What is 1/((cos(θ)+isin(θ))^2) ?

The solution to 1/((cos(θ)+isin(θ))^2) is cos(2θ)-sin(2θ)i

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simplify 1/((cos(θ)+isin(θ))^2)

simplify

\frac{1}{( cos ( θ ) + i sin ( θ ) ) ^{2}}

cos (2 θ) - sin (2 θ) i

Solution steps

\frac{1}{( cos ( θ ) + i sin ( θ ) ) ^{2}}

Expand

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

= \frac{1}{i sin ( 2 θ ) + cos ( 2 θ )}

Rewrite

i sin (2 θ) + cos (2 θ)

in standard complex form:

cos (2 θ) + sin (2 θ) i

= \frac{1}{cos ( 2 θ ) + sin ( 2 θ ) i}

Multiply by the conjugate

\frac{cos ( 2 θ ) - sin ( 2 θ ) i}{cos ( 2 θ ) - sin ( 2 θ ) i}

= \frac{1 \cdot ( cos ( 2 θ ) - sin ( 2 θ ) i )}{( cos ( 2 θ ) + sin ( 2 θ ) i ) ( cos ( 2 θ ) - sin ( 2 θ ) i )}

Simplify

\frac{1 \cdot ( cos ( 2 θ ) - sin ( 2 θ ) i )}{( cos ( 2 θ ) + sin ( 2 θ ) i ) ( cos ( 2 θ ) - sin ( 2 θ ) i )} : \frac{cos ( 2 θ ) - sin ( 2 θ ) i}{1}

= \frac{cos ( 2 θ ) - sin ( 2 θ ) i}{1}

Apply the fraction rule:

\frac{a + bi}{c} = \frac{a}{c} + \frac{b}{c} i

= \frac{cos ( 2 θ )}{1} + \frac{- sin ( 2 θ )}{1} i

Simplify

\frac{cos ( 2 θ )}{1} + \frac{- sin ( 2 θ )}{1} i : cos (2 θ) - sin (2 θ) i

= cos (2 θ) - sin (2 θ) i

(4 x^{3} - 3 x)^{4}

\frac{5}{2} cos (\frac{1}{2}) (x + \frac{π}{2})

- 42 (- 14 e^{- 7 x^{2}} x^{2} + e^{- 7 x^{2}})

\frac{1}{( n + 1 ) n}

(e^{x} + 2 x)^{2}

What is 1/((cos(θ)+isin(θ))^2) ?
The solution to 1/((cos(θ)+isin(θ))^2) is cos(2θ)-sin(2θ)i