1. 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

2. Prove that:

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

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

$\displaystyle \left[ \frac{\cos (A)+i \sin (A)}{\sin(A)+i\cos(A)} \right]^4 =$$\displaystyle \left[ \frac{(\cos (A)+i \sin (A))(\sin(A)-i\cos(A))}{(\sin(A)+i\cos(A))(\sin(A)-i\cos(A))} \right]^4 \displaystyle =\left[ \frac{2\sin(A)\cos (A)-i (2 (\cos(A))^2-1)}{1} \right]^4 \displaystyle =\left[ \sin(2A)-i \cos(2A) \right]^4 multiply top and bottom inside the bracket by \displaystyle i: \displaystyle =\left[ \frac{\cos(2A)+i \sin(2A)}{i} \right]^4 \displaystyle =\frac{(\cos(2A)+i \sin(2A))^4}{i^4} Then DeMoivre's theorem gives: \displaystyle =\frac{\cos(8A)+i \sin(8A)}{i^4}=\cos(8A)+i \sin(8A) RonL 3. CaptainBlack likes to complain how I make some solutions too complicated, well now I can complain he makes the solution too complicated. ------- Remember that, \displaystyle \mbox{cis} A=\cos A+i\sin A You have, \displaystyle \left( \frac{\cos A+i\sin A}{\sin A+i\cos A} \right)^4 Thus, using co-function identities, \displaystyle \left( \frac{\cos A+i\sin A}{\cos (\pi/2-A)+i\sin (\pi/2-A)} \right)^4 Thus, \displaystyle \left( \frac{\mbox{cis}A}{\mbox{cis}(\pi/2-A)}\right)^2 Using the rule, \displaystyle \frac{\mbox{cis}x}{\mbox{cis}y}=\mbox{cis}(x-y) You have, (watch those signs ) \displaystyle (\mbox{cis}(A-(\pi/2-A)))^4 Thus, \displaystyle (\mbox{cis}(2A-\pi/2))^4 By de-Moiver's theorem, \displaystyle \mbox{cis}(8A-2\pi)=\mbox{cis}8A 4. 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 RonL 5. Hello, Lalit! Just another approach . . . same punchline. Prove that: .\displaystyle \left[ \frac{\cos (A)+i \sin (A)}{\sin(A)+i\cos(A)} \right]^4 = \cos(8A)+i\sin(8A) As TPHacker pointed out: .\displaystyle \sin(A) + i\cos(A)\;=\;\cos\left(\frac{\pi}{2}-A\right) + i\sin\left(\frac{\pi}{2} - A\right) So we have: .\displaystyle \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: . . \displaystyle \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: .\displaystyle \frac{\cos(4A) + i\sin(4A)}{\cos(4A) - i\sin(4A)}\cdot\frac{\cos(4A) + i\sin(4A)}{\cos(4A) + i\sin(4A)} . . \displaystyle = \;\frac{\cos^2(4A) + i\cdot2\sin(4A)\cos(4A) + i^2\sin^2(4A)}{\cos^2(4A) - i^2\sin^2(4A)} . . \displaystyle =\;\frac{\left[\cos^2(4A) - \sin^2(4A)\right] + i\left[2\sin(4A)\cos(4A)\right]}{\sin^2(4A) + \cos^2(4A)} . . \displaystyle =\;\cos(8A) + i\sin(8A) 6. 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. 7. Hello, killer baby! 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: .\displaystyle [\cos(A) + i\sin(A)]^4 . . \displaystyle =\:\cos^4(A) + 4i\cos^3(A)\sin(A) - 6\cos^2(A)$$\displaystyle \sin^2(A) - 4i\cos(A)\sin^3(A) + \sin^4(A)$

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

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

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

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

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

Now u wrk on da denominator . . .

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

8. 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 $\displaystyle e^{iA}=cosA + isinA$
it will be easier to prove

Keep Smiling
Malay

9. thnx

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

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

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

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

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

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

Now u wrk on da denominator . . .

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

10. Originally Posted by killer baby
if you can guyz pls explain me d theorem or try to find out a different way.
The theorem says that for any positive integer,
$\displaystyle n$ you have that,
$\displaystyle [\cos x+i\sin x]^n=\cos nx+i\sin nx$

11. $\displaystyle $\left(\frac {\cos {A}+\iota \sin {A}}{\sin {A}+\iota \cos {A}}\right)^4$$
$\displaystyle =$\left(\frac {(\iota)^3(\cos {A}+\iota \sin {A})}{-\iota\sin {A}+ \cos {A}}\right)^4$$
$\displaystyle =$\left(\frac {(\iota)^3(\cos {A}+\iota \sin {A})}{\iota\sin {(-A)}+ \cos {A}}\right)^4$$
$\displaystyle =\left(\frac{e^{iA}}{e^{-iA}}\right)^4$
$\displaystyle =(e^{i2A})^4$
$\displaystyle =(e^{i8A}) =$\cos{8A}+\iota\sin{8A}$$