What is (i(e^{it}-e^{-it})^3)/8 ?

The solution to (i(e^{it}-e^{-it})^3)/8 is sin^3(t)

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simplify (i(e^{it}-e^{-it})^3)/8

simplify

\frac{i ( e ^{i t} - e ^{- i t} ) ^{3}}{8}

sin^{3} (t)

Solution steps

\frac{i ( e ^{i t} - e ^{- i t} ) ^{3}}{8}

i (e^{i t} - e^{- i t})^{3} = i (cos (t) + i sin (t) - (cos (- t) + i sin (- t)))^{3}

= \frac{i ( cos ( t ) + i sin ( t ) - ( cos ( - t ) + i sin ( - t ) ) ) ^{3}}{8}

(cos (t) + sin (t) i - (cos (- t) + sin (- t) i))^{3} = 2^{3} i^{3} sin^{3} (t)

= \frac{2 ^{3} i ^{3} i sin ^{3} ( t )}{8}

i 2^{3} i^{3} sin^{3} (t) = 2^{3} i^{4} sin^{3} (t)

= \frac{2 ^{3} i ^{4} sin ^{3} ( t )}{8}

Factor

8 : 2^{3}

= \frac{2 ^{3} i ^{4} sin ^{3} ( t )}{2 ^{3}}

Cancel the common factor:

2^{3}

= i^{4} sin^{3} (t)

i^{4} = 1

= 1 \cdot sin^{3} (t)

Multiply:

1 \cdot sin^{3} (t) = sin^{3} (t)

= sin^{3} (t)

\frac{d}{d t} ((x) (t^{2} + 3))

\frac{5}{3 ( x + 3 )}

2^{k} \cdot 2

(2 e^{x} - 5)^{2}

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

What is (i(e^{it}-e^{-it})^3)/8 ?
The solution to (i(e^{it}-e^{-it})^3)/8 is sin^3(t)