anonymous
  • anonymous
given that cos (A+B) = cosAcosB-sinAsinB and f(Q)=8sin^4(θ)-2sin^2(θ)-2 show that f(θ)=cos4θ-3cos2θ Now solve, 8sin^4(θ)-2sin^2θ+3cos2θ=2.5 for 0<θ
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
oh geez...I'm not good at this but Sarkar is. Maybe you could msg him?
anonymous
  • anonymous
just go to the envelope on the top left and put Sarkar in the recipients box :) and the link to this problem!
anonymous
  • anonymous
Idk either...probably the person who created this site.

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myininaya
  • myininaya
@omar1131 I'm confused a little about your problem
myininaya
  • myininaya
f(Q)=8sin^4(θ)-2sin^2(θ)-2 show that f(θ)=cos4θ-3cos2θ
myininaya
  • myininaya
So Q or theta for the first function?
myininaya
  • myininaya
\[8 \sin^4(\theta)-2 \sin^2(\theta)-2=\cos(4 \theta) -3 \cos(2 \theta)\] So we want to show this is an identity right?
myininaya
  • myininaya
assuming it is anyways...
myininaya
  • myininaya
so notice that on on side we have are angles being double and 2*double lets try to right without the double thing going on
myininaya
  • myininaya
\[\cos(4 \theta) -3 \cos(2 \theta)= \cos(2 \theta+2 \theta)-3 \cos( \theta+\theta)\] \[=(\cos(2 \theta) \cos(2 \theta)-\sin(2 \theta) \sin(2 \theta))-3(\cos(\theta) \cos(\theta)-\sin(\theta) \sin(\theta))\] I used the thing you gave us okay ? :) \[=\cos( 2 \theta) \cos( 2 \theta) -\sin(2 \theta) \sin(2 \theta))-3 \cos^2(\theta)+3 \sin^2(\theta)\] ------------------------------------------------- Now we just found that \[\cos(2 \theta)=\cos(\theta+\theta)=\cos^2(\theta)-\sin^2(\theta)\] But we still have some double action going on with the \[\sin(2 \theta)\] Can we use anything else besides the thing given? ------------------------------------------------- \[(\cos^2(\theta)-\sin^2(\theta)(\cos^2(\theta)-\sin^2(\theta))-\sin(2\theta) \sin(2 \theta)-3 \cos^2(\theta)+3 \sin^2(\theta)\] \[=\cos^4(\theta)-2 \cos^2(\theta) \sin^2(\theta) +\sin^4(\theta)-\sin(2 \theta) \sin(2 \theta)-3 \cos^2(\theta)+3 \sin^2(\theta)\]
myininaya
  • myininaya
So can we use anything else besides the thing given?
myininaya
  • myininaya
Because I can think of another useful sum/difference formula
myininaya
  • myininaya
like can we use \[\sin(2 \theta)=2 \sin(\theta) \cos(\theta)\]
anonymous
  • anonymous
maybe use \[\cos(2\theta) = 1 - 2\sin^2(\theta)\]
anonymous
  • anonymous
rearranging to get \[ -2\sin^2(\theta) \] and \[8\sin^4(\theta)\] and substitute those in
anonymous
  • anonymous
\[\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) -1 = 1 - 2\sin^2(\theta)\]
anonymous
  • anonymous
substituting in those in my earlier comment leaves you with \[2(1 - \cos(2\theta))^2 + \cos(2\theta) -3\]
anonymous
  • anonymous
try multiplying out the first bracket, collecting like terms and seeing if you can spot the other double angle identity lurking around somewhere...
anonymous
  • anonymous
hope this is helping xD
anonymous
  • anonymous
i can post up a detailed solution for reference if you want
myininaya
  • myininaya
@omar1131 I understand your anger. But OpenStudy is a pg13 website. Please don't use the bad language here.
