# cos20*cos40*cos80

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#1
cos20*cos40*cos80 = 1/8 but I can't work out why (without using a calculator!!) I tried using trig identities with 40 = 60 - 20 and 80 = 60 + 20 and got to cos20*cos40*cos80 = (cos20)^3 - 3/4*cos20 but I have no idea where you can get the eighth from that. I expect there's some way of writing the LHS in terms of the 'easy' cosines - I just can't spot it!!
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15 years ago
#2
edit: Ya no, nevermind, I clearly can't read or something :| sorry!

I'll get back when I write something sensical.
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#3
(Original post by fishpaste)
edit: Ya no, nevermind, I clearly can't read or something :| sorry!

I'll get back when I write something sensical.
ok! I didn't see what you put before the edit - anything interesting?
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15 years ago
#4
(Original post by Bezza)
cos20*cos40*cos80 = 1/8 but I can't work out why (without using a calculator!!) I tried using trig identities with 40 = 60 - 20 and 80 = 60 + 20 and got to cos20*cos40*cos80 = (cos20)^3 - 3/4*cos20 but I have no idea where you can get the eighth from that. I expect there's some way of writing the LHS in terms of the 'easy' cosines - I just can't spot it!!
If you consider the roots of the equation cos(18x) = 1, there are 0,20,40,60,80...340.

Now a bit of clever algebraic manipulaton shows that their product is -1/2^(16) (use de movire and the theorem of roots of polynomials).

Now we have cos(0) = 1 and cos(180) = -1, so we have

c20 c40 c60 c80 c100 c120 c140 c160 c200 ... c340 = 1/2^16

Now using c60 = 1/2, c120 = -1/2, c240 = -1/2, c300 = 1/2.

c20 c40 c80 c100 c140 c160 c200 c220 ... = 1/2^12.

Now using c100 c140 c160 = c200 c220 c260 = - c20 c40 c80 = - c340 c320 c280 you get (c20 c40 c80)^4 = 1/2^12. Thus c20 c40 c80 = 1/2^3 = 1/8 as desired
0
15 years ago
#5
i used cosAcos2Acos4A, then expanded with demoivres to get
0.25(cos7A+cos5A+cos3A+cosA) then simplified it down to 1/8 * (sin8A/sinA)
then plug in A=20 so u get 1/8 *(sin160/sin20) which is 1/8*1 using identities
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15 years ago
#6
(Original post by s-man)
i used cosAcos2Acos4A, then expanded with demoivres to get
0.25(cos7A+cos5A+cos3A+cosA) then simplified it down to 1/8 * (sin8A/sinA)
then plug in A=20 so u get 1/8 *(sin160/sin20) which is 1/8*1 using identities
What a nice way, you can go straight from cos(x) cos(2x) cos (4x) to 1/8 (sin(8x)/sin(x) which trivialises the problem, no need for de movire, i'll have to remember that
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#7
ok, thanks you two! I can't imagine why I didn't spot beauford's way 0
15 years ago
#8
(Original post by beauford)
What a nice way, you can go straight from cos(x) cos(2x) cos (4x) to 1/8 (sin(8x)/sin(x) which trivialises the problem, no need for de movire, i'll have to remember that
In fact that method leads to a neat generalisation, that

cosx cos2x ... cos(2^nx) = sin(2^(n+1)x)/2^(n+1).sinx
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