# *HELP* Rearrange Equation Using Trig Identities

Hi Helpers , the following is my problem:
I have to rearrange Equation(1) to make Equation(2) using trigonometric identities

(1): E(θ)=2Eo + (ΔR)^2 (c(x)cos^2(θ-θo) + c(y)sin^2(θ-θo))

(2): E(θ)=2Eo + ½(ΔR)^2 ((c(x)-c(y))cos(2(θ-θo)) + (c(x) + c(y))

I was able to get everything withing the bracket in Equation(2) but i cannot figure out how it changes from (ΔR)^2 to ½(ΔR)^2 ? (btw: when it says Eo I mean E sub 0 and θ sub 0, when I say c(x) or c(y) I really mean c sub x, also by ^2 i mean squared of course))
I really hope someone can help me out here!!!!:shy:

Also I used the Identity:
cos^2(θ-θo) --> ½(1+cos2(θ-θo))
and
sin^2(θ-θo) --> ½(1-cos2(θ-θo))

for '(θ-θo)' I just use something easier like A for example, anyways like i said i cannot understand the ½(ΔR)^2 in the 2nd equation!

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tiny-tim
Homework Helper
Welcome to PF!

Hi AnnaSuxCalc! Welcome to PF! (use the X2 and X2 tags just above the reply box )
I have to rearrange Equation(1) to make Equation(2) using trigonometric identities

(1): E(θ)=2Eo + (ΔR)^2 (c(x)cos^2(θ-θo) + c(y)sin^2(θ-θo))

(2): E(θ)=2Eo + ½(ΔR)^2 ((c(x)-c(y))cos(2(θ-θo)) + (c(x) + c(y))

cos^2(θ-θo) --> ½(1+cos2(θ-θo))
and
sin^2(θ-θo) --> ½(1-cos2(θ-θo))
Yes, you have exactly the correct trigonometric identities …

the 1/2 in both of them should still be in equation (2) …

i dunno why it disappears for you  Hey Timmy, thanks I noticed the sub and superscript after I posted the message, I tried to edit but it didn't work haha
I'm starting to think that maybe my Prof posted the question wrong :grumpy:

So this is my FINAL Solution: here x = (θ-θ0) --> to make it a bit simpler for me
Starting with Equation(1)
E(θ)=2E0 + (∆R)2(cx(½(1+cos2x)) + cy(½(1-cos2x)))
=2E0 + (∆R)2(cx(½+½cos2x) + cy(½-½cos2x))
=2E0 + (∆R)2(½cx + cx(½cos2x) + ½cy - cy(½cos2x))
=2E0 +½(∆R)2cx + (∆R)2(cx(½cos2x)) + ½(∆R)2cy - (∆R)2(cy(½cos2x)) --> factor out the ½(∆R)2
=2E0 + ½(∆R)2((cx - cy)cos(2x) + (cx+cy))

does this look like the correct way???:uhh:

tiny-tim
Homework Helper
So this is my FINAL Solution: here x = (θ-θ0) --> to make it a bit simpler for me
Starting with Equation(1)
E(θ)=2E0 + (∆R)2(cx(½(1+cos2x)) + cy(½(1-cos2x)))
=2E0 + (∆R)2(cx(½+½cos2x) + cy(½-½cos2x))
=2E0 + (∆R)2(½cx + cx(½cos2x) + ½cy - cy(½cos2x))
=2E0 +½(∆R)2cx + (∆R)2(cx(½cos2x)) + ½(∆R)2cy - (∆R)2(cy(½cos2x)) --> factor out the ½(∆R)2
=2E0 + ½(∆R)2((cx - cy)cos(2x) + (cx+cy))

does this look like the correct way???:uhh:
Yes, that's it except you could cut out a few lines … just do:

E(θ)=2E0 + (∆R)2(cx(½(1+cos2x)) + cy(½(1-cos2x)))
=2E0 + ½(∆R)2((cx - cy)cos(2x) + (cx+cy)) 