((1+sinx+icosx)/(1+sinx-icos))^n

**nylon** · November 28th 2011, 09:04 AM

((1+sinx+icosx)/(1+sinx-icos))^n
can any help me about this problem

**mr fantastic** · November 28th 2011, 10:40 AM

Originally Posted by nylon

((1+sinx+icosx)/(1+sinx-icos))^n
can any help me about this problem

What exactly is the question??

**princeps** · November 28th 2011, 10:47 AM

Use Euler's formula :

$e^{ix}=\cos (x )+ i\cdot \sin(x)$ ,so you can rewrite expression into :

$\left(\frac{1+e^{ix}}{1+e^{-ix}}\right)^{n}=\left(\frac{1+e^{ix}}{1+\frac{1}{e ^{ix}}}\right)^{n}$ ...etc.

Final result should be :

$\cos (nx) + i\cdot \sin (nx)$

**Soroban** · November 28th 2011, 12:15 PM

Hello, nylon!

I did it "the long way" . . .

$\text{Simplify: }\:\left(\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}\right)^n$

We have: . $\frac{(1 + \sin x) + i\cos x}{(1+\sin x) - i\cos x}$

Multiply numerator and denominator by the conjugate of the denominator:

. . $\frac{(1+\sin x) + i\cos x}{(1+\sin x) - i\cos x}\cdot\frac{(1+\sin x) + i\cos x}{(1+\sin x) + i\cos x)}$

. . $=\;\frac{(1+\sin x)^2 + 2i\cos x(1+\sin x) - \cos^2\!x}{(1+\sin x)^2 + \cos^2x}$

. . $=\; \frac{(1+\sin x)^2 + 2i\cos x(1 + \sin x) - \cos^2\!x}{1 + 2\sin x + \underbrace{\sin^2x + \cos^2x}_{\text{This is 1}}}$

. . $=\;\frac{(1+\sin x)^2 + 2i\cos x(1+\sin x) - \cos^2\!x}{2 + 2\sin x}$

. . $=\;\frac{(1+\sin x)^2 + 2i\cos x(1+\sin x) - \cos^2\!x}{2(1 + \sin x)}$

. . $=\;\frac{(1+\sin x)^2}{2(1+\sin x)} + \frac{2i\cos x(1+\sin x)}{2(1+\sin x)} - \frac{\cos^2\!x}{2(1+\sin x)}$

. . $=\;\frac{1+\sin x}{2} + i\cos x - \frac{\cos^2\!x}{2(1+\sin x)}$

. . $=\;\frac{1+\sin x}{2} + i\cos x - \frac{\cos^2\!x}{2(1+\sin x)}\cdot\frac{1-\sin x}{1-\sin x}$

. . $=\;\frac{1+\sin x}{2} + i\cos x - \frac{\cos^2\!x(1-\sin x)}{2(1-\sin^2\!x)}$

. . $=\;\frac{1+\sin x}{2} + i\cos x - \frac{\cos^2\!x(1-\sin x)}{2\cos^2\!x}$

. . $=\;\frac{1+\sin x}{2} + i\cos x - \frac{1-\sin x}{2}$

. . $=\;\tfrac{1}{2} + \tfrac{1}{2}\sin x + i\cos x - \tfrac{1}{2} + \tfrac{1}{2}\sin x$

. . $=\;\sin x + i\cos x$

Hence: . $\frac{1 + \sin x + i\cos x}{1 + \sin x - i\cos x} \;=\;\sin x + i\cos x$

Can you finish up?

**Prove It** · November 28th 2011, 03:38 PM

Originally Posted by princeps

Use Euler's formula :

$e^{ix}=\cos (x )+ i\cdot \sin(x)$ ,so you can rewrite expression into :

$\left(\frac{1+e^{ix}}{1+e^{-ix}}\right)^{n}=\left(\frac{1+e^{ix}}{1+\frac{1}{e ^{ix}}}\right)^{n}$ ...etc.

Final result should be :

$\cos (nx) + i\cdot \sin (nx)$

Except the expression was $\displaystyle \left(\frac{1 + \sin{x} + i\cos{x}}{1 + \sin{x} - i\cos{x}}\right)^n$ , not $\displaystyle \left(\frac{1 + \cos{x} + i\sin{x}}{1 + \cos{x} - i\sin{x}}\right)^n$ ...

**princeps** · November 28th 2011, 07:06 PM

Originally Posted by Prove It

Except the expression was $\displaystyle \left(\frac{1 + \sin{x} + i\cos{x}}{1 + \sin{x} - i\cos{x}}\right)^n$ , not $\displaystyle \left(\frac{1 + \cos{x} + i\sin{x}}{1 + \cos{x} - i\sin{x}}\right)^n$ ...

You are absolutely right. I have misread the question...So, let me try again

$\left(\frac{1+i(\cos x-i\sin x}{1-i(\cos x+i\sin x)}\right)^n=\left(\frac{1+ie^{-ix}}{1-ie^{ix}}\cdot \frac{1+ie^{ix}}{1+ie^{ix}}\right)^n =\left(\frac{1+ie^{ix}+ie^{-ix}-1}{1+e^{i2x}}\right)^n=$

$=\left(\frac{i(e^{ix}+\frac{1}{e^{ix}})}{1+e^{i2x} }\right)^n=\left(\frac {i(e^{i2x}+1)}{e^{ix}\cdot (1+e^{i2x})}\right)^n=\left(\frac{i}{e^{ix}}\right )^n=\left(ie^{-ix}\right)^n=$

$=i^{n}\cdot e^{-inx}=i^n\cdot \left(\cos(nx)-i\sin(nx)\right)$

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