# Can someone help ASAP

• Jun 19th 2006, 09:31 AM
lalitchugh
Can someone help ASAP
Hi,

Can someone tell me the step by step solution of question in the attachment ASAP.

Note that i am a beginner and want to learn it before going to next standard.

Step by step solution will help me in understanding.

Lalit Chugh
• Jun 20th 2006, 08:09 AM
CaptainBlack
Prove that:

$
\left[ \frac{\cos (A)+i \sin (A)}{\sin(A)+i\cos(A)} \right]^4 = \cos(8A)+i\sin(8A)
$

Multiply top and bottom inside the bracket on the left hand side by the
complex conjugate of what is on the bottom:

$
\left[ \frac{\cos (A)+i \sin (A)}{\sin(A)+i\cos(A)} \right]^4 =$
$\left[ \frac{(\cos (A)+i \sin (A))(\sin(A)-i\cos(A))}{(\sin(A)+i\cos(A))(\sin(A)-i\cos(A))} \right]^4$

$
=\left[ \frac{2\sin(A)\cos (A)-i (2 (\cos(A))^2-1)}{1} \right]^4
$

$
=\left[ \sin(2A)-i \cos(2A) \right]^4
$

multiply top and bottom inside the bracket by $i$:

$
=\left[ \frac{\cos(2A)+i \sin(2A)}{i} \right]^4
$

$
=\frac{(\cos(2A)+i \sin(2A))^4}{i^4}
$

Then DeMoivre's theorem gives:

$
=\frac{\cos(8A)+i \sin(8A)}{i^4}=\cos(8A)+i \sin(8A)
$

RonL
• Jun 20th 2006, 08:23 AM
ThePerfectHacker
CaptainBlack likes to complain how I make some solutions too complicated, well now I can complain he makes the solution too complicated.
-------
Remember that,
$\mbox{cis} A=\cos A+i\sin A$
You have,
$\left( \frac{\cos A+i\sin A}{\sin A+i\cos A} \right)^4$
Thus, using co-function identities,
$\left( \frac{\cos A+i\sin A}{\cos (\pi/2-A)+i\sin (\pi/2-A)} \right)^4$
Thus,
$\left( \frac{\mbox{cis}A}{\mbox{cis}(\pi/2-A)}\right)^2$
Using the rule,
$\frac{\mbox{cis}x}{\mbox{cis}y}=\mbox{cis}(x-y)$
You have, (watch those signs :eek: )
$(\mbox{cis}(A-(\pi/2-A)))^4$
Thus,
$(\mbox{cis}(2A-\pi/2))^4$
By de-Moiver's theorem,
$\mbox{cis}(8A-2\pi)=\mbox{cis}8A$
• Jun 20th 2006, 09:23 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
CaptainBlack likes to complain how I make some solutions too complicated, well now I can complain he makes the solution too complicated.
-------

Measured in lines of TeX I make the count 6 lines in my post (I'm
not counting the first its just the statement of the problem so I
can see where I am going - I hate problems in attachments I like
to be able to see them without clicking).

The count in PH's post is 8, so by that measure mine is less complex :cool:

RonL
• Jun 20th 2006, 02:25 PM
Soroban
Hello, Lalit!

Just another approach . . . same punchline.

Quote:

Prove that: . $\left[ \frac{\cos (A)+i \sin (A)}{\sin(A)+i\cos(A)} \right]^4 = \cos(8A)+i\sin(8A)$

As TPHacker pointed out: . $\sin(A) + i\cos(A)\;=\;\cos\left(\frac{\pi}{2}-A\right) + i\sin\left(\frac{\pi}{2} - A\right)$

So we have: . $\frac{[\cos(A) + i\sin(A)]^4}{[\cos(\frac{\pi}{2} - A) + i\sin(\frac{\pi}{2}-A)]^4}$

Apply DeMoivre's Theorem to the numerator and denominator:

. . $\frac{\cos(4A) + i\sin(4A)}{\cos(2\pi - 4A) + i\sin(2\pi - 4A)} \;= \;\frac{\cos(4A) + i\sin(4A)}{\cos(4A) - i\sin(4A)}
$

Rationalize: . $\frac{\cos(4A) + i\sin(4A)}{\cos(4A) - i\sin(4A)}\cdot\frac{\cos(4A) + i\sin(4A)}{\cos(4A) + i\sin(4A)}
$

. . $= \;\frac{\cos^2(4A) + i\cdot2\sin(4A)\cos(4A) + i^2\sin^2(4A)}{\cos^2(4A) - i^2\sin^2(4A)}$

. . $=\;\frac{\left[\cos^2(4A) - \sin^2(4A)\right] + i\left[2\sin(4A)\cos(4A)\right]}{\sin^2(4A) + \cos^2(4A)}$

. . $=\;\cos(8A) + i\sin(8A)$

• Jun 22nd 2006, 11:48 PM
killer baby
hey
is dere any other method widout using demoivre's theroem as i dnt knw dis theorem . if you can guyz pls explain me d theorem or try to find out a different way. :o
• Jun 23rd 2006, 05:01 AM
Soroban
Hello, killer baby!

