Question

((cos theta + isin theta)^4)/(icos theta+sin theta)^5 is equal to

Answer 1

Answer:

The given expression is equal to $sin9\theta-icos9\theta$

Step-by-step explanation:

Concept:

Euler's formula

$e^{i\theta}=cos\theta+i\:sin\theta$

Now,

$\frac{(cos\:\theta+i\:sin\:\theta)^4}{(sin\:\theta+i\:cos\:\theta)^5}$

$=\frac{(cos\:\theta+i\:sin\:\theta)^4}{i^5(cos\:\theta-i\:sin\:\theta)^5}$

$=\frac{(e^{i\theta})^4}{i(e^{-i\theta})^5}\\\\=(-i)\frac{e^{i4\theta}}{e^{-i5\theta}}\\\\=(-i)e^{i9\theta}$

$=-i[cos9\theta+i\:sin9\theta]\\\\=-icos9\theta-i^2sin9\theta\\\\=sin9\theta-icos9\theta$

Answer 2

Answer:

The given expression is equal to sin9\theta-icos9\thetasin9θ−icos9θ

Step-by-step explanation:

Concept:

Euler's formula

e^{i\theta}=cos\theta+i\:sin\thetae

iθ

=cosθ+isinθ

Now,

\frac{(cos\:\theta+i\:sin\:\theta)^4}{(sin\:\theta+i\:cos\:\theta)^5}

(sinθ+icosθ)

5

(cosθ+isinθ)

4

=\frac{(cos\:\theta+i\:sin\:\theta)^4}{i^5(cos\:\theta-i\:sin\:\theta)^5}=

i

5

(cosθ−isinθ)

5

(cosθ+isinθ)

4

\begin{lgathered}=\frac{(e^{i\theta})^4}{i(e^{-i\theta})^5}\\\\=(-i)\frac{e^{i4\theta}}{e^{-i5\theta}}\\\\=(-i)e^{i9\theta}\end{lgathered}

=

i(e

−iθ

)

5

(e

iθ

)

4

=(−i)

e

−i5θ

e

i4θ

=(−i)e

i9θ

\begin{lgathered}=-i[cos9\theta+i\:sin9\theta]\\\\=-icos9\theta-i^2sin9\theta\\\\=sin9\theta-icos9\theta\end{lgathered}

=−i[cos9θ+isin9θ]

=−icos9θ−i

2

sin9θ

=sin9θ−icos9θ