Question

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 14 years ago

Answer 1

oh geez...I'm not good at this but Sarkar is. Maybe you could msg him?

Answer 2

just go to the envelope on the top left and put Sarkar in the recipients box :) and the link to this problem!

Answer 3

Idk either...probably the person who created this site.

Answer 4

@omar1131 I'm confused a little about your problem

Answer 5

f(Q)=8sin^4(θ)-2sin^2(θ)-2

show that f(θ)=cos4θ-3cos2θ

Answer 6

So Q or theta for the first function?

Answer 7

\[8 \sin^4(\theta)-2 \sin^2(\theta)-2=\cos(4 \theta) -3 \cos(2 \theta)\]
So we want to show this is an identity right?

Answer 8

assuming it is anyways...

Answer 9

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

Answer 10

\[\cos(4 \theta) -3 \cos(2 \theta)= \cos(2 \theta+2 \theta)-3 \cos( \theta+\theta)\]
\[=(\cos(2 \theta) \cos(2 \theta)-\sin(2 \theta) \sin(2 \theta))-3(\cos(\theta) \cos(\theta)-\sin(\theta) \sin(\theta))\]
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?
-------------------------------------------------
\[(\cos^2(\theta)-\sin^2(\theta)(\cos^2(\theta)-\sin^2(\theta))-\sin(2\theta) \sin(2 \theta)-3 \cos^2(\theta)+3 \sin^2(\theta)\]
\[=\cos^4(\theta)-2 \cos^2(\theta) \sin^2(\theta) +\sin^4(\theta)-\sin(2 \theta) \sin(2 \theta)-3 \cos^2(\theta)+3 \sin^2(\theta)\]

Answer 11

So can we use anything else besides the thing given?

Answer 12

Because I can think of another useful sum/difference formula

Answer 13

like can we use
\[\sin(2 \theta)=2 \sin(\theta) \cos(\theta)\]

Answer 14

maybe use \[\cos(2\theta) = 1 - 2\sin^2(\theta)\]

Answer 15

rearranging to get \[ -2\sin^2(\theta) \]
and
\[8\sin^4(\theta)\]
and substitute those in

Answer 16

\[\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) -1 = 1 - 2\sin^2(\theta)\]

Answer 17

substituting in those in my earlier comment leaves you with \[2(1 - \cos(2\theta))^2 + \cos(2\theta) -3\]

Answer 18

try multiplying out the first bracket, collecting like terms and seeing if you can spot the other double angle identity lurking around somewhere...

Answer 19

hope this is helping xD

Answer 20

i can post up a detailed solution for reference if you want

Answer 21

@omar1131 I understand your anger. But OpenStudy is a pg13 website. Please don't use the bad language here.

Answer 22

I would still like to know the answer to my one question
Can we use
\[\sin(2 \theta)=2 \sin(\theta) \cos(\theta)\]

Answer 23

Ok great! :)

Answer 24

\[ =\cos^4(\theta)-2 \cos^2(\theta) \sin^2(\theta) +\sin^4(\theta)-\sin(2 \theta) \sin(2 \theta)-3 \cos^2(\theta)+3 \sin^2(\theta) \]
\[=\cos^4(\theta)-2 \cos^2(\theta) \sin^2(\theta)+\sin^4(\theta)-2 \sin(\theta) \cos(\theta) 2 \sin(\theta) \cos(\theta)-3 \cos^2(\theta)+3 \sin^2(\theta)\]
\[=\cos^4(\theta)-2 \cos^2(\theta) \sin^2(\theta)+\sin^4(\theta)-2 \sin^2(\theta)\cos^2(\theta)-3 \cos^2(\theta)+3 \sin^2(\theta)\]
\[=\cos^4(\theta)-4 \sin^2(\theta) \cos^2(\theta)-3\cos^2(\theta) +3 \sin^2(\theta)+\sin^4(\theta)\]
Now the other side is in terms of sine so lets write this side in terms of sine

Answer 25

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 \[\cos(2\theta)= 1 - 2sin^2(\theta)\]

Answer 26

ill let myininaya finish, thats a good way (although i prefer my solution :p)

Answer 27

cya guys

Answer 28

\[=(\cos^2(\theta))^2-4\sin^2(\theta)(1-\sin^2(\theta))-3(1-\sin^2(\theta))+3 \sin^2(\theta)+\sin^4(\theta)\]
\[=(1-\sin^2(\theta))^2-4\sin^2(\theta)+4 \sin^4(\theta)-3 +3 \sin^2(\theta)+3 \sin^2(\theta)+\sin^4(\theta)\]
\[=1-2\sin^2(\theta)+\sin^4(\theta)-4\sin^2(\theta)+4 \sin^4(\theta)-3 +3 \sin^2(\theta)+3 \sin^2(\theta)+\sin^4(\theta)\]
Like terms :)
\[=6 \sin^4(\theta) +0\sin^2(\theta)-2 =6 \sin^4(\theta)-2 \]
I think I made a mistake somewhere :(

Answer 29

I need to do this by hand lol

Answer 30

lol ok @omar1131 but let me know if you run into any more trouble
This one is crazy long

Answer 31

Crazy long the way I'm approaching it lol

Answer 32

ok i got it using the the identities i was talking about

Answer 33

I think I will just scan this and let you look at it
is that okay?

Answer 34

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

Answer 35

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 :)

Answer 36

It is nice to meet you to and I'm sorry you are confused. What part confuses you exactly?