find the value of sin1*sin2*sin3*...*sin179 (note: all angles are in degrees)

Dear Student,

The general product is

     P[n] = Product[Sin[k Pi/n], {k,1,n-1}] = n/2^(n-1),

and the special case is 

     P[180] = 180*2^(-179) 

To prove this, let

     v = Exp[Pi i/n]

be a primitve (2n)th root of unity.  Then we have

     Sin[k Pi/n] = (v^k - v^-k)/(2i) 

                 = (1 - v^(-2k))(v^k/(2i)).

Therefore

     P[n] = Product[1 - v^(-2k), {k,1,n-1}] Product[v^k/(2i), {k,1,n-1}].

But the first product is simply the factorization of

     (z^(2n) - 1)/(z^2 - 1) = z^(2n-2) + z^(2n-4) + ... + 1

with z = 1.  Consequently, this first product has value n.  The
second product is found by summing the exponents, giving

     P[n] = n v^Sum[k, {k,1,n-1}]/(2i)^(n-1)

          = n v^((n-1)n/2) /(2i)^(n-1)

          = n Exp[Pi i (n-1)/2]/(2 Exp[Pi i/2])^(n-1)

          = n/2^(n-1) Exp[Pi i (n-1)/2 - Pi i(n-1)/2]

          = n/2^(n-1),

Best Of luck

Plz Approve the answer...!!!!

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Thanks

Aman Bansal

Askiitian Expert

Approved

The answer is ZERO!!

As the series is in increasing order... therefore, there is sin90° in between... which is 0 and multiplying with 0 is alway 0...

plz approve my answer!! :)