Question

use identities to find values of the sine and cosine functions for the angle measure.
θ, given that cos 2θ = 40/41 and 0° < θ < 90°
sinθ = √82/82
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
cosθ = □
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)

Explanation:

Step1: Use the double - angle identity for cosine

The double - angle identity for cosine is \(\cos2\theta = 1 - 2\sin^{2}\theta=2\cos^{2}\theta - 1\). We can use \(\cos2\theta=2\cos^{2}\theta - 1\) to find \(\cos\theta\).
Given \(\cos2\theta=\frac{40}{41}\), then \(2\cos^{2}\theta-1 = \frac{40}{41}\).

Step2: Solve for \(\cos^{2}\theta\)

Add \(1\) to both sides of the equation \(2\cos^{2}\theta-1=\frac{40}{41}\):
\(2\cos^{2}\theta=\frac{40}{41}+1=\frac{40 + 41}{41}=\frac{81}{41}\).
Divide both sides by \(2\): \(\cos^{2}\theta=\frac{81}{82}\).

Step3: Solve for \(\cos\theta\)

Since \(0^{\circ}<\theta<90^{\circ}\), \(\cos\theta>0\). Take the square root of \(\cos^{2}\theta=\frac{81}{82}\), so \(\cos\theta=\sqrt{\frac{81}{82}}=\frac{9}{\sqrt{82}}=\frac{9\sqrt{82}}{82}\).

Answer:

\(\cos\theta=\frac{9\sqrt{82}}{82}\)