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anonymous

  • 5 years ago

Can someone show me why cos^2(x) is equal to (1/2)cos(2x) + 1/2 ?

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  1. anonymous
    • 5 years ago
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    no one?

  2. shadowfiend
    • 5 years ago
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    Workin' it.

  3. anonymous
    • 5 years ago
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    Gimme a mo

  4. shadowfiend
    • 5 years ago
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    *sigh* Ignore RodneyF.

  5. anonymous
    • 5 years ago
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    wolfram says it's true... don't know how in the world that is true.

  6. anonymous
    • 5 years ago
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    There's a few reasons you can show depending on your understanding. There's the \(e^{i\theta}\) argument, or there's the angle addition arguement. Which would you prefer?

  7. anonymous
    • 5 years ago
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    Its basically a derivative of the double angle formula for sin and cos. I can have a go at writing the proof out, but you may be better to google it

  8. anonymous
    • 5 years ago
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    There is no algebraic way? hmm..

  9. anonymous
    • 5 years ago
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    There is. Its just quite fiddly

  10. anonymous
    • 5 years ago
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    There is, using the sum of angles formula.

  11. anonymous
    • 5 years ago
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    cos (2x) = cos^2(x) - sin^2(x) cos(2x) + sin^2(x) = cos^2(x) cos (2x) + 1 - cos^2(x) = cos^(x) cos(2x) +1 = 2cos^2(x) cos(2x)/2 + 1/2 = cos^2(x)

  12. anonymous
    • 5 years ago
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    Got it thanks.

  13. myininaya
    • 5 years ago
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  14. anonymous
    • 5 years ago
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    Yep.

  15. anonymous
    • 5 years ago
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    But doesn't that assume the prior knowledge of that trig identity?

  16. anonymous
    • 5 years ago
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    Was just about finished with mine.

  17. anonymous
    • 5 years ago
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    Or doesnt that matter?

  18. anonymous
    • 5 years ago
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    It assumes you know the addition of angles identity yes.

  19. myininaya
    • 5 years ago
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    lol

  20. anonymous
    • 5 years ago
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    I'm in calc.. should have known that identity.. Always forget important stuff...

  21. anonymous
    • 5 years ago
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    oh well..

  22. anonymous
    • 5 years ago
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    sin 2a= sin ( a+a ) = sin a cos a + cos a sin a = 2 sin a cos a. cos 2 a= cos ( a+a ) = cos a cos a − sin a sin a cos 2a = cos² a− sin²a. . . . . . . (1) This is the first of the three versions of cos 2a. To derive the second version, in line (1) use this Pythagorean identity: sin²a = 1 − cos²a. Line (1) then becomes cos 2a = cos²a − (1 − cos²a) = cos²a − 1 + cos²a. cos 2a = 2 cos²a − 1. . . . . . . . . . (2) To derive the third version, in line (1) use this Pythagorean identity: cos² a= 1 − sin²a. We have cos 2 a= 1 − sin² a− sin²a;. cos 2a = 1 − 2 sin²a. . . . . . . . . . (3) These are the three forms of cos 2a.

  23. myininaya
    • 5 years ago
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    cos(x+x)=cos(x)cos(x)-sin(x)sin(x)

  24. myininaya
    • 5 years ago
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    im always late

  25. anonymous
    • 5 years ago
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    \[cos(\theta + \phi) = cos(\theta)cos(\phi) - sin(\theta)sin(\phi)\] So if \(\theta = \phi\) you start getting things simplified very quickly.

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