4/7, what is cosθ?

Jun 20, 2018

cosθ= +- 33^(1/2)/7

#### Explanation:

Since

Sin^2θ xx cos^2θ =1

you have

(4/7)^2 xx cos^2θ=1

So

Cos^2θ= 1- 16/49

Cosθ= +- 33^(1/2)/7

Jun 20, 2018

$\cos \left(\theta\right) = \pm \frac{\sqrt{33}}{7}$

#### Explanation:

We want to find $\cos \left(\theta\right)$, when $\sin \left(\theta\right) = \frac{4}{7}$

By the pythagorean trig identity

color(blue)(cos^2(theta)+sin^2(theta)=1

$\implies {\cos}^{2} \left(\theta\right) = 1 - {\sin}^{2} \left(\theta\right)$

$\implies \cos \left(\theta\right) = \pm \sqrt{1 - {\sin}^{2} \left(\theta\right)}$

Substitute color(red)(sin(theta)=4/7

$\cos \left(\theta\right) = \pm \sqrt{1 - {\left(\frac{4}{7}\right)}^{2}}$

color(white)(cos(theta))=+-sqrt(1-(16/49)

$\textcolor{w h i t e}{\cos \left(\theta\right)} = \pm \sqrt{\frac{49 - 16}{49}}$

$\textcolor{w h i t e}{\cos \left(\theta\right)} = \pm \frac{\sqrt{33}}{7}$

Jun 20, 2018

$\cos \left(\theta\right) = \pm \frac{\sqrt{33}}{7}$

#### Explanation:

Using the relation
${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$ then we get
$\cos \left(\theta\right) = \pm \sqrt{1 - \frac{16}{49}}$