Help with de Moivre's problem!

Given that sin6x/sinx = a*cos^5(x)+b*cos^3(x)+c*cos(x), where sinx cant be equal to 0, use de Moivre's theorem with n = 6 to find the values of the constants a, b and c.

Hence deduce the value of lim (x->0) sin6x/sinx

Thank you guys!

(1) By DeMovire's Theorem:
[cos(x) + i*sin(x)]^6 = cos(6x) + i*sin(6x).

On the other hand, expanding the above using the Binomial Theorem gives:
[cos(x) + i*sin(x)]^6
= cos^6(x) + 6cos^5(x)[i*sin(x)] + 15cos^4(x)[i*sin(x)]^2 + 20cos^3(x)[i*sin(x)]^3 + 15cos^2(x)[i*sin(x)]^4 + 6cos(x)[i*sin(x)]^5 + sin^6(x)
= cos^6(x) + 6i*cos^5(x)sin(x) - 15cos^4(x)sin^2(x) - 20i*cos^3(x)sin^3(x) + 15cos^2(x)sin^4(x) + 6i*cos(x)sin^5(x) - sin^6(x)
= [cos^6(x) - 15cos^4(x)sin^2(x) + 15cos^2(x)sin^4(x) - sin^6(x)] + [6cos^5(x)sin(x) - 20cos^3(x)sin^3(x) + 6cos(x)sin^5(x)]i.

Since both of these expressions equal [cos(x) + i*sin(x)]^6, they must be equal. Since two complex numbers are the same, their real and imaginary parts must equal. By comparing the imaginary parts, we see that:
sin(6x) = 6cos^5(x)sin(x) - 20cos^3(x)sin^3(x) + 6cos(x)sin^5(x)
==> sin(6x)/sin(x) = 6cos^5(x) - 20cos^3(x)sin^2(x) + 6cos(x)sin^4(x).

Using sin^2(x) = 1 - cos^2(x), we can represent this in terms of cosine as follows:
sin(6x)/sin(x) = 6cos^5(x) - 20cos^3(x)sin^2(x) + 6cos(x)sin^4(x)
= 6cos^5(x) - 20cos^3(x)[1 - cos^2(x)] + 6cos(x)[1 - cos^2(x)]^2
= 32cos^5(x) - 32cos^3(x) + 6cos(x).

Thus, a = 32, b = -32, and c = 6.

(2) Using the result from part (1):
lim (x-->0) sin(6x)/sin(x) = lim (x-->0) [32cos^5(x) - 32cos^3(x) + 6cos(x)]
= 32 - 32 + 6, by evaluating at x = 0
= 6.

I hope this helps!
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Fixed.