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

10 years ago

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*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

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Thanks

Aman Bansal

10 years ago