Prove that cos20°cos40°cos60°cos80°= 1/16
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Answers
L.H.S.=> cos20°cos40°cos60°cos80°
As, cos60°=1/2
Therefore, (1/2)(2cos20°cos40°)(1/2)cos80°
=>(1/4)[cos(20°+40°)+cos(20°-40°)]cos80°
=>(1/4)(cos60°+cos20°)cos80°
=>(1/4)(cos60°cos80°+cos20°cos80°)
=>(1/4)(1/2)cos80°+(1/4)cos20°cos80°
=>(1/8)cos80°+(1/4)(1/2)(2cos20°cos80°)
=>(1/8)cos80°+(1/8)[cos(20°+80°)+cos(20°-80°)]
=>(1/8)cos80°+(1/8)(cos100°+cos60°)
=>(1/8)cos80°+(1/8)cos100°+(1/8)cos60°
=>(1/8)(cos80°+cos100°)+(1/8)×(1/2)
=>(1/8)[{2cos(80°+100°)/2}{cos(80°-100°)/2}]+(1/16)
=>(1/8)(2cos90°cos10°)+(1/16)
As, cos90°=0
Therefore, 0+(1/16)
=>1/16
R.H.S.=> 1/16
As, L.H.S.= R.H.S.
Hence, proved
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cos20°cos40°cos60°cos80°
=(1/2)(2cos20°cos40°)(1/2)cos80° [∵,cos60°=1/2]
=(1/4)[cos(20°+40°)+cos(20°-40°)]cos80°
=(1/4)(cos60°+cos20°)cos80°
=(1/4)(cos60°cos80°+cos20°cos80°)
=(1/4)(1/2)cos80°+(1/4)cos20°cos80°
=(1/8)cos80°+(1/4)(1/2)(2cos20°cos80°)
=(1/8)cos80°+(1/8)[cos(20°+80°)+cos(20°-80°)]
=(1/8)cos80°+(1/8)(cos100°+cos60°)
=(1/8)cos80°+(1/8)cos100°+(1/8)cos60°
=(1/8)(cos80°+cos100°)+(1/8)×(1/2)
=(1/8)[{2cos(80°+100°)/2}{cos(80°-100°)/2}]+(1/16)
=(1/8)(2cos90°cos10°)+(1/16)
=0+(1/16) [cos90°=0]
=1/16 (proved).