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# C3 - Confusing question watch

1. http://www.edexcel.org.uk/VirtualCon...r_MkScheme.pdf

c3, question 7iii)

I got this far

dy/dx = [e^-x].[2cos2x - sin2x]

How do you get R and alpha?
2. let [e^-x].[2cos2x - sin2x] = R[e^-x]cos(2x + a)

now expand the rhs and compare coefficients
3. let [e^-x].[2cos2x - sin2x] = R[e^-x]cos(2x + a)
RHS = R[e^-x]cos2x + R[e^-x]a

what do i do now, i'm not sure how i can get it in a good form to compare coefficients.
4. divide both sides by e^-1, and what does the eqn look like?

edit: you haven't expanded the cos(2x+a) properly!!
5. (Original post by nas7232)
let [e^-x].[2cos2x - sin2x] = R[e^-x]cos(2x + a)
RHS = R[e^-x]cos2x + R[e^-x]a

what do i do now, i'm not sure how i can get it in a good form to compare coefficients.
No no no!

The bold line is an illegal move. Completely illegal.

More like this:

[ a = alpha ]

y = e-xsin2x

dy/dx = [e-x].[2cos2x - sin2x] = 2e-xcos2x - e-xsin2x

Re-xcos(2x + a) = Re-xcos2xcosa - Re-xsin2xsina

Compare coefficients: Rcosa = 2 and RsinA = 1

So: R2 = 5 ==> R = 2.24 (3sf)

and tana = 1/2 ==> a = 0.464 rad (3sf)

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