anonymous
  • anonymous
Rewrite with only sin x and cos x. cos 3x cos x - 4 cos x sin2x -sin3x + 2 sin x cos x -sin2x + 2 sin x cos x 2 sin2x cos x - 2 sin x cos x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zepdrix
  • zepdrix
Oh boy this one is a doozy :u
anonymous
  • anonymous
haha does that mean its tough?
zepdrix
  • zepdrix
Yah it's a bit of work to widdle it down :) \[\Large \cos(3x) \quad=\quad \cos(2x+x)\]Let's apply the `Angle Sum Identity for Cosine` \[\large \color{royalblue}{\cos(\alpha+\beta) \quad=\quad \cos \alpha \cos \beta - \sin \alpha \sin \beta}\]

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anonymous
  • anonymous
i honestly know didly squat about this one
zepdrix
  • zepdrix
+_+
zepdrix
  • zepdrix
Well look at the blue identity a sec. We'll apply it and our function will break down like this,\[\Large \cos(2x+x) \quad=\quad \cos(2x)\cos(x)-\sin(2x)\sin(x)\]Look at da blueee +_+ Understand how I applied that identity?
zepdrix
  • zepdrix
That's not the answer by the way U: We have a long ways to go still.
anonymous
  • anonymous
so we are just using identities?
zepdrix
  • zepdrix
Yes
anonymous
  • anonymous
ok yup i get that
zepdrix
  • zepdrix
See how we have a cos(x) and a sin(x)? That's what we're trying to achieve. We want all of the angles to be in terms of just x. So we still need to deal with the cos(2x) and sin(2x). To do so, we can apply our `Double Angle Identities`:\[\large \color{ #CC0033}{\cos(2x) \quad=\quad 2\cos^2x-1}\]\[\large \color{#FF11AA}{\sin(2x)\quad=\quad 2\sin(x)\cos(x)}\]
anonymous
  • anonymous
okay!
anonymous
  • anonymous
:)
zepdrix
  • zepdrix
\[\Large =\color{#CC0033}{\cos(2x)}\cos x-\color{#FF11AA}{\sin(2x)}\sin x\]So this becomes,\[\Large =\color{#CC0033}{\left(2\cos^2x-1\right)}\cos x-\color{#FF11AA}{\left(2\sin x \cos x\right)}\sin x\]
anonymous
  • anonymous
ok cool got that
anonymous
  • anonymous
then do i just facter those in?
zepdrix
  • zepdrix
Yes, distribute the cosx and sinx respectively. We might have to fiddle with it a bit after that. There are `several` forms for the cosine double angle. So depending on which one THEY expected us to use, our answer might look a little drifferent. Multiply that all out, then we'll decide whether or not we need to use our Square Identity to fix it.
anonymous
  • anonymous
ok one min im gunna east some dinner
anonymous
  • anonymous
ok im back
anonymous
  • anonymous
do you wanna see the answer choices that i have?
zepdrix
  • zepdrix
No, I want you to do some of the work :3 lol Multiply that last step out.
anonymous
  • anonymous
ok so i can replace the trig functions with a variable like u correct? while i do the algerbra
zepdrix
  • zepdrix
\[\Large =\color{#CC0033}{\left(2\cos^2x-1\right)}\cos x-\color{#FF11AA}{\left(2\sin x \cos x\right)}\sin x\]Oh do something like this?\[\Large =\color{#CC0033}{\left(2u^2-1\right)}u-\color{#FF11AA}{\left(2vu\right)}v\]Yah I guess you could do that +_+
anonymous
  • anonymous
ok so then i just take the 2u^2-1 and foil it to u correct? for the first step?
zepdrix
  • zepdrix
mmmmmm ya :u
anonymous
  • anonymous
ok so is it 2u^3-1u-2v^2u

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