<s>[INST] values of basic trig function, given values in terms of trig functions Get the function value if $\displaystyle \sin x = \frac{2}{\sqrt{13}}$ and $\displaystyle \cos x = -\frac{3}{\sqrt{13}}$ and the function value is going to be calculated from $\displaystyle \sin\left(2x-\frac{7\pi}{6}\right)$.
 [/INST] $$\sin(2x-\frac{7\pi}{6})=\sin2x\cos(7\pi/6)-\cos2x\sin(7\pi/6)=$$
$$=2\sin x\cos x\cos(7\pi/6)-(\cos^2x-\sin^2x)\sin(7\pi/6)=$$
$$=2(2/\sqrt 13)(-3/\sqrt13)\cos(7\pi/6)-(4/13-9/13)\sin(7\pi/6)=$$
$$=-12/13\cos(7\pi/6)+5/13\sin(7\pi/6)=$$
$$=-12/13(-\sqrt 3/2)+5/13(-1/2)=$$
$$=12\sqrt 3/26-5/26=(12\sqrt 3-5)/26$$
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