Question

Proof (sin7α×sin3α = sin²5α-sin²2α)

Answer 1

Answer:

Since we know that,

sinC+sinD=2.sin(C+D)/2.cos(C-D)/2

& sinC-sinD=2.cos(C+D)/2.sin(C-D)/2

& 2sinA/2.cosA/2= sinA

Step-by-step explanation:

Given,

R.H.S.= sin²5a-sin²2a

=(sin5a+sin2a).(sin5a-sin2a)

=[2.sin(5a+2a)/2.cos(5a-2a)/2].[2.cos(5a+2a)/2.sin(5a-2a)/2]

=[2sin(7a/2).cos(3a/2)].[2cos(7a/2).sin(3a/2)]

=[2sin(7a/2).cos(7a/2)].[2sin(3a/2).cos(3a/2)]

=[sin2.(7a/2)].[sin2.(3a/2)]

=sin7a.sin3a

=L.H.S.

Hence proved.