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anonymous
 one year ago
Use Pythagorean identity cos^2(theta) + sine^2(thetaof theta cosine
anonymous
 one year ago
Use Pythagorean identity cos^2(theta) + sine^2(thetaof theta cosine

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, you can certainly isolate \(\cos\theta\), but you need more than just that identity to find it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sin (theta) = sqrt(7)/7

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \sin^2\theta + \cos^2\theta = 1\implies \cos\theta = \sqrt{1\sin^2\theta} \]Since we know that \(\theta\) is acute, we know that \(\cos\theta > 0\), and so \(\cos\theta = \cos\theta\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Given the equation you just gave, we have: \[ \cos\theta = \sqrt{1 \left(\frac{\sqrt 7}{7}\right)^2} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You just need to simplify.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm having trouble with the simplification, sorry.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Start simple: \[ \left(\frac{\sqrt 7}{7}\right)^2=\left(\frac{\sqrt 7}{7}\right)\left(\frac{\sqrt 7}{7}\right)=\frac{\sqrt 7\cdot \sqrt 7}{7\cdot 7} =\ldots \]Can you simplify this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm getting sqrt(6)/7

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the answer to your last question is 1/7

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is \(1  1/7\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ 1\frac 17 = \frac77\frac17 = \ldots \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so: \[ \sqrt{\frac 67} = \frac{\sqrt 6}{\sqrt7} \]To rationalize the denominator, you multiply top and bottom by it: \[ \frac{\sqrt 6}{\sqrt7}= \frac{\sqrt 6\cdot\sqrt7}{\sqrt7\cdot\sqrt7} = \ldots \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \left(\frac{\sqrt{7}}7\right)^2+\left(\frac{\sqrt{42}}7\right)^2 = 1 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Also, \(\sqrt{42}/2 < 1 < \pi /2\), so it is acute.
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