Question

use identities to find the values of the sine and cosine functions for the following angle measure.\\(\theta\\), given that \\(\cos 2\theta = \frac{40}{41}\\) and \\(\theta\\) terminates in quadrant i\\\\\\(\sin \theta = \square\\)\\(\text{(type an exact answer, using radicals as needed. rationalize all denominators.)}\\)\\(\cos \theta = \square\\)\\(\text{(type an exact answer, using radicals as needed. rationalize all denominators.)}\\)

Explanation:

Step1: Recall the double - angle identity for cosine

The double - angle identity for cosine is $\cos(2\theta)=2\cos^{2}\theta - 1=1 - 2\sin^{2}\theta$. We know that $\cos(2\theta)=\frac{40}{41}$ and $\theta$ is in quadrant I, so $\cos\theta>0$ and $\sin\theta>0$. We will first use the identity $\cos(2\theta)=2\cos^{2}\theta - 1$ to find $\cos\theta$.

We start with $\cos(2\theta)=2\cos^{2}\theta - 1$. Substitute $\cos(2\theta)=\frac{40}{41}$ into the identity:
$\frac{40}{41}=2\cos^{2}\theta - 1$

Step2: Solve for $\cos^{2}\theta$

Add 1 to both sides of the equation:
$2\cos^{2}\theta=\frac{40}{41}+1=\frac{40 + 41}{41}=\frac{81}{41}$

Then divide both sides by 2:
$\cos^{2}\theta=\frac{81}{82}$

Since $\theta$ is in quadrant I, $\cos\theta=\sqrt{\frac{81}{82}}=\frac{9}{\sqrt{82}}=\frac{9\sqrt{82}}{82}$ (rationalizing the denominator by multiplying numerator and denominator by $\sqrt{82}$)

Step3: Use the Pythagorean identity to find $\sin\theta$

The Pythagorean identity is $\sin^{2}\theta+\cos^{2}\theta = 1$. We know that $\cos^{2}\theta=\frac{81}{82}$, so:
$\sin^{2}\theta=1-\cos^{2}\theta=1 - \frac{81}{82}=\frac{82 - 81}{82}=\frac{1}{82}$

Since $\theta$ is in quadrant I, $\sin\theta=\sqrt{\frac{1}{82}}=\frac{1}{\sqrt{82}}=\frac{\sqrt{82}}{82}$ (rationalizing the denominator by multiplying numerator and denominator by $\sqrt{82}$)

Answer:

$\sin\theta=\frac{\sqrt{82}}{82}$, $\cos\theta=\frac{9\sqrt{82}}{82}$