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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 $\displaystyle \sin\left(2x-\frac{7\pi}{6}\right)$.

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$$\sin(2x-\frac{7\pi}{6})=\sin2x\cos(7\pi/6)-\cos2x\sin(7\pi/6)=$$ $$=2\sin x\cos x\cos(7\pi/6)-(\cos^2x-\sin^2x)\sin(7\pi/6)=$$ $$=2(2/\sqrt 13)(-3/\sqrt13)\cos(7\pi/6)-(4/13-9/13)\sin(7\pi/6)=$$ $$=-12/13\cos(7\pi/6)+5/13\sin(7\pi/6)=$$ $$=-12/13(-\sqrt 3/2)+5/13(-1/2)=$$ $$=12\sqrt 3/26-5/26=(12\sqrt 3-5)/26$$

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What is the ending of that? –  user1838781 Mar 4 '13 at 17:27
    
Answer is edited now –  Adi Dani Mar 4 '13 at 17:56
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You can use the sum and difference formulas to split up $\sin\left(2x-\frac{7\pi}{6}\right)$. You will then get expressions involving $\sin(x)$, $\cos(x)$, $\sin\frac{7\pi}{6}$, and $\cos \frac{7\pi}{6}.$

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