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Question

Prove that cos20°×cos40°×cos80°=18


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Solution

L.H.S.=cos20°×cos40°×cos80°

Multiplying and dividing by 2

=12(2cos20°×cos40°×cos80°)

We know that, 2cosacosb=cosa+b+cosa-b

=12cos20°+80°+cos20°-80°cos40°

=12cos100°+cos-60°.cos40°

=12cos100°+12.cos40°

=14cos40°+12(cos40°.cos100°)

Multiplying and dividing by 2.

=22×14cos40°+12×2(2cos40°.cos100°)

=14cos40°+14(cos40°+100°+cos40°-100°

=14cos40°+14(cos140°+cos-60°)…………………..(As, 2cosacosb=cosa+b+cosa-b)

=14cos40°+14cos140°+14×12…………………….(As, cos-60°=12)

=14cos40°+cos140°+18

We know that 2cosacosb=cosa+b+cosa-b,

So, 90°+50°=140°,90°-50°=40°

=142cos90°.cos-50°+18

=1420.cos-50°+18…………………………(cos90°=0)

=0+18

=18.

Therefore, it is proved that cos20°×cos40°×cos80°=18.


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