Question

proving de moivre’s theorem
proof of de moivres theorem
cos(θ) + i sin(θ)^{k + 1}
a = cos(θ) + i sin(θ)^k · cos(θ) + i sin(θ)^1
b = cos(kθ) + i sin(kθ) · cos(θ) + i sin(θ)
c = cos(kθ)cos(θ) − sin(kθ)sin(θ) + i sin(kθ)cos(θ) + cos(kθ)sin(θ)
d = cos(kθ + θ) + i sin(kθ + θ)
e = cos(k + 1)θ + i sin(k + 1)θ
what is the correct justification for the indicated steps?
step c:
step d:
step e:

Explanation:

• Step C: Expand the product of two complex numbers in rectangular form by distributing each term, using the fact that $i^2=-1$ to rewrite $i\sin(k\theta) \cdot i\sin(\theta)$ as $-\sin(k\theta)\sin(\theta)$.
• Step D: Apply the cosine and sine angle addition formulas: $\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$ and $\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)$ where $A=k\theta$ and $B=\theta$.
• Step E: Simplify the angle expression by factoring out the common term, combining $k\theta + \theta$ into $(k+1)\theta$.

Answer:

Step C: Expand the product of complex numbers (using $i^2=-1$)
Step D: Apply angle addition identities for sine/cosine
Step E: Factor and simplify the angle expression