# Solve: (cos(5pi/2-x)-cos(3pi+x))^2 + (cos(x-9pi/2)-cos(8pi-x))^2

You should convert the differences of cosines into products, using the following formula, such that:

`cos a - cos b = 2 sin((a + b)/2)sin((b - a)/2)`

Reasoning by analogy, yields:

`cos((5pi)/2-x)-cos(3pi+x) = 2 sin ((5pi/2 - x + 3pi + x)/2) sin ((3pi + x - 5pi/2 + x)/2)`

`cos((5pi)/2-x)-cos(3pi+x) = 2 sin ((11pi)/4) sin (pi/4 + x)`

`cos(x-9pi/2) - cos(8pi-x) = 2 sin ((x - (9pi)/2 + 8pi - x)/2) sin ((8pi - x - x + (9pi)/2)/2)`

`cos(x-9pi/2) - cos(8pi-x) = 2 sin ((x - (9pi)/2 + 8pi - x)/2) sin ((8pi - x - x + (9pi)/2)/2)`

`cos(x-9pi/2) - cos(8pi-x) = 2 sin ((7pi)/4) sin (25pi/4 - x)`

`sin (25pi/4) = sin (24pi/4 - pi/4) = sin 6pi cos (pi/4) - sin (pi/4) cos 6pi`

`sin (25pi/4) = -sin (pi/4)`

`cos (25pi/4) = cos (24pi/4 - pi/4) = cos 6pi*cos(pi/4) + sin 6p*sin(pi/4)`

`cos (25pi/4) = cos(pi/4)`

`sin (25pi/4 - x) = -sin (pi/4)*cos x - sin x*cos(pi/4) = -sin (pi/4 + x)`

Hence, substituting `2 sin ((11pi)/4) sin (pi/4 + x)` for `cos((5pi)/2-x)-cos(3pi+x)` and -`2 sin ((7pi)/4)sin (pi/4 + x)` for` cos(x-9pi/2) - cos(8pi-x)` yields:

`(2 sin ((11pi)/4) sin (pi/4 + x))^2 + (-2 sin ((7pi)/4)sin (pi/4 + x))^2 = 4sin^2 (pi/4 + x)(sin^2 ((11pi)/4) + sin^2 ((7pi)/4)) `

`sin (11pi/4) = sin(pi + 7pi/4) = sin pi*cos(7pi/4) + sin(7pi/4)*cos pi`

`sin (11pi/4) = - sin(7pi/4)`

`(2 sin ((11pi)/4) sin (pi/4 + x))^2 + (-2 sin (7pi/4)*sin (pi/4 + x))^2 = 4sin^2 (pi/4 + x)(2 sin^2 (7pi/4))`

`(2 sin ((11pi)/4) sin (pi/4 + x))^2 + (-2 sin ((7pi)/4)sin (pi/4 + x))^2 = 8sin^2 (pi/4 + x)(sin^2 (7pi/4))`

`sin (3pi/4) = sin(pi - pi/4) = sin (pi/4)`

`sin (7pi/4) = sin (pi + 3pi/4) = -sin(3pi/4) = -sin(pi/4)`

`(-sin(pi/4))^2 = 2/4 = 1/2`

`(2 sin ((11pi)/4) sin (pi/4 + x))^2 + (-2 sin ((7pi)/4)sin (pi/4 + x))^2 = 4sin^2 (pi/4 + x)`

`(2 sin (11pi/4) *sin (pi/4 + x) )^2 + (-2 sin ((7pi)/4)sin (pi/4 + x))^2 = 4(sqrt2/2*sin x + sqrt2/2*cos x)^2`

`(2 sin ((11pi)/4) sin (pi/4 + x))^2 + (-2 sin (7pi/4)*sin (pi/4 + x))^2 = 2(sin x + cos x)^2`

Expanding the square yields:

`(sin x + cos x)^2 = sin^2 x + 2 sin x*cos x + cos^2 x`

Using the fundamental formula of trigonometry yields:

`sin^2 x + cos^2 x = 1`

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

`(2 sin ((11pi)/4) sin (pi/4 + x))^2 + (-2 sin ((7pi)/4)sin (pi/4 + x))^2 = 2(1 + sin 2x)`

Hence, evaluating the given trigonometric expression yields `(2 sin ((11pi)/4) sin (pi/4 + x))^2 + (-2 sin ((7pi)/4)sin (pi/4 + x))^2 = 2(1 + sin 2x).`

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