Question

What is the value of $\sin 600{}^\circ \cos 330{}^\circ -\cos 120{}^\circ \sin 150{}^\circ $ .

Answer 1

Hint: Use the property that trigonometric ratios are periodic functions and the knowledge of complementary angles related to trigonometric ratios.

Complete step-by-step answer:
Before moving to the solution, let us discuss the periodicity of sine and cosine function, which we would be using in the solution. All the trigonometric ratios, including sine and cosine, are periodic functions. We can better understand this using the graph of sine and cosine.
First, let us start with the graph of sinx.

Next, let us see the graph of cosx.

Looking at both the graphs, we can say that the graphs are repeating after a fixed period i.e. $2{{\pi }^{c}}$ . So, we can say that the fundamental period of the cosine function and the sine function is $2{{\pi }^{c}}=360{}^\circ $ . So, we can mathematically show it as:
$\sin (2{{\pi }^{c}}+x)=\sin (360{}^\circ +x)=\sin x$
$\cos (2{{\pi }^{c}}+x)=cos(360{}^\circ +x)=\cos x$
Now solving the expression given in the question.
$\sin 600{}^\circ \cos 330{}^\circ -\cos 120{}^\circ \sin 150{}^\circ $
$\Rightarrow \sin (360{}^\circ +240{}^\circ )\cos (360{}^\circ -30{}^\circ )-\cos 120{}^\circ \sin 150{}^\circ $
We know $\sin (2{{\pi }^{c}}+x)=\sin (360{}^\circ +x)=\sin x$ , so, our expression becomes:
$\sin 240{}^\circ \cos (360{}^\circ +(-30{}^\circ ))-\cos 120{}^\circ \sin 150{}^\circ $
Now using the formula $\cos (2{{\pi }^{c}}+x)=cos(360{}^\circ +x)=\cos x$, we get
$\sin 240{}^\circ \cos (-30{}^\circ )-\cos 120{}^\circ \sin 150{}^\circ $
$\sin (180{}^\circ +60{}^\circ )\cos (-30{}^\circ )-\cos (180{}^\circ -60{}^\circ )\sin (180{}^\circ -30{}^\circ )$
Now the other properties of sine and cosine function include:
$\sin (180{}^\circ -x)=\sin x$
$\sin (180{}^\circ +x)=-\sin x$
$cos(180{}^\circ -x)=-\cos x$
$cos(-x)=\cos x$
So, using the above properties to our expression, we get
$-\sin 60{}^\circ \cos 30{}^\circ -(-\cos 60{}^\circ )\sin 30{}^\circ $
$=-\sin 60{}^\circ \cos 30{}^\circ +\cos 60{}^\circ \sin 30{}^\circ $
We know sin(A-B) = sinAcosB - cosAsinB, applying this to our expression, we get
 $\sin (30{}^\circ -60{}^\circ )$
$=\sin (-30{}^\circ )$
We know sin(-x) is equal to –sinx, so our expression becomes: $-\sin 30{}^\circ $. And the value of $-\sin 30{}^\circ $ is $-\dfrac{1}{2}$ . Therefore, the answer is $-\sin 30{}^\circ $ which is equal to $-\dfrac{1}{2}$ .

Note: It is useful to remember the graph of the trigonometric ratios along with the signs of their values in different quadrants. For example: sine is always positive in the first and the second quadrant while negative in the other two.