1. ## Induction Trig Proof

Prove by induction:

For every integer n > 1

[cos(x) + i * sin(x)]^n = cos(nx) + i * sin(nx)

2. ## Re: Induction Trig Proof

Originally Posted by thamathkid1729
Prove by induction:

For every integer n > 1

[cos(x) + i * sin(x)]^n = cos(nx) + i * sin(nx)
Base Step: n = 1

\displaystyle \displaystyle \begin{align*} LHS &= \left(\cos{x} + i\sin{x}\right)^1 \\ &= \cos{x} + i\sin{x} \end{align*}

\displaystyle \displaystyle \begin{align*} RHS &= \cos{\left(1 \cdot x \right)} + i\sin{\left( 1 \cdot x \right)} \\ &= \cos{x} + i\sin{x} \\ &= LHS \end{align*}

Inductive step, assume the statement is true for n = k, i.e. \displaystyle \displaystyle \begin{align*} \left(\cos{x} + i\sin{x}\right)^k &= \cos{\left(k\,x\right)} + i\sin{\left(k\,x\right)} \end{align*}, try to show the statement is true for n = k + 1, i.e. \displaystyle \displaystyle \begin{align*} \left(\cos{x} + i\sin{x}\right)^{k + 1} = \cos{\left[\left(k + 1\right)x\right]} + i\sin{\left[\left(k + 1\right)x\right]} \end{align*}

\displaystyle \displaystyle \begin{align*} LHS &= \left(\cos{x} + i\sin{x}\right)^{k + 1} \\ &= \left(\cos{x} + i\sin{x}\right)^k\left(\cos{x} + i\sin{x}\right) \\ &= \left[\cos{\left(k\,x\right)} + i\sin{\left(k\,x\right)}\right]\left(\cos{x} + i\sin{x}\right) \\ &= \cos{\left(k\,x\right)}\cos{x} + i\sin{x}\cos{\left(k\,x\right)} + i\cos{x}\sin{\left(k\,x\right)} + i^2\sin{\left(k\,x\right)}\sin{x} \\ &= \cos{\left(k\,x\right)}\cos{x} - \sin{\left(k\,x\right)}\sin{x} + i\left[\sin{x}\cos{\left(k\,x\right)} + \cos{x}\sin{\left(k\,x\right)}\right] \\ &= \cos{\left(k\,x + x\right)} + i\sin{\left(k\,x + x\right)} \\ &= \cos{\left[\left(k + 1\right)x\right]} + i\sin{\left[\left(k + 1\right)x\right]} \\ &= RHS \end{align*}

Q.E.D.