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cos pie/a.cos2pie/a.cos3pie/a....coskpie/a=1/2^k.Prove this if a=2k+1

cos pie/a.cos2pie/a.cos3pie/a....coskpie/a=1/2^k.Prove this if a=2k+1


2 Answers

Prajwal Kavad
603 Points
6 years ago
plz post the question properly or plz upload a picture to get a clear form of d question
185 Points
6 years ago
Products of cosine series is done by n-th root of unity equation.
xn-1=0 has roots X=1,ei2π/n, ei4π/n ,ei6π/n ,ei8π/n ,ei10π/n, ……… ei2(n-1)π/n.
So, xn-1 = (x-1)*(x - ei2π/n)* (x - ei4π/n)* (x - ei6π/n)*……(x - ei2(n-1)π/n),
Now e2π-x = e.e-x= e-x, …… eq (1)
Here I have assumed n is even, u can also derive for odd n.
So, xn-1 = (x-1) *(x - ei2π/n)* (x - ei4π/n)* (x - ei6π/n)……..(x- ei(n-2)π/n).(x- ei(n)π/n) (x- ei(n+2)π/n)…….. (x - ei2(n-1)π/n),
Using (1),
xn-1 = (x-1) *(x - ei2π/n)* (x - ei4π/n)* (x - ei6π/n)……..(x- ei(n-2)π/n).(x+1) (x- e-i(n-2)π/n)…….. (x – e-i2π/n),
Rearranging terms,
xn-1 = (x2-1) *[(x - ei2π/n)* (x - e-i2π/n)]*[(x - ei4π/n)* (x - e-i4π/n)]*[(x - ei6π/n)*(x - ei6π/n)]………..[(x- ei(n-2)π/n).(x- e-i(n-2)π/n)],
Multiplying adjascent brackets,
xn-1 = (x2-1) *[(x2 +1 – 2*cos(2π/n)x)]*[(x2 +1 – 2*cos(4π/n)x)]*[(x2 +1 – 2*cos(6π/n)x)]…… [(x2 +1 – 2*cos((n-2)π/n)x)],
This is the general equation one gets for n-th root of unity problem,
To get rid of x2+1, x is put equal to i,
So, (in – 1) = (-1-1) * (-2i)(n-2)/2* (product of cosines) ,
Since n is even, put n=2a,
(i2a - 1) = = (-1-1) * (-2i)(a-1) * (cos(π/a) * cos(2π/a) * cos(3π/a)…… cos((a-1)π/a)),
As a is odd (as you have mentioned that a=2k+1), i2a = -1, so -2 gets cancelled and (-1)a-1 = 1
(cos(π/a) * cos(2π/a) * cos(3π/a)…… cos((a-1)π/a)) = 2-(a-1)* (i)(-a+1)………eq(3)
((i)(-a+1) = +1 if a-1 is of the form 4Z) and ((i)(-a+1) = -1 if a-1 is of the form 4Z+2)………eq(2)
Now in your question you are doing product in first half ie. 0 to 90 deg. To get that equation from above equation, put cos(π –x) = - cos x.
(cos(π/a) * cos(2π/a) * cos(3π/a)…… cos((a-1)π/a)) = (equation in your problem)2 * (-1)(a-1)/2,……eq(4)
((-1)(a-1)/2=1 if (a-1)/2 is even that is a-1 is of form 4Z) and ((-1)(a-1)/2=-1 if (a-1)/2 is even that is a-1 is of form 4Z+2). Hence (-ve) term, if it’s there will get canceled out with eq (2).
Putting eq(4) in eq(3),
equation in your problem = 2-(a-1)/2

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