what does ( e2ixe^{2ix} + e−2ixe^{-2ix} )/4 + e2ln(sin(x))e^{2ln(sin(x))} +1/2 equal to?

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



You asked:

Evaluate the expression: e2ix+e−(2ix)4+e2ln⁡(sin⁡(x))+12\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



e2ix+e−(2ix)4+e2ln⁡(sin⁡(x))+12=e2xi4+sin⁡2(x)+12+e−2xi4\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

e2ix+e−(2ix)4+e2ln⁡(sin⁡(x))+12=e2xi4+sin⁡2(x)+12+e−2xi4\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

e2ix+e−(2ix)4+e2ln⁡(sin⁡(x))+12=(e4xi+4e2xisin⁡2(x)+2e2xi+1)e−2xi4\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}

ii is the imaginary unit, defined as i2=−1i^2 = -1.

ee is Euler's number, a mathematical constant that is approximately 2.718282.71828.

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