Question

Find the value of the trigonometric expression:$\sin (79^\circ )\cos (49^\circ ) - \cos (79^\circ )\sin (49^\circ )$

Answer 1

Hint:We are given an expression with sine and cosine functions and two different angles in degrees. We have to simply find the value of the trigonometric expression by using the standard identity of $\sin \left( {A - B} \right)$ to find its value and then evaluate the sine value for the resultant angle.

Complete solution step by step:
Firstly we write down the trigonometric expression given in the question i.e.
$\sin (79^\circ )\cos (49^\circ ) - \cos (79^\circ )\sin (49^\circ )\,{\text{ - - - - - - (1)}}$

Now as we can see that expression (1) has sine and cosine functions with two different angles in degree

so we compare this with standard trigonometric identity i.e.
$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$

Comparing expression (1) with the above formula we have
$
A = 79^\circ \\
B = 49^\circ \\
$

Now we put the values of (1) in the formula we have
$\sin (79^\circ )\cos (49^\circ ) - \cos (79^\circ )\sin (49^\circ ) = \sin (79^\circ - 49^\circ ) = \sin (30^\circ )$

So we have found the result after simplifying the expression to be
$\sin 30^\circ $

And we know it is a standard trigonometric function whose value is known to us i.e.
$\sin 30^\circ = \dfrac{1}{2}$

So we have obtained the required value of the given trigonometric expression.

Note: With the help of right trigonometric identity and value of sine function we solve the question and it is always helpful when you memorize these identities alongside with the values of all six functions described in the table.