myininaya
  • myininaya
I would still like to know the answer to my one question Can we use \[\sin(2 \theta)=2 \sin(\theta) \cos(\theta)\]
myininaya
  • myininaya
Ok great! :)
myininaya
  • myininaya
\[ =\cos^4(\theta)-2 \cos^2(\theta) \sin^2(\theta) +\sin^4(\theta)-\sin(2 \theta) \sin(2 \theta)-3 \cos^2(\theta)+3 \sin^2(\theta) \] \[=\cos^4(\theta)-2 \cos^2(\theta) \sin^2(\theta)+\sin^4(\theta)-2 \sin(\theta) \cos(\theta) 2 \sin(\theta) \cos(\theta)-3 \cos^2(\theta)+3 \sin^2(\theta)\] \[=\cos^4(\theta)-2 \cos^2(\theta) \sin^2(\theta)+\sin^4(\theta)-2 \sin^2(\theta)\cos^2(\theta)-3 \cos^2(\theta)+3 \sin^2(\theta)\] \[=\cos^4(\theta)-4 \sin^2(\theta) \cos^2(\theta)-3\cos^2(\theta) +3 \sin^2(\theta)+\sin^4(\theta)\] Now the other side is in terms of sine so lets write this side in terms of sine
anonymous
  • anonymous
we have \[\cos(A + B) = cosAcosB - sinAsinB\] so\[\cos(2\theta) = \cos(\theta + \theta) = \cos(\theta)\cos(\theta) - \sin(\theta)\sin(\theta)\]\[= \cos^2(\theta) -\sin^2(\theta) [1]\] now using\[\cos^2(\theta) +\sin^2(\theta) = 1[2]\]\[\sin^2(\theta) = (1-\cos^2(\theta)) [3]\] substitute equation [3] into equation [1]\[\cos(2\theta) = \cos^2(\theta) - (1-\cos^2(\theta))\]\[\cos(2\theta)= 2\cos^2(\theta) -1\] doing the same but rearranging [2] to get cos^2 and subbing we get \[\cos(2\theta)= 1 - 2sin^2(\theta)\]
anonymous
  • anonymous
ill let myininaya finish, thats a good way (although i prefer my solution :p)
anonymous
  • anonymous
cya guys
myininaya
  • myininaya
\[=(\cos^2(\theta))^2-4\sin^2(\theta)(1-\sin^2(\theta))-3(1-\sin^2(\theta))+3 \sin^2(\theta)+\sin^4(\theta)\] \[=(1-\sin^2(\theta))^2-4\sin^2(\theta)+4 \sin^4(\theta)-3 +3 \sin^2(\theta)+3 \sin^2(\theta)+\sin^4(\theta)\] \[=1-2\sin^2(\theta)+\sin^4(\theta)-4\sin^2(\theta)+4 \sin^4(\theta)-3 +3 \sin^2(\theta)+3 \sin^2(\theta)+\sin^4(\theta)\] Like terms :) \[=6 \sin^4(\theta) +0\sin^2(\theta)-2 =6 \sin^4(\theta)-2 \] I think I made a mistake somewhere :(
myininaya
  • myininaya
I need to do this by hand lol
myininaya
  • myininaya
lol ok @omar1131 but let me know if you run into any more trouble This one is crazy long
myininaya
  • myininaya
Crazy long the way I'm approaching it lol
myininaya
  • myininaya
|dw:1332876900546:dw|
myininaya
  • myininaya
ok i got it using the the identities i was talking about
myininaya
  • myininaya
I think I will just scan this and let you look at it is that okay?
myininaya
  • myininaya
1 Attachment
myininaya
  • myininaya
I guess you can say that lol But we are just suppose to show they are equal and that is what I did Or that is what I'm assume from asking earlier
myininaya
  • myininaya
Well we showed that \[\cos(4 \theta)-3 \cos(2 \theta)=8 \sin^4(\theta)-2 \sin^2(\theta)-2\] But we know that \[f( \theta)=8 \sin^4(\theta)-2 \sin^2(\theta)-2 \] And since \[\cos(4 \theta)-3 \cos(2 \theta)=8 \sin^4(\theta)-2 \sin^2(\theta)-2\] then we also have \[f(\theta)=\cos(4 \theta)-3 \cos(2 \theta)\] And we are done with this part of your problem(s) lol well @omar1131 Some people may feel uncomfortable about give their private info out(e.x their name) so it is against the CoC. But my name is Christy :)
myininaya
  • myininaya
It is nice to meet you to and I'm sorry you are confused. What part confuses you exactly?
myininaya
  • myininaya
Let me turn my paper the other way and try rewriting this it may help for it to be on one line
myininaya
  • myininaya
I see nothing wrong with my way honestly
myininaya
  • myininaya
But we can try to go the other way if you really want One sec...
myininaya
  • myininaya
\[8 \sin^4(\theta)-2 \sin^2(\theta)-2 \] ----------------------------------------------- First are here are some identities we may need: \[\cos(2a)=\cos^2(a)-\sin^2(a)=1-2\sin^2(a)\] \[\sin^2(a)=\frac{1}{2}(1-\cos(2a))\] \[\sin^4(a)=\frac{1}{4}(1-2 \cos(2a)+\cos^2(2a))\] ------------------------------------------------ So using some of the identities above we get: \[8(\frac{1}{4}-\frac{1}{2}\cos(2 \theta)+\frac{1}{4} \cos^2(2 \theta))-2 (\frac{1}{2}(1-\cos(2 \theta))-2 \] \[2-4 \cos(2 \theta)+2 \cos^2(2 \theta)-1+\cos(2 \theta)-3\] \[=2 \cos^2 (2 \theta)-3 \cos(2 \theta)-1 \] Or \[2 \cos^2( 2 \theta)-1 -3 \cos(2 \theta)\] \[\text{ ( Note :So we need to show those first two terms } =\cos(4 \theta) \text{)}\]
myininaya
  • myininaya
\[\text{ Well } \cos(2a)=2 \cos^2(a)-1 => \text{ if } a= 2 \theta \text{ then we have } \cos( 4 \theta)=2 \cos^2(2 \theta)-1\]
myininaya
  • myininaya
lol I prefer you to call me woman :p
myininaya
  • myininaya
Anyways you got it? No questions?
myininaya
  • myininaya
lol Just be on team myininaya
myininaya
  • myininaya
No don't worry about it though. No worries. I like to help people. :)
myininaya
  • myininaya
It is in a different dimension. You have to find the time warp that gets to that team.
myininaya
  • myininaya
lol I just say crazy things.
myininaya
  • myininaya
Did you delete something?
myininaya
  • myininaya
hey omar1131 i have bad news

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