*HELP* Rearrange Equation Using Trig Identities

[AnnaSuxCalc]

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!

 

[tiny-tim]

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 don't know why it disappears for you

 

[AnnaSuxCalc]

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:

 

[AnnaSuxCalc]

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(c_x(½(1+cos2x)) + c_y(½(1-cos2x)))
=2E₀ + (∆R)²(c_x(½+½cos2x) + c_y(½-½cos2x))
=2E₀ + (∆R)²(½c_x + c_x(½cos2x) + ½cy - c_y(½cos2x))
=2E₀ +½(∆R)²c_x + (∆R)²(c_x(½cos2x)) + ½(∆R)²c_y - (∆R)²(c_y(½cos2x)) --> factor out the ½(∆R)²
=2E₀ + ½(∆R)²((c_x - c_y)cos(2x) + (c_x+c_y))

does this look like the correct way?:uhh:

 

[tiny-tim]

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(c_x(½(1+cos2x)) + c_y(½(1-cos2x)))
=2E₀ + (∆R)²(c_x(½+½cos2x) + c_y(½-½cos2x))
=2E₀ + (∆R)²(½c_x + c_x(½cos2x) + ½cy - c_y(½cos2x))
=2E₀ +½(∆R)²c_x + (∆R)²(c_x(½cos2x)) + ½(∆R)²c_y - (∆R)²(c_y(½cos2x)) --> factor out the ½(∆R)²
=2E₀ + ½(∆R)²((c_x - c_y)cos(2x) + (c_x+c_y))

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(c_x(½(1+cos2x)) + c_y(½(1-cos2x)))
=2E₀ + ½(∆R)²((c_x - c_y)cos(2x) + (c_x+c_y))