Question

Sin 3x . Cos 5x = ??

Please give detailed explanation!

Answer 1

____________________________________________________________


♦ Basic Trigonometry ♦

◘ Recall : 
  → sin [ A + B ] = sin A cos B + sin B cos A 

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

◘ Substitute : 
  → A = 5x ; B = 3x 

=> [ 2 sin 3x . cos 5x ] = [ sin ( 8x ) - sin ( 2x ) ]

=> ( sin 3x . cos 5x ) = ( 1/2 )[ sin ( 8x ) - sin ( 2x ) ]
                                 = ( 0.5 sin( 8x ) - 0.5 sin( 2x ) )
_____________________________________________________________

→ Hence, ( sin 3x . cos 5x ) can be expressed as difference of sine functions as : [ 0.5 sin( 8x ) - 0.5 sin( 2x ) ]
_____________________________________________________________

♦ Hope this nelps