What is simplify-(4(e^{2it}-e^{-2it}))/5 ?

The solution to simplify-(4(e^{2it}-e^{-2it}))/5 is -(8isin(2t))/5

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simplify-(4(e^{2it}-e^{-2it}))/5

simplify

- \frac{4 ( e ^{2 i t} - e ^{- 2 i t} )}{5}

- \frac{8 i sin ( 2 t )}{5}

Solution steps

- \frac{4 ( e ^{2 i t} - e ^{- 2 i t} )}{5}

4 (e^{2 i t} - e^{- 2 i t}) = 4 (cos (2 t) + i sin (2 t) - (cos (- 2 t) + i sin (- 2 t)))

= - \frac{4 ( cos ( 2 t ) + i sin ( 2 t ) - ( cos ( - 2 t ) + i sin ( - 2 t ) ) )}{5}

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

= - \frac{4 ( cos ( 2 t ) + i sin ( 2 t ) - ( cos ( 2 t ) - i sin ( 2 t ) ) )}{5}

Expand

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

= - \frac{4 \cdot 2 i sin ( 2 t )}{5}

Multiply the numbers:

4 \cdot 2 = 8

= - \frac{8 i sin ( 2 t )}{5}

\frac{1}{25} (- 5 e^{- 5 x} x - e^{- 5 x})

\frac{1}{( 2 x + 7 ) ^{6}}

l im x \Rightarrow 2 (x^{2} - 3 x + 1)

\frac{( 20 s + 40 )}{( 4 s ^{2} - 4 s + 5 ) ^{2}}

\frac{300}{( q + 6 ) ^{2}}

What is simplify-(4(e^{2it}-e^{-2it}))/5 ?
The solution to simplify-(4(e^{2it}-e^{-2it}))/5 is -(8isin(2t))/5