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0.20 ^ { \circ } \cdot \operatorname { cos } 40 ^ { \circ } \cdot \operatorname { cos } 60 ^ { \circ } \cdot \operatorname { cos } 80 ^ { \circ } = \frac { 1 } { 16 } |
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Answer» 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) what abt cos20° |
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