Quote:

is dere any other method widout using demoivre's theroem as i dnt knw dis theorem.
if you can guyz pls explain me d theorem or try to find out a different way.
If u dnt knw DeMoivre's Theorem, u shudn't b workn on dis problem.

The only alternative is to expand everything . . . and simplify
. . . but it's very very long and tedious.

The numerator is: . $[\cos(A) + i\sin(A)]^4$
. . $=\:\cos^4(A) + 4i\cos^3(A)\sin(A) - 6\cos^2(A)$ $\sin^2(A) - 4i\cos(A)\sin^3(A) + \sin^4(A)$

. . $= \;[\cos^4(A) - 6\cos^2(A)\sin^2(A) + \sin^4(A)] +$ $4i[\cos^3(A)\sin(A) - \cos(A)\sin^3(A)]$

. . $=\;\left[\cos^4(A) - 2\cos^2(A)\sin^2(A) + \sin^4(A) - 4\cos^2(A)\sin^2(A)\right]$ $+ 4i\cos(A)\sin(A)\left[\cos^2(A) - \sin^2(A)\right]$

. . $= \;\left[\cos^2(A) - \sin^2(A)\right]^2 -$ $\left[4\sin^2(A)\cos^2(A)\right] +$ $2i\left[2\sin(A)\cos(A)\right]\left[\cos^2(A) - \sin^2(A)\right]$

. . $= \;\underbrace{\cos^2(2A) - \sin^2(2A)} \:+ \:i\underbrace{2\sin(2A)\cos(2a)}$

. . $= \;\cos(4a) + i\sin(4a)$

Now u wrk on da denominator . . .

If u say, "Howja du dat?", Ive totally wasted my time . . .
• Jun 23rd 2006, 05:21 AM
malaygoel
Quote:

Originally Posted by ThePerfectHacker
CaptainBlack likes to complain how I make some solutions too complicated, well now I can complain he makes the solution too complicated.

If we use $e^{iA}=cosA + isinA$
it will be easier to prove

Keep Smiling
Malay
• Jun 23rd 2006, 07:44 AM
killer baby
thnx
Quote:

Originally Posted by Soroban
Hello, killer baby!

If u dnt knw DeMoivre's Theorem, u shudn't b workn on dis problem.

The only alternative is to expand everything . . . and simplify
. . . but it's very very long and tedious.

The numerator is: . $[\cos(A) + i\sin(A)]^4$
. . $=\:\cos^4(A) + 4i\cos^3(A)\sin(A) - 6\cos^2(A)$ $\sin^2(A) - 4i\cos(A)\sin^3(A) + \sin^4(A)$

. . $= \;[\cos^4(A) - 6\cos^2(A)\sin^2(A) + \sin^4(A)] +$ $4i[\cos^3(A)\sin(A) - \cos(A)\sin^3(A)]$

. . $=\;\left[\cos^4(A) - 2\cos^2(A)\sin^2(A) + \sin^4(A) - 4\cos^2(A)\sin^2(A)\right]$ $+ 4i\cos(A)\sin(A)\left[\cos^2(A) - \sin^2(A)\right]$

. . $= \;\left[\cos^2(A) - \sin^2(A)\right]^2 -$ $\left[4\sin^2(A)\cos^2(A)\right] +$ $2i\left[2\sin(A)\cos(A)\right]\left[\cos^2(A) - \sin^2(A)\right]$

. . $= \;\underbrace{\cos^2(2A) - \sin^2(2A)} \:+ \:i\underbrace{2\sin(2A)\cos(2a)}$

. . $= \;\cos(4a) + i\sin(4a)$

Now u wrk on da denominator . . .

If u say, "Howja du dat?", Ive totally wasted my time . . .

• Jun 23rd 2006, 09:04 AM
ThePerfectHacker
Quote:

Originally Posted by killer baby
if you can guyz pls explain me d theorem or try to find out a different way. :o

The theorem says that for any positive integer,
$n$ you have that,
$[\cos x+i\sin x]^n=\cos nx+i\sin nx$
• Jun 23rd 2006, 11:35 PM
Navesh
$
$\left(\frac {\cos {A}+\iota \sin {A}}{\sin {A}+\iota \cos {A}}\right)^4$
$

$
=$\left(\frac {(\iota)^3(\cos {A}+\iota \sin {A})}{-\iota\sin {A}+ \cos {A}}\right)^4$
$

$
=$\left(\frac {(\iota)^3(\cos {A}+\iota \sin {A})}{\iota\sin {(-A)}+ \cos {A}}\right)^4$
$

$
=\left(\frac{e^{iA}}{e^{-iA}}\right)^4
$

$
=(e^{i2A})^4
$

$
=(e^{i8A})
=$\cos{8A}+\iota\sin{8A}$
$