Express $8\sin\theta \cos\theta - 6 \sin^2 \theta$ in the form $R \sin(2\theta + \alpha) + k$

Edit: I am sorry, I thought it was a somewhat interesting question. I shall let you know of the progress I have made. I first tried to rewrite it in the form $R \sin (\theta -\alpha)$ by rewriting it as $10 (\frac{8}{10} \sin \theta \cos \theta - \frac {6}{10} \sin ^2 \theta)$. For this to have a similar form to the sine subtraction formula, I noted that therefore $\frac{6}{10}= \cot \theta$, however, I am not sure of this assertion and do not know how to proceed. I tried the standard identities already, as someone indicated below.