Question

11. sin 0 + sin 30 + sin 50 + sin 70 = 4 cos 0 cos 20 sin 40.
(N.C.E.R.T.; H.P.B. 2011​

Answer 1

We know that

i) sin 2A = 2sinAcosA

ii) sin(90-A)= cosA

iii)sin30=1/2

LHS = sin 10 sin30 sin50sin 70

=sin 10 × 1/2 × sin (90-40)× sin(90-20)

=1/2[sin 10 ×cos 40×cos 20]

We multiply with[2 cos 10/2cos 10]

=(1/2×1/2cos10)[2sin10cos10×cos20×cos40]

=(1/4cos10)[sin20cos20×cos40]

=(1/8cos10)[2sin20cos20×cos40]

=(1/8cos10)[sin40cos40]

=(1/16cos10)[2sin40co40]

=(1/16cos10)×sin80

=(1/16cos10)×sin(90-10)

=(1/16cos10)×cos10

=1/16

= RHS