What is e^{iθ}(e^{iθ}+e^{-iθ}-2) ?

The solution to e^{iθ}(e^{iθ}+e^{-iθ}-2) is (2cos^2(θ)-2cos(θ))+i(-2sin(θ)+2cos(θ)sin(θ))

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simplify e^{iθ}(e^{iθ}+e^{-iθ}-2)

simplify

e^{i θ} (e^{i θ} + e^{- i θ} - 2)

(2 cos^{2} (θ) - 2 cos (θ)) + i (- 2 sin (θ) + 2 cos (θ) sin (θ))

Solution steps

e^{i θ} (e^{i θ} + e^{- i θ} - 2)

e^{i θ} = cos (θ) + i sin (θ)

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

e^{i θ} = cos (θ) + i sin (θ)

e^{- i θ} = cos (θ) - i sin (θ)

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

Simplify

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

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

Rewrite

(cos (θ) + sin (θ) i) (2 cos (θ) - 2)

in standard complex form:

(2 cos^{2} (θ) - 2 cos (θ)) + (2 sin (θ) cos (θ) - 2 sin (θ)) i

= (2 cos^{2} (θ) - 2 cos (θ)) + (2 sin (θ) cos (θ) - 2 sin (θ)) i

\frac{( 1 )}{( x + a )}

L t (- cos (t) + 1)

\frac{2 ( x - 2 )}{( 4 - x ) \cdot x \cdot ( x + 2 )}

t or i z a r (\frac{x ^{2}}{2} - 5 x) \cdot \frac{1}{x}

\frac{d}{d t} ((x) (3 x - 2 y))

What is e^{iθ}(e^{iθ}+e^{-iθ}-2) ?
The solution to e^{iθ}(e^{iθ}+e^{-iθ}-2) is (2cos^2(θ)-2cos(θ))+i(-2sin(θ)+2cos(θ)sin(θ))