Question

16cos20.cos40.cos60.cos80 = 1

Answer 1

L.H.S. 1/2 .16 cos20(cos40cos80)
=1/4.16cos20(2cos40cos80)
=4cos20[cos(40+80)+cos(40-80)]
=4cos20[cos120+cos(-40)]
=4cos20[-cos60+cos40]
=4cos20[-1/2+cos40]
=4.1/2cos20+4cos20cos40
=2cos20+1/2.4[2cos20cos40]
=2cos20+2[cos(20+40)+cos(20-40)

=2cos20+2[cos60+cos(-20)]
=2cos20+2[1/2 -cos20]
=2cos20+1 -2cos20
=1
=R.H.S proved
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