Expand Using De Moivre's Theorem tan(x)

Problem

tan(n*x)

Solution

1. Relate the tangent function to sine and cosine using the identity tan(n*x)=sin(n*x)/cos(n*x)

2. Apply De Moivre's Theorem, which states (cos(x)+i*sin(x))n=cos(n*x)+i*sin(n*x)

3. Expand the left side of the equation (cos(x)+i*sin(x))n using the Binomial Theorem.

(cos(x)+i*sin(x))n=(∑_k=0^n)((n/k))*cos(x)(n−k)*(i*sin(x))k)

4. Separate the real and imaginary parts of the binomial expansion to find expressions for cos(n*x) and sin(n*x)

cos(n*x)=(∑_k* even^)((n/k))*(−1)(k/2)*cos(x)(n−k)*sink(x))

sin(n*x)=(∑_k* odd^)((n/k))*(−1)((k−1)/2)*cos(x)(n−k)*sink(x))

5. Divide the expression for sin(n*x) by the expression for cos(n*x) and divide both numerator and denominator by cosn(x) to express the result in terms of tan(x)

tan(n*x)=(n/1)tan(x)−(n/3)tan3(x)+(n/5)tan5(x)−…)))/(1−(n/2)tan2(x)+(n/4)tan4(x)−…)))

Final Answer

tan(n*x)=((∑_k=1,3,5,…^)(−1)*(n/k)tank(x)))/((∑_k=0,2,4,…^)(−1)*(n/k)tank(x)))

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