What is (e^{ix}-e^{-ix})/(2i) ?

The solution to (e^{ix}-e^{-ix})/(2i) is sin(x)

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simplify (e^{ix}-e^{-ix})/(2i)

simplify

\frac{e ^{i x} - e ^{- i x}}{2 i}

sin (x)

Solution steps

\frac{e ^{i x} - e ^{- i x}}{2 i}

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

= \frac{cos ( x ) + i sin ( x ) - ( cos ( - x ) + i sin ( - x ) )}{2 i}

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

= \frac{cos ( x ) + i sin ( x ) - ( cos ( x ) - i sin ( x ) )}{2 i}

Expand

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

= \frac{2 i sin ( x )}{2 i}

Divide the numbers:

\frac{2}{2} = 1

= \frac{i sin ( x )}{i}

Cancel the common factor:

i

= sin (x)

81 m^{5} n^{2} - 9 m^{6} n^{5}

1 0^{2} + 1 0^{2} + 5^{2}

7 a + 8 a^{2} + 7 a^{0} - 4 a^{2} - 9 a

(4 + 4 i)^{2}

5 x^{2} + 10 x + 10