 QUESTION

# Evaluate the following trigonometric expression: $\cos \alpha .\sin \left( \beta -\gamma \right)+\cos \beta .\sin \left( \gamma -\alpha \right)+\cos \gamma .\sin \left( \alpha -\beta \right)$(a) 0(b) $\dfrac{1}{2}$ (c) 1 (d) $\cos \alpha .\cos \beta .\cos \gamma$

Hint: First use the formula $\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y$ to expand the second part of each term. Then, multiply the first part (cosine part) to the expanded form of the second parts in each term. You will get six terms in total. Now, you will see that there are three pairs of terms which differ only in sign. Group these terms together. These terms cancel each other out and you will get 0 as the final answer.

In this question, we need to find the value of $\cos \alpha .\sin \left( \beta -\gamma \right)+\cos \beta .\sin \left( \gamma -\alpha \right)+\cos \gamma .\sin \left( \alpha -\beta \right)$ among the given options.

For this, we will do the following.

We know that the sine of the difference of two angles x and y is equal to the difference of the product of sine of x and cosine of y and the product of cosine of x and sine of y.

i.e. $\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y$

Using the above property in the given equation, we will get the following:

$\cos \alpha \sin \beta \cos \gamma -\cos \alpha \cos \beta \sin \gamma +\cos \beta \sin \gamma \cos \alpha -\cos \beta \cos \gamma \sin \alpha +\cos \gamma \sin \alpha \cos \beta -\cos \gamma \cos \alpha \sin \beta$

$\cos \alpha \sin \beta \cos \gamma -\cos \alpha \cos \beta \sin \gamma +\cos \beta \sin \gamma \cos \alpha -\cos \beta \cos \gamma \sin \alpha +\cos \gamma \sin \alpha \cos \beta -\cos \gamma \cos \alpha \sin \beta$We will rearrange this to get the following:

$\left( \cos \alpha \sin \beta \cos \gamma -\cos \gamma \cos \alpha \sin \beta \right)+\left( \cos \beta \sin \gamma \cos \alpha -\cos \alpha \cos \beta \sin \gamma \right)+\left( \cos \gamma \sin \alpha \cos \beta -\cos \beta \cos \gamma \sin \alpha \right)=0$As we can see, the terms in each of the brackets will get cancelled. So, all the terms are cancelled, and we will get 0.

So, $\cos \alpha .\sin \left( \beta -\gamma \right)+\cos \beta .\sin \left( \gamma -\alpha \right)+\cos \gamma .\sin \left( \alpha -\beta \right)=0$

Hence, (a) is the correct option.

Note: In this question, it is very important to know the property that that the sine of the difference of two angles x and y is equal to the difference of the product of sine of x and cosine of y and the product of cosine of x and sine of y. i.e. $\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y$. Without this the question cannot be solved.