Alternative form of a Trigonometrics Expression 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. 
 A: $$\begin{align}8\sin(\theta)\cos(\theta)-6\sin^2(\theta)&=4\sin(2\theta)-3(1-\cos(2\theta))\\&=5\left(\frac{4}{5}\sin(2\theta)-\frac{3}{5}\cos(2\theta)\right)-3\end{align}$$
Notice that $\left(\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2=1$, therefore there is an angle $s=\arccos\left(\frac{4}{5}\right)$ such that $\cos(s)=\frac{4}{5}$ and $\sin(s)=\frac{3}{5}$.
Therefore our expression can be written as
$$5\left(\sin(2\theta)\cos(s)-\cos(2\theta)\sin(s)\right)-3=5\sin(2\theta-s)-3$$
A: Write $8\sin\theta\cos\theta-6\sin^2\theta=4\sin2\theta+3\cos2\theta-3$ using $2\sin^2\theta=1-\cos2\theta$. So you want
$$
R\sin(\varphi+\alpha)+k=4\sin\varphi+3\cos\varphi-3
$$
for all $\varphi$. Differentiate both sides to get
$$
R\cos(\varphi+\alpha)=4\cos\varphi-3\sin\varphi
$$
Differentiate again to get
$$
-R\sin(\varphi+\alpha)=-4\sin\varphi-3\cos\varphi
$$
Evaluate both sides of the previous identities for $\varphi=0$:
$$
\begin{cases}
R\cos\alpha=4\\
R\sin\alpha=3
\end{cases}
$$
So $R^2=16+9=25$ and $R=5$. You can choose $\alpha=\arctan(3/4)$
From the first identity with $\varphi=0$, we get
$$
R\sin\alpha+k=0
$$
so $k=-3$.
A: Hopefully you can use these helpful identities to find the answer:
$\sin(2\theta) = 2\sin\theta\cos\theta$
$\sin^2\theta = 1-\cos^2\theta$
