Show that,(cos4x-cos2x)/2sin3x=-sinx for all x such that sin(3x) is not equal to zero

Answer 1

(cos(4x) - cos(2x))/2sin3x = - sin(x)

first work on Numerator

cos(4x) - cos(2x)

= cos(2(2x)) - cos(2x) => factoring cos4x into a double angle of 2x

= cos^2(2x) - sin^2(2x) - cos(2x) => cosine double angle

= cos^2(2x) - [1 - cos^2(2x)] - cos(2x)

= cos^2(2x) - 1 + cos^2(2x) - cos(2x) => write in terms of cos(2x)

= 2cos^2(2x) - cos(2x) - 1

let y = of cos(2x) then 2cos^2(2x) - cos(2x) - 1 becomes 2y^2 - y -1

2y^2 - y - 1 = 2y^2 - 2y + y -1

2y(y - 1) + (y - 1) = (2y + 1)(y - 1) => factoring the quadratic eq

= (2cos(2x) + 1)(cos(2x) - 1) substituting y = cos(2x)

= (2cos(2x) + 1)(cos(2x) + 1 - 2)

= (2cos(2x) + 1)(2cos^2(x) - 2)

= (2cos(2x) + 1)(2)(cos^2(x) - 1)

= (2cos(2x) + 1)(2)(-sin^2(x))

= -2(2cos(2x) + 1)(sin^2(x)) => Final Numerator

Now work on the Denominator

2sin3x = 2[sin(x + 2x)]

= 2[sin(x)cos(2x) + cos(x)sin(2x)] => expanding sin(x + 2x)

= 2sin(x)(cos(2x)) + 2cos(x)(2sin(x)cos(x)) => factoring out 2sin(x)

= 2sin(x)(cos(2x) + 2cos^2(x)) => from cos(2x) = 2cos^2(x) - 1

= 2sin(x)(cos(2x) + 1 + cos(2x))

= 2sin(x)(2cos(2x) + 1) => Final Denominator

=>[ -2(2cos(2x) + 1)(sin^2(x)) ] / [ 2sin(x)(2cos(2x) + 1)] => divide Final Numerator by Final Denominator

cancel like terms =>

(cos(4x) - cos(2x)) / 2sin3x = - sin(x)