Question

Evalute costheta if sintheta = (sqrt5)/3

Answer 1

Answer: $cos(\theta)=\±\frac{2}{3}$

Step-by-step explanation:

In this case we know that:

$sin(\theta) = \frac{\sqrt{5}}{3}$

To find the value of $cos(\theta)$ we use the following trigonometric identity

$cos^2(\theta)=1-sin^2(\theta)$

So

$sin^2(\theta) = (\frac{\sqrt{5}}{3})^2$

Therefore

$cos^2(\theta)=1-(\frac{\sqrt{5}}{3})^2$

$cos^2(\theta)=1-\frac{5}{9}$

$cos^2(\theta)=\frac{4}{9}$

$cos(\theta)=\±\sqrt{\frac{4}{9}}$

$cos(\theta)=\±\frac{2}{3}$

Answer 2

Answer:

$\large\boxed{\cos\theta=\pm\dfrac{2}{3}}$

Step-by-step explanation:

Use $\sin^2x+\cos^2x=1$.

We have

$\sin\theta=\dfrac{\sqrt5}{3}$

Substitute:

$\left(\dfrac{\sqrt5}{3}\right)^2+\cos^2\theta=1\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(\sqrt5)^2}{3^2}+\cos^2\theta=1\qquad\text{use}\ (\sqrt{a})^2=a\\\\\dfrac{5}{9}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{5}{9}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{9}{9}-\dfrac{5}{9}\\\\\cos^2\theta=\dfrac{4}{9}\to \cos\theta=\pm\sqrt{\dfrac{4}{9}}\\\\\cos\theta=\pm\dfrac{\sqrt4}{\sqrt9}\\\\\cos\theta=\pm\dfrac{2}{3}$