What is (e^{ix}-e^{-ix})^4 ?

The solution to (e^{ix}-e^{-ix})^4 is 16sin^4(x)

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simplify (e^{ix}-e^{-ix})^4

simplify

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

16 sin^{4} (x)

Solution steps

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

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

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

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

Expand

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

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

2^{4} = 16

= 16 i^{4} sin^{4} (x)

i^{4} = 1

= 16 \cdot 1 \cdot sin^{4} (x)

Multiply the numbers:

16 \cdot 1 = 16

= 16 sin^{4} (x)

(y^{3})^{4}

3 x (x + x^{4}) d x

(2 x^{2} - 3 x - 20)^{5}

\frac{( x ^{3} - 4 ) ^{6}}{64 x ^{6}}

\frac{( x + 4 ) ( x - 5 )}{( x + 6 ) ( 3 x - 2 )}