Question

# What is the formula of ${\mathrm{sin}}^{3}\left(x\right)$β?

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Solution

## Determine the formula for ${\mathrm{sin}}^{3}\left(x\right)$.Use the trigonometric identity, $\mathbit{s}\mathbit{i}\mathbit{n}\left(3x\right)\mathbf{=}\mathbf{3}\mathbit{s}\mathbit{i}\mathbit{n}\left(x\right)\mathbf{-}\mathbf{4}\mathbit{s}\mathbit{i}{\mathbit{n}}^{\mathbf{3}}\left(x\right)$ and simplify as follow:$\begin{array}{rcl}\mathrm{sin}\left(3x\right)-3\mathrm{sin}\left(x\right)& =& -4{\mathrm{sin}}^{3}\left(x\right)\\ & \beta & {\mathrm{sin}}^{3}\left(x\right)=-\frac{1}{4}\left[\mathrm{sin}\left(3x\right)-3\mathrm{sin}\left(x\right)\right]\\ & \beta & 4{\mathrm{sin}}^{3}\left(x\right)=-\frac{1}{4}\mathrm{sin}\left(3x\right)+\frac{3}{4}\mathrm{sin}\left(x\right)\\ & \beta & {\mathrm{sin}}^{3}\left(x\right)=\frac{3}{4}\mathrm{sin}\left(x\right)-\frac{1}{4}\mathrm{sin}\left(3x\right)\end{array}$Hence, the formula is $\begin{array}{rcl}{\mathrm{sin}}^{3}\left(x\right)& =& \frac{3}{4}\mathrm{sin}\left(x\right)-\frac{1}{4}\mathrm{sin}\left(3x\right)\end{array}$.

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