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anonymous

  • one year ago

If sine of x equals 1 over 2, what is cos(x) and tan(x)? Explain your steps in complete sentences.

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  1. jim_thompson5910
    • one year ago
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    Hint: Pythagorean Trig Identity \[\Large \sin^2(x) + \cos^2(x) = 1\]

  2. anonymous
    • one year ago
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    Ok, so how would I continue off that? May you guide me through the problem please?

  3. jim_thompson5910
    • one year ago
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    since sine is 1/2, squaring that gives you 1/4 you can replace all of sin^2 with 1/4 like this \[\Large \sin^2(x) + \cos^2(x) = 1\] \[\Large \frac{1}{4} + \cos^2(x) = 1\]

  4. jim_thompson5910
    • one year ago
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    now isolate cos(x)

  5. anonymous
    • one year ago
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    we subract 1/4 from both sides

  6. anonymous
    • one year ago
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    so now we square root both sides right?

  7. jim_thompson5910
    • one year ago
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    correct on both statements

  8. anonymous
    • one year ago
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    then how do we get tangent

  9. jim_thompson5910
    • one year ago
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    \[\Large \tan(x) = \frac{\sin(x)}{\cos(x)}\]

  10. anonymous
    • one year ago
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    Ok, hold on. Let me put everything together.

  11. anonymous
    • one year ago
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    cos(x) = \[\sqrt{.75}\] is that how it should be written

  12. jim_thompson5910
    • one year ago
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    or \[\Large \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2}\]

  13. jim_thompson5910
    • one year ago
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    most books I've seen use \[\Large \frac{\sqrt{3}}{2}\]

  14. anonymous
    • one year ago
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    oh ok, so \[\tan(x) = \frac{ 1 }{ 2 } /\frac{ \sqrt{3} }{ 2 }\]

  15. jim_thompson5910
    • one year ago
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    Yes correct. Now simplify

  16. anonymous
    • one year ago
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    How would I do that?

  17. jim_thompson5910
    • one year ago
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    |dw:1439850551781:dw|

  18. jim_thompson5910
    • one year ago
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    |dw:1439850577166:dw|

  19. anonymous
    • one year ago
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    so the answer is \[\frac{ 1 }{ \sqrt{3} }\]

  20. jim_thompson5910
    • one year ago
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    rationalizing the denominator makes that turn into \[\Large \frac{\sqrt{3}}{3}\]

  21. anonymous
    • one year ago
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    Thank you so much

  22. jim_thompson5910
    • one year ago
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    no problem

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