runddansm88e

2022-04-06

Finding the root mean square of a sum of trig functions
$v\left(t\right)=3-2\mathrm{sin}\left(t\right)+8{\mathrm{sin}}^{2}\left(t\right)$

regulerenes4w

Using
$v\left(t\right)=3-2\mathrm{sin}\left(t\right)+8{\mathrm{sin}}^{2}\left(t\right)$
and squaring (just brute force), you have (just set $x=\mathrm{sin}\left(t\right)$ and use binomial expansion)
${v}^{2}\left(t\right)=9-12\mathrm{sin}\left(t\right)+52{\mathrm{sin}}^{2}\left(t\right)-32{\mathrm{sin}}^{3}\left(t\right)+64{\mathrm{sin}}^{4}\left(t\right)$
Now, you can use power reduction formulae which give
${\mathrm{sin}}^{2}\left(x\right)=\frac{1-\mathrm{cos}\left(2x\right)}{2}$
${\mathrm{sin}}^{3}\left(x\right)=\frac{3\mathrm{sin}\left(x\right)-\mathrm{sin}\left(3x\right)}{4}$
${\mathrm{sin}}^{4}\left(x\right)=\frac{3-4\mathrm{cos}\left(2x\right)+\mathrm{cos}\left(4x\right)}{8}$
Replacing, you the have
${v}^{2}\left(t\right)=59-36\mathrm{sin}\left(t\right)-58\mathrm{cos}\left(2t\right)+8\mathrm{sin}\left(3t\right)+8\mathrm{cos}\left(4t\right)$

Do you have a similar question?