Trig Question (1 Viewer)

Skeptyks · Sep 16, 2012

Just need some help with this, my question is on the last line of the latex.

$\sqrt{3}cos2t = 1+sin2t \\ cos(2t+\frac{pi}{6}) = \frac{1}{2} \\ t = n(pi) + \frac{pi}{12}\ or\ t=n(pi) - \frac{pi}{4}\ is\ the\ general\ solution\\ Later\ on\ for\ another\ part\ of\ the\ question\ when\ solving\ this,\\ t=0+\frac{pi}{12}\ or\ t=pi - \frac{pi}{4}\\ I\ do\ not\ understand\ this\ last\ line\ of\ working\ out.\ How\ does\ n\ become\ 0\ and\ 1?$

bii) Find the general solution to 3^(1/2)cos2t - sin2t = 1
c) Two particles, A and B start simultaneously from points on the x axis. At time t, their positions are given by
x(a) = 3^(1/2)cos2t and x(b) = 1+sin2t
i) state the amplitudes and periods of the motions
ii) Sketch on one diagram
iii) Find the times in the interval o<=t<=pi at which the particles occupy the same position and find their velocities at these times.

D94 · Sep 16, 2012

Post the whole question up.

Skeptyks · Sep 17, 2012

D94 said:
Post the whole question up.

Edited.

Drongoski · Sep 17, 2012

$\sqrt 3 cos2t - 1 sin2t = 1$

$\therefore \frac {\sqrt 3}{2} cos2t - \frac {1}{2} sin 2t = \frac {1}{2} $ ( 2 being $ \sqrt {(\sqrt 3)^2 + 1^2} )$

$\therefore cos \frac {\pi}{6} cos 2t - sin \frac {\pi}{6} sin 2t = cos ( \frac {\pi}{3} + 2k \pi )$ where $k = \pm 0,1,2, \cdots$

i.e. $cos (2t + \frac {\pi}{6}) = cos ( \pm \frac {\pi}{3} + 2k \pi)$

$\therefore 2t + \frac {\pi}{6} = \pm \frac {\pi}{3} + 2k \pi$

$\therefore t = \frac {\pi}{12} + k \pi$ or $k\pi - \frac { \pi}{4}$

etc etc

edit: amended

Skeptyks · Sep 17, 2012

Drongoski said:
$\sqrt 3 cos2t - 1 sin2t = 1$

$\therefore \frac {\sqrt 3}{2} cos2t - \frac {1}{2} sin 2t = \frac {1}{2} $ 2 being $ \sqrt {(\sqrt 3)^2 + 1^2}$

$\therefore cos \frac {\pi}{6} cos 2t - sin \frac {\pi}{6} sin 2t = cos ( \frac {\pi}{3} + 2k \pi )$ where $k = \pm 0,1,2, \cdots$

i.e. $cos (2t + \frac {\pi}{6}) = cos (\frac {\pi}{3} + 2k \pi)$

$\therefore 2t + \frac {\pi}{6}{ = \frac {\pi}{3} + 2k \pi$

$\therefore t = \frac {\pi}{12} + k \pi$

etc etc

Thanks dragonski but I'm still confused as to how to reach the answer of t = 3pi/4

Drongoski · Sep 17, 2012

Oh - my solution is incomplete. Now amended.

Skeptyks · Sep 17, 2012

Drongoski said:
Oh - my solution is incomplete. Now amended.

Thanks heaps drongoski. I have one more type of question that I'm having trouble with if you could possibly help me out with it.Considering the function f(x) = (x^3 + 8x)/8, evaluate the integral of f^-1(x) dx with limits 0 - 3.This is after graphing f(x) and its inverse. The answers say, the integrals measure the area of the shaded region = 6 - integral of f(x)dx with limits 0-2. I don't understand why it is " 6 - .... " and for most questions like this when they ask you to integrate the inverse function, I end up unable to answer them.Thanks again.Anyone :] ?edit:solved

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