What is (e^{-i2t}-e^{-i3t})(e^{+i2t}-e^{+i3t}) ?

The solution to (e^{-i2t}-e^{-i3t})(e^{+i2t}-e^{+i3t}) is (cos(2t)-cos(3t))^2+(-sin(2t)+sin(3t))^2

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simplify (e^{-i2t}-e^{-i3t})(e^{+i2t}-e^{+i3t})

simplify

(e^{- i 2 t} - e^{- i 3 t}) (e^{+ i 2 t} - e^{+ i 3 t})

(cos (2 t) - cos (3 t))^{2} + (- sin (2 t) + sin (3 t))^{2}

Solution steps

(e^{- i 2 t} - e^{- i 3 t}) (e^{i 2 t} - e^{i 3 t})

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

e^{- i 3 t} = cos (3 t) - i sin (3 t)

= (cos (2 t) - i sin (2 t) - (cos (3 t) - i sin (3 t))) (e^{2 i t} - e^{3 i t})

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= (cos (2 t) - i sin (2 t) - (cos (3 t) - i sin (3 t))) (cos (2 t) + i sin (2 t) - e^{3 i t})

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= (cos (2 t) - i sin (2 t) - (cos (3 t) - i sin (3 t))) (cos (2 t) + i sin (2 t) - (cos (3 t) + i sin (3 t)))

Apply complex arithmetic rule:

(a + bi) (a - bi) = a^{2} + b^{2}

a = cos (2 t) - cos (3 t), b = - sin (2 t) + sin (3 t)

= (cos (2 t) - cos (3 t))^{2} + (- sin (2 t) + sin (3 t))^{2}

\frac{( x - 1 ) ^{3} + 1}{x}

(x^{2} - 1) \frac{d y}{d x} + 2 y

\frac{2 x + 6}{( s - 1 ) ^{3}}

\frac{p ( x )}{( x ^{2} + 1 ) ^{2} ( x - 1 ) ^{2}}

\frac{( 12 x + 18 )}{x ( x ^{2} + 9 )}

What is (e^{-i2t}-e^{-i3t})(e^{+i2t}-e^{+i3t}) ?
The solution to (e^{-i2t}-e^{-i3t})(e^{+i2t}-e^{+i3t}) is (cos(2t)-cos(3t))^2+(-sin(2t)+sin(3t))^2