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Further trigonometry FP3

Hi, I'm stuck on the first question (1a) of excersize 7C. The question is:
Show that:
1+4cos2x+6cos4x+4cos6x+cos8x=16cos4xcos4x 1+4cos2x + 6cos4x + 4cos6x + cos8x = 16cos4x cos^4 x
The examples on the pages before are all about sums of infinite geometric series which I understood, but I'm not really sure how this question relates to the example in any way.

Help would be appreciated :smile:
Reply 1
Avatar for nuodai
nuodai
17
Well you can write the LHS as Re(1+4e2ix+6e4ix+4e6ix+e8ix)\text{Re}(1+4e^{2ix} + 6e^{4ix} + 4e^{6ix} + e^{8ix}). As a hint, think Pascal's triangle... how can you factorise this expression?
Reply 2
Avatar for jamie092
jamie092
OP
11
Ah ye, I'm thinking its something to the power 4, because of those binomial coefficients. Is it (1+e^2ix)^4?
Reply 3
Avatar for nuodai
nuodai
17
Original post by jamie092
Ah ye, I'm thinking its something to the power 4, because of those binomial coefficients. Is it (1+e^2ix)^4?


Yup, so you have to find the real part of (1+e2ix)4(1+e^{2ix})^4.
Reply 4
Avatar for jamie092
jamie092
OP
11
Original post by nuodai
Yup, so you have to find the real part of (1+e2ix)4(1+e^{2ix})^4.


Surely thats just (1+cos2x)4(1+cos2x)^4?
Reply 5
Avatar for nuodai
nuodai
17
Original post by jamie092
Surely thats just (1+cos2x)4(1+cos2x)^4?


Nope, in general it's not true that Re(zw)=Re(z)Re(w)\text{Re}(zw) = \text{Re}(z) \text{Re}(w) so you can't just take the "Re" inside the brackets.
Reply 6
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jamie092
OP
11
Original post by nuodai
Nope, in general it's not true that Re(zw)=Re(z)Re(w)\text{Re}(zw) = \text{Re}(z) \text{Re}(w) so you can't just take the "Re" inside the brackets.


Oh ye ofcourse ;<
Well now I'm stumped =(
Reply 7
Avatar for nuodai
nuodai
17
Original post by jamie092
Oh ye ofcourse ;<
Well now I'm stumped =(


Well if z=x+iyz=x+iy then z4=x4+4ix3y−6x2y2−4ixy3+y4z^4 = x^4 + 4ix^3y - 6x^2y^2 - 4ixy^3 + y^4. You'll notice that the terms with even powers of y are real, and the terms with odd powers of y are imaginary (because of how the powers of i work). So set z=1+e2iθ=(1+cos⁡2θ)+isin⁡2θ≡x+iyz=1+e^{2i\theta} = (1+\cos 2\theta) + i\sin 2\theta \equiv x+iy and expand as above. A bit of tedious algebraic manipulation should give you the right answer.
Reply 8
Avatar for jamie092
jamie092
OP
11
Original post by nuodai
Well if z=x+iyz=x+iy then z4=x4+4ix3y−6x2y2−4ixy3+y4z^4 = x^4 + 4ix^3y - 6x^2y^2 - 4ixy^3 + y^4. You'll notice that the terms with even powers of y are real, and the terms with odd powers of y are imaginary (because of how the powers of i work). So set z=1+e2iθ=(1+cos⁡2θ)+isin⁡2θ≡x+iyz=1+e^{2i\theta} = (1+\cos 2\theta) + i\sin 2\theta \equiv x+iy and expand as above. A bit of tedious algebraic manipulation should give you the right answer.


ok thanks I'll give it a go :smile:

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