Q. Use the sum identity and double identity for sine to find $\sin 3x$. $$ \begin{align} \sin 3x &= \sin (2x + x)\\ &=\sin 2x \cos x + \cos 2x \sin x \\ &= (2\sin x \cos x) \cos x + (1 - 2\sin^2 x) \sin x\\ &=2\sin x \cos^2 x + \sin x - 2\sin^3 x \\ &=2\sin x (1 - \sin^2 x) + \sin x - 2\sin^3 x\\ &= " 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x \\ &=3\sin x - 4\sin^3 x" \end{align} $$ The part of the problem I'm having trouble with is in quotations.
My question:
is how does $\sin x - 2\sin^3 x = 4\sin^3 x$?
I see it as this $\sin x - 2\sin^3 x = 2\sin^{3-1} x = 4\sin x$.
$\endgroup$3 Answers
$\begingroup$Observe it's rather: $$ \begin{align} \color{blue}{2\sin x} - 2\sin^3 x + \color{blue}{\sin x} - 2\sin^3 x &=\color{blue}{2\sin x} + \color{blue}{\sin x}- 2\sin^3 x - 2\sin^3 x \\\\&=3\sin x - 4\sin^3 x. \end{align} $$
$\endgroup$ 1 $\begingroup$With a hopefully obvious notation,
$$s_3=s_2c+c_2s=2sc^2+(1-2s^2)s=2s(1-s^2)+(1-2s^2)s=3s-4s^3.$$
$\endgroup$ $\begingroup$You seem to have forgotten the first two terms inside the quotation marks.
$$2 \sin x - 2 \sin^3 x + \sin x - 2 \sin^3 x = 3 \sin x - 4 \sin^3 x$$
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