anonymous
  • anonymous
Can someone show me why cos^2(x) is equal to (1/2)cos(2x) + 1/2 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
no one?
shadowfiend
  • shadowfiend
Workin' it.
anonymous
  • anonymous
Gimme a mo

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shadowfiend
  • shadowfiend
*sigh* Ignore RodneyF.
anonymous
  • anonymous
wolfram says it's true... don't know how in the world that is true.
anonymous
  • anonymous
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?
anonymous
  • anonymous
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
anonymous
  • anonymous
There is no algebraic way? hmm..
anonymous
  • anonymous
There is. Its just quite fiddly
anonymous
  • anonymous
There is, using the sum of angles formula.
anonymous
  • anonymous
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)
anonymous
  • anonymous
Got it thanks.
myininaya
  • myininaya
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anonymous
  • anonymous
Yep.
anonymous
  • anonymous
But doesn't that assume the prior knowledge of that trig identity?
anonymous
  • anonymous
Was just about finished with mine.
anonymous
  • anonymous
Or doesnt that matter?
anonymous
  • anonymous
It assumes you know the addition of angles identity yes.
myininaya
  • myininaya
lol
anonymous
  • anonymous
I'm in calc.. should have known that identity.. Always forget important stuff...
anonymous
  • anonymous
oh well..
anonymous
  • anonymous
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.
myininaya
  • myininaya
cos(x+x)=cos(x)cos(x)-sin(x)sin(x)
myininaya
  • myininaya
im always late
anonymous
  • anonymous
\[cos(\theta + \phi) = cos(\theta)cos(\phi) - sin(\theta)sin(\phi)\] So if \(\theta = \phi\) you start getting things simplified very quickly.

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