Question

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

Answer 1

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.

Explanation:

Answer 2

Answer:
`(√(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((x)/(2)\right) = \pm \sqrt{(1-\cos(x))/(2)}\\\\\sin\left((2\theta)/(2)\right) = \pm \sqrt{(1-\cos(2\theta))/(2)}\\\\\sin(\theta) = \sqrt{(1-\cos(2\theta))/(2)} \ \ \text{since } \ \sin(\theta) > 0\\\\\sin(\theta) = \sqrt{(1-(12)/(13))/(2)}\\\\`

`\sin(\theta) = \sqrt{(1)/(26)}\\\\\sin(\theta) = (√(1))/(√(26))\\\\\sin(\theta) = (1)/(√(26))\\\\\sin(\theta) = (1*√(26))/(√(26)*√(26)) \ \ \text{ rationalizing denominator}\\\\\sin(\theta) = (√(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) = √(1-\sin^2(\theta))\ \ \text{cosine is positive in Q1}\\\\\cos(\theta) = \sqrt{1-\left((1)/(√(26))\right)^2}\\\\\cos(\theta) = \sqrt{1-(1)/(26)}\\\\\cos(\theta) = \sqrt{(25)/(26)}\\\\\cos(\theta) = (√(25))/(√(26))\\\\\cos(\theta) = (5)/(√(26))\\\\\cos(\theta) = (5√(26))/(√(26)*√(26))\\\\\cos(\theta) = (5√(26))/(26)\\\\`