Question

use the pythagorean identity sin²θ + cos²θ = 1 to find cosθ. the angle θ is an acute angle. sinθ = $\frac{sqrt{3}}{3}$ cosθ = (simplify your answer. type an exact answer, using radicals as needed. use integers or fractions for any numbers in the expression. rationalize all denominators.)

Explanation:

Step1: Substitute sinθ into identity

Given $\sin\theta=\frac{\sqrt{3}}{3}$, substitute into $\sin^{2}\theta+\cos^{2}\theta = 1$. So we have $(\frac{\sqrt{3}}{3})^{2}+\cos^{2}\theta=1$.

Step2: Calculate $(\frac{\sqrt{3}}{3})^{2}$

$(\frac{\sqrt{3}}{3})^{2}=\frac{(\sqrt{3})^{2}}{3^{2}}=\frac{3}{9}=\frac{1}{3}$. Then the equation becomes $\frac{1}{3}+\cos^{2}\theta = 1$.

Step3: Solve for $\cos^{2}\theta$

Subtract $\frac{1}{3}$ from both sides: $\cos^{2}\theta=1 - \frac{1}{3}=\frac{3 - 1}{3}=\frac{2}{3}$.

Step4: Find $\cos\theta$

Since $\theta$ is acute, $\cos\theta>0$. So $\cos\theta=\sqrt{\frac{2}{3}}=\frac{\sqrt{2}}{\sqrt{3}}$. Rationalize the denominator: $\cos\theta=\frac{\sqrt{2}\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\frac{\sqrt{6}}{3}$.

Answer:

$\frac{\sqrt{6}}{3}$