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=a^2+2*a*b+b^2 cos^2(x)+2*i*sin(x)*cos(x)+i^2*sin^2(x)

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

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

Final Answer

cos(2*x)=cos^2(x)−sin^2(x)

Want more problems? Check here!