what does ( $e^{2ix}$ + $e^{-2ix}$ )/4 + $e^{2ln(sin(x))}$ +1/2 equal to?

asked by guest
on Jan 11, 2025 at 4:16 pm



You asked:

Evaluate the expression: \(\frac{{e}^{2 i x} + {e}^{-\left( 2 i x \right)}}{4} + {e}^{2 \ln\left( \sin\left( x \right) \right)} + \frac{1}{2}\)

MathBot Answer:

Evaluated



\(\displaystyle \frac{{e}^{2 i x} + {e}^{-\left( 2 i x \right)}}{4} + {e}^{2 \ln\left( \sin\left( x \right) \right)} + \frac{1}{2} = \frac{e^{2 x i}}{4} + \sin^{2}{\left(x \right)} + \frac{1}{2} + \frac{e^{- 2 x i}}{4} \)

Expanded

\[\frac{{e}^{2 i x} + {e}^{-\left( 2 i x \right)}}{4} + {e}^{2 \ln\left( \sin\left( x \right) \right)} + \frac{1}{2} = \frac{e^{2 x i}}{4} + \sin^{2}{\left(x \right)} + \frac{1}{2} + \frac{e^{- 2 x i}}{4}\]

Factored

\[\frac{{e}^{2 i x} + {e}^{-\left( 2 i x \right)}}{4} + {e}^{2 \ln\left( \sin\left( x \right) \right)} + \frac{1}{2} = \frac{\left(e^{4 x i} + 4 e^{2 x i} \sin^{2}{\left(x \right)} + 2 e^{2 x i} + 1\right) e^{- 2 x i}}{4}\]

\(i\) is the imaginary unit, defined as \(i^2 = -1\).

\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).

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