Find the value of sin78° + cos132°?

Correct option is A

Given:

sin78° + cos132°

​Formula Used:​

$$\cos \theta = \sin (90^\circ - \theta) = \sin\theta$$

$$\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$$​​

Solution:

$$\sin 78^\circ + \cos 132^\circ$$​​

$$= \sin 78^\circ + \sin (90^\circ - 132^\circ)$$

​​$$=\sin 78^\circ + \sin (-42^\circ)$$​​

= $$\sin 78^\circ - \sin 42^\circ$$

Using the identity:

$$\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$$

$$A = 78^\circ$$​$$B = 42^\circ​​​$$​ and,

= $$2 \cos \left(\frac{78^\circ + 42^\circ}{2}\right) \sin \left(\frac{78^\circ - 42^\circ}{2}\right)$$

= $$2 \cos (60^\circ) \sin (18^\circ)$$

Since $$cos 60^\circ = \frac{1}{2}​​$$​

​$$= 2 \times \frac{1}{2} \times \sin 18^\circ$$

$$= \sin 18^\circ$$

Thus, the correct answer is (a).