Question

sin^2(A+B) - sin^2(A-B) = sin2A sin2B
i need to prove this

Answer 1

Step-by-step explanation:

$let \: \: a = x \: \: \: b \: \: = y \: \: \: \\ \\ \\ { \sin}^{2} (x + y) - { \sin }^{2} (x - y) \\ \\ wkt \\ \\ \sin(x + y) = \sin(x) \cos(y) + \cos(x) \sin(y) \\ \\ \sin(x - y) = \sin(x) \cos(y) - \cos(x) \sin(y) \\ \\ {(x + y)}^{2} - {(x - y)}^{2} = 4xy \\ \\ 2 \sin(x) \cos(x) = \sin(2x) \\ \\ \\ now \\ \\ { \sin}^{2} (x + y) - { \sin }^{2} (x - y) \\ \\ = ( \sin(x) \cos(y) + \cos(x) \sin(y)) {}^{2} - {( \sin(x) \cos(y) - \cos(x) \sin(y))}^{2} \\ \\ = 4 \sin(x) \cos(x) \sin(y) \cos(y) \\ \\ = (2 \sin(x) \cos(x) )(2 \sin(y) \cos(y) ) \\ \\ = \sin(2x) \sin(2y) \\ \\ = \sin(2a) \sin(2b)$

HOPE IT HELPS

Answer 2

Step-by-step explanation:

$\begin{lgathered}let \: \: a = x \: \: \: b \: \: = y \: \: \: \\ \\ \\ { \sin}^{2} (x + y) - { \sin }^{2} (x - y) \\ \\ wkt \\ \\ \sin(x + y) = \sin(x) \cos(y) + \cos(x) \sin(y) \\ \\ \sin(x - y) = \sin(x) \cos(y) - \cos(x) \sin(y) \\ \\ {(x + y)}^{2} - {(x - y)}^{2} = 4xy \\ \\ 2 \sin(x) \cos(x) = \sin(2x) \\ \\ \\ now \\ \\ { \sin}^{2} (x + y) - { \sin }^{2} (x - y) \\ \\ = ( \sin(x) \cos(y) + \cos(x) \sin(y)) {}^{2} - {( \sin(x) \cos(y) - \cos(x) \sin(y))}^{2} \\ \\ = 4 \sin(x) \cos(x) \sin(y) \cos(y) \\ \\ = (2 \sin(x) \cos(x) )(2 \sin(y) \cos(y) ) \\ \\ = \sin(2x) \sin(2y) \\ \\ = \sin(2a) \sin(2b) \\ \\\end{lgathered}$