Question

Verify the identity $\sin (A-B)=\sin A\cos B-\cos A\sin B$ for $A=B=60{}^\circ $ .

Answer 1

Hint: To solve the above question, you need to individually solve the right-hand side and left-hand side of the equation given in the question by putting the value $A=B=60{}^\circ $ followed by using the trigonometric table to get the values of $\sin 60{}^\circ ,\cos 60{}^\circ \text{ and sin0}{}^\circ $ and show that the final values of both the sides are equal to verify the identity.

Complete step by step solution:
Before moving to the solution, let us discuss the nature of sine and cosine function, which we would be using in the solution. 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, and using the relations between the different trigonometric ratios, we get

Let us start the solution to the above question by simplifying the left-hand side of the equation $\sin (A-B)=\sin A\cos B-\cos A\sin B$ by putting the values $A=60{}^\circ $ and $B=60{}^\circ $ .
$\sin (A-B)$
$=\sin (60{}^\circ -60{}^\circ )$
$=\sin 0{}^\circ $
And we know that the value of $\sin 0{}^\circ $ is equal to 0. So, we get
$=\sin 0{}^\circ =0$
So, the left-hand side of the equation $\sin (A-B)=\sin A\cos B-\cos A\sin B$ is equal to $0$ .
Now let us simplify the right-hand side of the equation by putting the values $A=60{}^\circ $ and $B=60{}^\circ $ .
$\sin 60{}^\circ \cos 60{}^\circ -\cos 60{}^\circ \sin 60{}^\circ $
$=0$
Therefore, we can say that the left-hand side of the equation $\sin (A-B)=\sin A\cos B-\cos A\sin B$ is equal to the right-hand side. So, we have verified the above equation.

Note: Be careful while putting the values of the different trigonometric ratios in the expression for solving the question as they are very confusing and need to be learned. Also, remember that proving and verifying are two different things, when you prove an equation, the equation becomes valid for all the values of variables in the domain of the variable, while verifying makes the equation true only for the value that you have verified the equation for.