+2 votes
in JEE by kratos

Show that sin8 A/2 sin A = cos A + cos3A + cos 5A + cos 7A

1 Answer

+4 votes
by kratos
 
Best answer

R.H.*. = (cosA + cos3A) + (cos5A + cos7A)

= [cos(2A − A) + cos(2A + A)] + [cos(6A − A) + cos(6A + A)]

= 2cosA⋅cos2A + 2cosA⋅cos6A

= 2cosA[cos(4A −2A) + cos(4A +2A)]

= 2cosA⋅2cos2A⋅cos4A

=(2sinAcosA)/sinA . 2cosA.cos4A

=(2sin2A.cos2A).cos4A/sinA

=sin4A.cos4A/sinA = sin8 A/2 sinA

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