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