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Math Help - Induction Trig Proof

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    Induction Trig Proof

    Prove by induction:

    For every integer n > 1

    [cos(x) + i * sin(x)]^n = cos(nx) + i * sin(nx)
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  2. #2
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    Re: Induction Trig Proof

    Quote Originally Posted by thamathkid1729 View Post
    Prove by induction:

    For every integer n > 1

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

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


    \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 \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 \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 \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.
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