Prove that `cos(pi/7)*cos(2pi/7)*cos(4pi/7)= -1/8`

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justaguide | College Teacher | (Level 2) Distinguished Educator

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It has to be proved that `cos(pi/7)*cos(2pi/7)*cos(4pi/7)= -1/8`

Use the property that the product `2*cos a*sin a = sin (2a)`

`2*sin (pi/7)*cos(pi/7) = sin (2*pi/7)` ...(1)

`2*sin(2*pi/7)*cos(2*pi/7) = sin(4*pi/7)` ...(2)

`2*sin(4*pi/7)*cos(4*pi/7)` = `sin(8*pi/7)` = `sin (pi + pi/7)` = `-sin(pi/7)` ...(3)

Multiply the three equations derived above:

(1)*(2)*(3)

=> `2*sin (pi/7)*cos(pi/7)*2*sin(2*pi/7)*cos(2*pi/7)*2*sin(4*pi/7)*cos(4*pi/7)` = `-sin(2*pi/7)*sin(4*pi/7)*sin(pi/7)`

Cancel the sine terms from both the sides.

=> `2*cos(pi/7)*2*cos(2*pi/7)*2*cos(4*pi/7) = -1`

=> `8*cos(pi/7)*cos(2*pi/7)*cos(4*pi/7) = -1`

=> `cos(pi/7)*cos(2*pi/7)*cos(4*pi/7) = -1/8`

This proves that `cos(pi/7)*cos(2*pi/7)*cos(4*pi/7) = -1/8`

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