<s>[INST] 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 $-\dfrac{1}{\sqrt{2}}$
 [/INST] Use fact that
$$
\sin(a - b) = \sin a\cos b - \sin b \cos a
$$
$$\text{Therefore, }\,
-\sin (65 - 20) = -\sin 65 \cos 20 + \sin 20 \cos 65 = -\left(\frac{1}{\sqrt{2}}\right)
$$
$$\text{That is },\,
\sin 45 = \frac{1}{\sqrt{2}}
$$
</s>