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We have to prove that cos 4x - sin 4x * cot 2x = -1
cos 4x - sin 4x * cot 2x = -1
use cos 2x = (cos x)^2 - (sin x)^2 and sin 2x = 2 sin x cos x and cot x = (cos x)/(sin x)
=> (cos 2x)^2 - (sin x)^2 - 2*(sin 2x)*(cos 2x)*(cos 2x)/(sin 2x)
=>(cos 2x)^2 - (sin x)^2 - 2*(cos 2x)*(cos 2x)
=> ( cos 2x)^2 - 2 ( cos 2x)^2 - ( sin 2x)^2
=> - (cos 2x)^2 - (sin 2x)^2
=> -1*[(cos 2x)^2 + (sin 2x)^2]
As (cos x)^2 + (sin x)^2 = 1
=> -1
Therefore we proved that
cos 4x - sin 4x * cot 2x = -1
We know that cos 4x = cos 2*(2x) = 2(cos 2x)^2 - 1
We'll write sin 4x = sin 2(2x) = 2 sin (2x)*cos (2x)
We'll write cot 2x = cos 2x/sin 2x
We'll re-write the identity:
2(cos 2x)^2 - 1 - 2 sin (2x)*cos (2x)*cos 2x/sin 2x = -1
We'll simplify and we'll get:
2(cos 2x)^2 - 1 - 2*cos 2x*cos 2x = -1
2(cos 2x)^2 - 1 - 2(cos 2x)^2 = -1
We'll add 2(cos 2x)^2 both sides:
2(cos 2x)^2 - 1 = 2(cos 2x)^2 - 1 q.e.d.
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