# Rewrite sin(x+7pi/6) in terms of sin(x) and cos(x)

Rewrite $\mathrm{sin}\left(x+\frac{7\pi }{6}\right)$ in terms of sin(x) and cos(x)
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Solution:
It is asked to write $\mathrm{sin}\left(x+\frac{7\mathrm{\Pi }}{6}\right)$ in terms of sin(x) and cos(x).
Using the identity,
$\mathrm{sin}\left(a+b\right)=\mathrm{sin}\left(a\right)\mathrm{cos}\left(b\right)+\mathrm{cos}\left(a\right)\mathrm{sin}\left(b\right),$
Where
$a=x\phantom{\rule{0ex}{0ex}}b=\frac{7\mathrm{\Pi }}{6}\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(x+\frac{7\mathrm{\Pi }}{6}\right)=\mathrm{sin}\left(x\right)\mathrm{cos}\left(\frac{7\mathrm{\Pi }}{6}\right)+\mathrm{cos}\left(x\right)\mathrm{sin}\left(\frac{7\mathrm{\Pi }}{6}\right)\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(\frac{7\mathrm{\Pi }}{6}\right)=\mathrm{sin}\left(\mathrm{\Pi }+\frac{\mathrm{\Pi }}{6}\right)$
Using identity $\mathrm{sin}\left(\mathrm{\Pi }+\mathrm{\Theta }\right)=-\mathrm{sin}\left(\mathrm{\Theta }\right)\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(\frac{7\mathrm{\Pi }}{6}\right)=-\mathrm{sin}\left(\frac{\mathrm{\Pi }}{6}\right)\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(\frac{7\mathrm{\Pi }}{6}\right)=-\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(\frac{7\mathrm{\Pi }}{6}\right)=\mathrm{cos}\left(\mathrm{\Pi }+\frac{\mathrm{\Pi }}{6}\right)$
Using identity $\mathrm{cos}\left(\mathrm{\Pi }+\mathrm{\Theta }\right)=-\mathrm{cos}\left(\mathrm{\Theta }\right)\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(\frac{7\mathrm{\Pi }}{6}\right)=-\mathrm{cos}\left(\frac{\mathrm{\Pi }}{6}\right)\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(\frac{7\mathrm{\Pi }}{6}\right)=-\frac{\sqrt{3}}{2}$
Our expression becomes
$\mathrm{sin}\left(x+\frac{7\mathrm{\Pi }}{6}\right)=\mathrm{sin}\left(x\right)\left(-\frac{\sqrt{3}}{2}\right)+\mathrm{cos}\left(x\right)\left(-\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(x+\frac{7\mathrm{\Pi }}{6}\right)=-\frac{\sqrt{3}}{2}\mathrm{sin}\left(x\right)-\frac{1}{2}\mathrm{cos}\left(x\right)\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(x+\frac{7\mathrm{\Pi }}{6}\right)=-\frac{1}{2}\left[\sqrt{3}\mathrm{sin}\left(x\right)+\mathrm{cos}\left(x\right)\right]$