Proving the trig identity $\sin(20^\circ)\cos(65^\circ)-\cos( 20^\circ)\sin(65^\circ)$ Using the trigonometric identities I have to prove that: 
$$
\sin(20^\circ)\cos(65^\circ)-\cos(20^\circ)\sin(65^\circ)=-\frac{1}{\sqrt{2}}
$$
I solved 
$$
\sin(20^\circ)\cos(65^\circ) = \frac{\sin(-45^\circ)+\sin(85^\circ)}{2}$$ 
and 
$$
-\cos(20^\circ)\sin(65^\circ)=\frac{\sin(45^\circ)+\sin(85^\circ)}{2}
$$
How do I continue from here to prove that it is equal $-\dfrac{1}{\sqrt{2}}$
 A: Use fact that
$$
\sin(a - b) = \sin a\cos b - \sin b \cos a
$$
$$\text{Therefore, }\,
-\sin (65 - 20) = -\sin 65 \cos 20 + \sin 20 \cos 65 = -\left(\frac{1}{\sqrt{2}}\right)
$$
$$\text{That is },\,
\sin 45 = \frac{1}{\sqrt{2}}
$$
A: Hint: Use the sine of differences:
$$\sin(a-b) = \sin(a)\cos(b) - \cos(a)\sin(b)$$
Using this you can show that the expression equals $\sin(-45^{\circ})$ by inspection.
A: This answer seems to match the OP's original intent.
Using the product to sum formula:
$$
\sin(a)\cos(b)=\frac{1}{2}(\sin(a+b)+\sin(a-b)),
$$
it follows that
\begin{align*}
\sin(20)\cos(65)&=\frac{1}{2}(\sin(85)+\sin(-45))\\
\sin(65)\cos(20)&=\frac{1}{2}(\sin(85)+\sin(45)).
\end{align*}
Then, 
$$
\sin(20)\cos(65)-\sin(65)\cos(20)=\frac{1}{2}(\sin(85)+\sin(-45))-\frac{1}{2}(\sin(85)+\sin(45)).
$$
Since $\sin$ is an odd function, $\sin(-45)=-\sin(45)$.  Therefore, the $\sin(85)$'s cancel and you're left with
$$
\sin(20)\cos(65)-\sin(65)\cos(20)=-\sin(45).
$$
By evaluating this directly, you have the value you're looking for.
