What is simplify-(e^{ix}+e^{-ix})i(e^{ix}-e^{-ix}) ?

The solution to simplify-(e^{ix}+e^{-ix})i(e^{ix}-e^{-ix}) is 4cos(x)sin(x)

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simplify-(e^{ix}+e^{-ix})i(e^{ix}-e^{-ix})

simplify

- (e^{i x} + e^{- i x}) i (e^{i x} - e^{- i x})

4 cos (x) sin (x)

Solution steps

- (e^{i x} + e^{- i x}) i (e^{i x} - e^{- i x})

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

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

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

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

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

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

Simplify

cos (x) + sin (x) i + cos (x) - i sin (x) : 2 cos (x)

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

Expand

cos (x) + sin (x) i - (cos (x) - i sin (x)) : 2 i sin (x)

= - 2 \cdot 2 ii cos (x) sin (x)

Multiply the numbers:

2 \cdot 2 = 4

= - 4 ii cos (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

ii = i^{1 + 1}

= - 4 cos (x) i^{1 + 1} sin (x)

Add the numbers:

1 + 1 = 2

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

i^{2} = - 1

= - 4 (- 1) cos (x) sin (x)

Refine

= 4 cos (x) sin (x)

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

(2 + x cos (y)) d

(e^{6 (3 + 5 t)}) (6 t^{5}) + (t^{6} e^{6 (3 + 5 t)}) (5)

\frac{n \cdot ( n + 1 )}{2}

(e^{i θ} + e^{- i θ})^{4}

What is simplify-(e^{ix}+e^{-ix})i(e^{ix}-e^{-ix}) ?
The solution to simplify-(e^{ix}+e^{-ix})i(e^{ix}-e^{-ix}) is 4cos(x)sin(x)