# 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 $-\dfrac{1}{\sqrt{2}}$

• From where you are, you're almost done. Recall that $\sin$ is an odd function, so $\sin(-45)=-\sin(45)$, add the expressions together and you'll get the answer you're looking for (although you might have a sign backwards somewhere...) – Michael Burr May 8 '17 at 21:34
• the trig functions are in degrees – super95 May 8 '17 at 21:35

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.

Use fact that

$$\sin(a - b) = \sin a\cos b - \sin b \cos a$$

$$\text{Therefore, }\, -\sin (65 - 20) = -\sin 65 \cos 20 + \sin 20 \cos 65 = -\left(\frac{1}{\sqrt{2}}\right)$$

$$\text{That is },\, \sin 45 = \frac{1}{\sqrt{2}}$$

• @vrugtehagel Absolutely, it was just a typo – caverac May 8 '17 at 21:50

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.