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anonymous

  • one year ago

Use Pythagorean identity cos^2(theta) + sine^2(thetaof theta cosine

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  1. anonymous
    • one year ago
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    Well, you can certainly isolate \(\cos\theta\), but you need more than just that identity to find it.

  2. anonymous
    • one year ago
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    sin (theta) = sqrt(7)/7

  3. anonymous
    • one year ago
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    \[ \sin^2\theta + \cos^2\theta = 1\implies |\cos\theta| = \sqrt{1-\sin^2\theta} \]Since we know that \(\theta\) is acute, we know that \(\cos\theta > 0\), and so \(|\cos\theta| = \cos\theta\).

  4. anonymous
    • one year ago
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    Given the equation you just gave, we have: \[ \cos\theta = \sqrt{1- \left(\frac{\sqrt 7}{7}\right)^2} \]

  5. anonymous
    • one year ago
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    You just need to simplify.

  6. anonymous
    • one year ago
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    I'm having trouble with the simplification, sorry.

  7. anonymous
    • one year ago
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    Start 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?

  8. anonymous
    • one year ago
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    I'm getting sqrt(6)/7

  9. anonymous
    • one year ago
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    Not quite it

  10. anonymous
    • one year ago
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    the answer to your last question is 1/7

  11. anonymous
    • one year ago
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    What is \(1 - 1/7\)?

  12. anonymous
    • one year ago
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    0

  13. anonymous
    • one year ago
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    No, \[ 1-\frac17 \]

  14. anonymous
    • one year ago
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    \[ 1-\frac 17 = \frac77-\frac17 = \ldots \]

  15. anonymous
    • one year ago
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    I'm getting 6/7

  16. anonymous
    • one year ago
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    Okay, so: \[ \sqrt{\frac 67} = \frac{\sqrt 6}{\sqrt7} \]To rationalize the denominator, you multiply top and bottom by it: \[ \frac{\sqrt 6}{\sqrt7}= \frac{\sqrt 6\cdot\sqrt7}{\sqrt7\cdot\sqrt7} = \ldots \]

  17. anonymous
    • one year ago
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    sqrt(42)/7

  18. anonymous
    • one year ago
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    That looks right.

  19. anonymous
    • one year ago
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    \[ \left(\frac{\sqrt{7}}7\right)^2+\left(\frac{\sqrt{42}}7\right)^2 = 1 \]

  20. anonymous
    • one year ago
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    Also, \(\sqrt{42}/2 < 1 < \pi /2\), so it is acute.

  21. anonymous
    • one year ago
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    thank you

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