Question

use identities to find values of the sine and cosine functions of the function for the angle measure.
2θ, given sinθ = \frac{\sqrt{6}}{7} and cosθ > 0
cos2θ = □
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
sin2θ = □
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find the value of \(\cos\theta\)

Use the Pythagorean identity \(\sin^{2}\theta+\cos^{2}\theta = 1\).
Given \(\sin\theta=\frac{\sqrt{6}}{7}\), then \(\cos^{2}\theta=1-\sin^{2}\theta\).
Substitute \(\sin\theta\) into the formula: \(\cos^{2}\theta=1 - (\frac{\sqrt{6}}{7})^{2}=1-\frac{6}{49}=\frac{49 - 6}{49}=\frac{43}{49}\).
Since \(\cos\theta>0\), then \(\cos\theta=\frac{\sqrt{43}}{7}\).

Step2: Find the value of \(\cos2\theta\)

Use the double - angle formula \(\cos2\theta=1 - 2\sin^{2}\theta\).
Substitute \(\sin\theta=\frac{\sqrt{6}}{7}\) into the formula:
\(\cos2\theta=1-2\times(\frac{\sqrt{6}}{7})^{2}=1 - 2\times\frac{6}{49}=1-\frac{12}{49}=\frac{49-12}{49}=\frac{37}{49}\).

Step3: Find the value of \(\sin2\theta\)

Use the double - angle formula \(\sin2\theta = 2\sin\theta\cos\theta\).
Substitute \(\sin\theta=\frac{\sqrt{6}}{7}\) and \(\cos\theta=\frac{\sqrt{43}}{7}\) into the formula:
\(\sin2\theta=2\times\frac{\sqrt{6}}{7}\times\frac{\sqrt{43}}{7}=\frac{2\sqrt{258}}{49}\).

Answer:

\(\cos2\theta=\frac{37}{49}\), \(\sin2\theta=\frac{2\sqrt{258}}{49}\)