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# find out the following (solve) cos36.cos72.cos108.cos144

Grade:12th pass

## 1 Answers

Shahid
74 Points
4 years ago
cos36.cos72.cos108.cos144
= cos36.cos(90-18).cos(90+18).cos(180-36)
{As Cos(90-x) = sin(x)   , cos(180-x)=(-cosx)  &  Cos(90+x)=(-sinx)}
= cos36.sin18.(-sin18).(-cos36)
=cos236.sin218
=$\left \{ \right.(\sqrt5 -1) / 4\left. \right \} \left \{ \right.(\sqrt5 +1) / 4\left. \right \}$ $\left \{ \right.(\sqrt5 -1) / 4\left. \right \} \left \{ \right.(\sqrt5 +1) / 4\left. \right \}$
$\left \{ 5^2 - 1^2 / 16 \left. \right \} \right.$  $\left \{ 5^2 - 1^2 / 16 \left. \right \} \right.$
=( 24 x 24 ) / (16 x 16 )
= 9/4

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