What is (e^{-i2x}-e^{-i4x})(e^{i2x}-e^{i4x}) ?

The solution to (e^{-i2x}-e^{-i4x})(e^{i2x}-e^{i4x}) is (cos(2x)-cos(4x))^2+4cos^2(3x)sin^2(x)

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simplify (e^{-i2x}-e^{-i4x})(e^{i2x}-e^{i4x})

simplify

(e^{- i 2 x} - e^{- i 4 x}) (e^{i 2 x} - e^{i 4 x})

(cos (2 x) - cos (4 x))^{2} + 4 cos^{2} (3 x) sin^{2} (x)

Solution steps

(e^{- i 2 x} - e^{- i 4 x}) (e^{i 2 x} - e^{i 4 x})

e^{- i 2 x} = cos (2 x) - i sin (2 x)

e^{- i 4 x} = cos (4 x) - i sin (4 x)

= (cos (2 x) - i sin (2 x) - (cos (4 x) - i sin (4 x))) (e^{2 i x} - e^{4 i x})

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= (cos (2 x) - i sin (2 x) - (cos (4 x) - i sin (4 x))) (cos (2 x) + i sin (2 x) - e^{4 i x})

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= (cos (2 x) - i sin (2 x) - (cos (4 x) - i sin (4 x))) (cos (2 x) + i sin (2 x) - (cos (4 x) + i sin (4 x)))

Apply complex arithmetic rule:

(a + bi) (a - bi) = a^{2} + b^{2}

a = cos (2 x) - cos (4 x), b = - sin (2 x) + sin (4 x)

= (cos (2 x) - cos (4 x))^{2} + (- sin (2 x) + sin (4 x))^{2}

Simplify

(cos (2 x) - cos (4 x))^{2} + (- sin (2 x) + sin (4 x))^{2} : (cos (2 x) - cos (4 x))^{2} + 4 cos^{2} (3 x) sin^{2} (x)

= (cos (2 x) - cos (4 x))^{2} + 4 cos^{2} (3 x) sin^{2} (x)

\frac{4}{( 1 - 2 x ) ^{2}}

(x^{2} - x y + y^{2}) d x - x y d y

2 y \cdot \frac{d}{d x} ((y) (x^{3} (x - 2)))

\frac{( 1 + x ) ( 1 + 2 x ) ( 1 + 3 x )}{x}

π^{8} (19 - 2 - y^{2})^{2}

What is (e^{-i2x}-e^{-i4x})(e^{i2x}-e^{i4x}) ?
The solution to (e^{-i2x}-e^{-i4x})(e^{i2x}-e^{i4x}) is (cos(2x)-cos(4x))^2+4cos^2(3x)sin^2(x)