Expand Using De Moivre's Theorem cos(2x)

Problem

cos(2*x)

Solution

1. State De Moivre's Theorem for an integer n which relates complex numbers in polar form to powers: (cos(x)+i*sin(x))n=cos(n*x)+i*sin(n*x)

2. Set the exponent to n=2 to match the argument of the given function: (cos(x)+i*sin(x))2=cos(2*x)+i*sin(2*x)

3. Expand the left side using the binomial expansion (a+b)2=a2+2*a*b+b2 cos2(x)+2*i*sin(x)*cos(x)+i2*sin2(x)

4. Simplify the imaginary unit using the property i2=−1 cos2(x)+2*i*sin(x)*cos(x)−sin2(x)

5. Group the real and imaginary parts of the expanded expression: (cos2(x)−sin2(x))+i*(2*sin(x)*cos(x))

6. Equate the real parts from both sides of the equation cos(2*x)+i*sin(2*x)=(cos2(x)−sin2(x))+i*(2*sin(x)*cos(x))

Final Answer

cos(2*x)=cos2(x)−sin2(x)

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