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

The solution to (e^{ipix}-e^{-ipix})^4 is 16sin^4(pix)

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

simplify

(e^{iπ x} - e^{- iπ x})^{4}

16 sin^{4} (π x)

Solution steps

(e^{iπ x} - e^{- iπ x})^{4}

e^{iπ x} = cos (π x) + i sin (π x)

e^{- iπ x} = cos (π x) - i sin (π x)

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

Expand

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

= (2 i sin (π x))^{4}

Apply exponent rule:

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

= 2^{4} i^{4} sin^{4} (π x)

2^{4} = 16

= 16 i^{4} sin^{4} (π x)

i^{4} = 1

= 16 \cdot 1 \cdot sin^{4} (π x)

Multiply the numbers:

16 \cdot 1 = 16

= 16 sin^{4} (π x)

- \frac{12 ( - x ^{2} + 1 )}{( x ^{2} + 3 ) ^{3}}

(2 y - x)^{3} + e^{2 x + 3 y} - 5 ln (y)

(tan (x) + 3) (cos (x) + 2)

(2 x 2 + y) (2 x 2 - y)

\frac{a ^{2} + 2 a s + s ^{2} - b ^{2}}{( a + s ) ^{3}}

What is (e^{ipix}-e^{-ipix})^4 ?
The solution to (e^{ipix}-e^{-ipix})^4 is 16sin^4(pix)