Current Affairs JEE Main & Advanced

Category : JEE Main & Advanced

(1) \[2\sin A\cos B=\sin (A+B)+\sin (A-B)\]

(2) \[2\cos A\sin B=\sin (A+B)-\sin (A-B)\]

(3) \[2\cos A\cos B=\cos (A+B)+\cos (A-B)\]

(4) \[2\sin A\sin B=\cos (A-B)-\cos (A+B)\]

Let \[A+B=C\] and \[A-B=D\]

Then, \[A=\frac{C+D}{2}\] and \[B=\frac{C-D}{2}\]

Therefore, we find out the formulae to transform the sum or difference into product.

(1) \[\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}\]

(2) \[\sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\]

(3) \[\cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}\]

(4) \[\cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}=-2\sin \frac{C+D}{2}\sin \frac{C-D}{2}\].