What is (Ae^{ix}+Be^{-ix})^2 ?

The solution to (Ae^{ix}+Be^{-ix})^2 is (A(cos(x)+isin(x))+B(cos(x)-isin(x)))^2

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simplify (Ae^{ix}+Be^{-ix})^2

simplify

(A e^{i x} + B e^{- i x})^{2}

(A (cos (x) + i sin (x)) + B (cos (x) - i sin (x)))^{2}

Solution steps

(A e^{i x} + B e^{- i x})^{2}

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

= (A (cos (x) + i sin (x)) + B e^{- i x})^{2}

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

= (A (cos (x) + i sin (x)) + B (cos (x) - i sin (x)))^{2}

(e^{iπ θ} - e^{- iπ θ})^{6}

\frac{1}{( s + 2 ) ^{5}}

(2 - 3) 3 π

\frac{c ( - D a c + 2 B a c + D c ^{2} )}{( a + c ) ^{3}}

2 \cdot sin (x) (x - 6 y)

What is (Ae^{ix}+Be^{-ix})^2 ?
The solution to (Ae^{ix}+Be^{-ix})^2 is (A(cos(x)+isin(x))+B(cos(x)-isin(x)))^2