`16(costheta-cospi/8)(costheta-cos(3pi)/8)(costheta-cos(5pi)/8)(costheta-cos(7pi)/8)=lambdacos4theta,` then the value of `lambda` is _____.

Answer

2

Explanation

`16(cos theta - cos ""(pi)/(8)) ( cos theta - cos ""(3pi)/(8)) ( cos theta - cos ""(5pi)/(8)) ( cos theta -cos ""(7pi)/(8))`
`= 16 ( cos theta - cos ""(pi)/(8)) ( cos theta - cos ""(7pi)/(8))xx ( cos theta - cos ""(3pi)/(8)) ( cos theta - cos ""(5pi )/(8)) `
`= 16 ( cos theta - cos ""(pi) /(8)) (cos theta + cos ""(pi)/(8)) xx ( cos theta - cos ""(3pi)/(8)) ( cos theta + cos ""(3pi)/(8))`
`= 16 ( cos ^(2) theta - cos ^(2)""(pi)/(8)) ( cos ^(2) theta - cos ^(2) ""(3pi)/(8))`
`= 16 (cos ^(2) theta - cos ^(2)""(pi)/(8)) ( cos ^(2) theta - sin ^(2) ""(pi)/(8))`
`= 16 ( cos ^(4) theta - cos ^(2) theta + sin ^(2) ""(pi)/(8) cos ^(2)""(pi)/(8))`
`= 16 ( cos ^(4) theta - cos ^(2) theta + (1)/(8)) `
` = 16 ( - cos ^(2) theta sin^(2) theta + (1)/(8))`
`= 16 ((-sin ^(2) 2 theta)/( 4) + (1)/(8))`
`= 16 ((1-2 sin^(2) 2 theta)/( 8))`
` = (16 cos 4 theta)/(8) `