help on solving....

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August 11th 2006, 05:56 AM
watsonmath

help on solving....

1) prove cos3A=4cos^3A-3cosA

2) prove 4cos2AsinAcosA =sin4A

6. find real number of x and y such that (2x-j3y)-(1+j5)x=3+j2
ans y=-17/3,x=3

Can anyone provide me the explainations for the above question. i tried hard but i am lost. Help needed.
August 11th 2006, 06:38 AM
galactus

For #6:

$x-3yj-5jx=3+2j$

Obviously, x must equal 3

$3-3yj-5j(3)=3+2j$

$3-3yj-15j=3+2j$

$-3y-15=2$

Solve for y.
August 11th 2006, 07:16 AM
ThePerfectHacker

Quote:

Originally Posted by watsonmath

1) prove cos3A=4cos^3A-3cosA

Express as,
$\cos (x+2x)$
Expand,
$\cos x\cos 2x-\sin x\sin 2x$
Double angle identity,
$\cos x(\cos^2 x-\sin^2 x)-\sin x (2\sin x\cos x)$
Thus,
$\cos^3 x-\cos x\sin^2 x-2\sin ^2x\cos x$
$\cos ^3x-\cos x(\sin^2 x+2\sin^2x)$
$\cos^3x-\cos x(3\sin ^2 x)$
$\cos^3x-\cos x(3(1-\cos^2x))$
$\cos^3x-\cos x(3-3\cos^2x)$
$\cos^3x-3\cos x+3\cos^3x$
$4\cos^3x-3\cos x$
August 11th 2006, 08:57 AM
Soroban

Hello, watsonmath!

You need these identities:
. . $\cos(A + B) \:=\:\cos(A)\cos(B) - \sin(A)\sin(B)$
. . $\cos(2A) \:=\:2\cos^2A - 1$
. . $\sin(2A) \:=\:2\sin(A)\cos(A)$

Quote:

1) Prove: . $\cos3\theta \;= \;4\cos^3\theta - 3\cos\theta$

We have: . $\cos3\theta \:=\:\cos(2\theta + \theta)$

. . . . . . . . . . . . $= \qquad\cos2\theta\cos\theta \quad- \quad\sin2\theta\sin\theta$

. . . . . . . . . . . . $=\:\overbrace{(2\cos^2\theta - 1)}\cos\theta - \overbrace{2\sin\theta\cos\theta}\cdot\sin\theta$

. . . . . . . . . . . . $= \;2\cos^3\theta - \cos\theta \;\;- \;\;2\sin^2\theta\cos\theta$

. . . . . . . . . . . . $= \;2\cos^3\theta - \cos\theta - 2\overbrace{(1-\cos^2\theta)}\cos\theta$

. . . . . . . . . . . . $= \;2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta$

. . . . . . . . . . . . $= \;4\cos^3\theta - 3\cos\theta$

Quote:

2) Prove: . $4\cos2\theta\sin\theta\cos\theta \: = \:\sin4\theta$

We have: . $2\cos2\theta\cdot\underbrace{2\sin\theta\cos\theta }$

. . . . . . $= \quad\underbrace{2\cos2\theta\cdot\sin2\theta}$

. . . . . . $=\qquad\sin4\theta$
August 11th 2006, 02:58 PM
watsonmath

BIG thanks guys...