Aubrey Hendricks

2022-01-26

Precalc Trig Identity, verify: $1+\mathrm{cos}\left(x\right)+\mathrm{cos}\left(2x\right)=\frac{12}{+}\frac{\mathrm{sin}\left(5\frac{x}{2}\right)}{2\mathrm{sin}\left(\frac{x}{2}\right)}$

Souticexi

For $\mathrm{sin}\frac{x}{2}\ne 0$ we obtain:
$1+\mathrm{cos}\left\{x\right\}+\mathrm{cos}2x=\frac{2\mathrm{sin}\frac{x}{2}+2\mathrm{sin}\frac{x}{2}\mathrm{cos}\left\{x\right\}+2\mathrm{sin}\frac{x}{2}\mathrm{cos}2x}{2\mathrm{sin}\frac{x}{2}}=$
$=\frac{2\mathrm{sin}\frac{x}{2}+\mathrm{sin}\frac{3x}{2}-\mathrm{sin}\frac{x}{2}+\mathrm{sin}\frac{5x}{2}-\mathrm{sin}\frac{3x}{2}}{2\mathrm{sin}\frac{x}{2}}=\frac{1}{2}+\frac{\mathrm{sin}\frac{5x}{2}}{2\mathrm{sin}\frac{x}{2}}$
I used the following formula.
$\mathrm{sin}\alpha \mathrm{cos}\beta =\frac{1}{2}\left(\mathrm{sin}\left(\alpha +\beta \right)+\mathrm{sin}\left(\alpha -\beta \right)\right)$
For example,
$2\mathrm{sin}\frac{x}{2}\mathrm{cos}\left\{x\right\}=2\cdot \frac{1}{2}\left(\mathrm{sin}\left(\frac{x}{2}+x\right)+\mathrm{sin}\left(\frac{x}{2}-x\right)\right)=\mathrm{sin}\frac{3x}{2}-\mathrm{sin}\frac{x}{2}$

Do you have a similar question?