Question

40 points!!!!!PLEASE HELP!!!!

Answer 1

Answer:  $\frac{\sqrt{26}}{26}$

This is the square root of 26 over 26. The second 26 is not inside the square root. You can type it in as sqrt(26)/26.

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$\theta$ is in Q1 so $\sin(\theta) > 0$

Use the trig identity below and plug in the given value to get...

$\sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1-\cos(x)}{2}}\\\\\sin\left(\frac{2\theta}{2}\right) = \pm \sqrt{\frac{1-\cos(2\theta)}{2}}\\\\\sin(\theta) = \sqrt{\frac{1-\cos(2\theta)}{2}} \ \ \text{since } \ \sin(\theta) > 0\\\\\sin(\theta) = \sqrt{\frac{1-\frac{12}{13}}{2}}$

$\sin(\theta) = \sqrt{\frac{1}{26}}\\\\\sin(\theta) = \frac{\sqrt{1}}{\sqrt{26}}\\\\\sin(\theta) = \frac{1}{\sqrt{26}}\\\\\sin(\theta) = \frac{1*\sqrt{26}}{\sqrt{26}*\sqrt{26}} \ \ \text{ rationalizing denominator}\\\\\sin(\theta) = \frac{\sqrt{26}}{26}$

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Edit:

To find cos(theta), we use the pythagorean identity below

$\sin^2(\theta) + \cos^2(\theta) = 1\\\\\cos^2(\theta) = 1-\sin^2(\theta)\\\\\cos(\theta) = \sqrt{1-\sin^2(\theta)}\ \ \text{cosine is positive in Q1}\\\\\cos(\theta) = \sqrt{1-\left(\frac{1}{\sqrt{26}}\right)^2}\\\\\cos(\theta) = \sqrt{1-\frac{1}{26}}\\\\\cos(\theta) = \sqrt{\frac{25}{26}}\\\\\cos(\theta) = \frac{\sqrt{25}}{\sqrt{26}}\\\\\cos(\theta) = \frac{5}{\sqrt{26}}\\\\\cos(\theta) = \frac{5\sqrt{26}}{\sqrt{26}*\sqrt{26}}\\\\\cos(\theta) = \frac{5\sqrt{26}}{26}$

Answer 2

Answer:

This is the square root of 26 over 26. The second 26 is not inside the square root. You can type it in as sqrt(26)/26.

Step-by-step explanation: