Author Topic: Trig functions (Read 3958 times) Tweet Share

m@tty · « **Reply #15 on:** April 09, 2010, 01:42:15 pm »

He made an error on the first line, where he said the difference of those two thing was equal to " $\cos\left(2x+\frac{\pi}{6}\right)$ ". Then he used the cosine compound angle formula ( $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ ).

Stroodle · « **Reply #16 on:** April 09, 2010, 02:32:50 pm »

Crap. Sorry; we did trig at the start of the year, and I seem to have forgotten how to do it already

I thought $cos(2x+\frac{\pi}{6})=cos(2x)cos(\frac{\pi}{6})-sin(2x)sin(\frac{\pi}{6})$ ??

*edit: were my solution incorrect? :

$x=\frac{\pi}{12}$ , $x=\frac{3\pi}{4}$ , $x=\frac{7\pi}{4}$ and $x=\frac{13\pi}{12}$

m@tty · « **Reply #17 on:** April 09, 2010, 02:57:42 pm »

I see what you did now :buck2: I thought in the first line you were writing out the equation darkphoenix gave to solve...

Now that I see that is not that, it is all correct. Just a little unclear because of my misunderstanding...

$\cos(2x+\frac{\pi}{6})$ is indeed equal to $\cos(2x)\cos\left(\frac{\pi}{6}\right)-\sin(2x)\sin\left(\frac{\pi}{6}\right)$

Stroodle · « **Reply #18 on:** April 09, 2010, 03:11:35 pm »

Oh, cool. Thought it was correct; spun out for a min there - thought I musta missed something really basic...

darkphoenix · « **Reply #19 on:** April 09, 2010, 04:29:07 pm »

Alright cool yeah i get it now. Thanks guys.

EDIT: One more:

$cos^2\theta = 2cos^2\frac{\theta}{2}, [0,360]$ (In degrees)

Stroodle · « **Reply #20 on:** April 09, 2010, 05:23:20 pm »

This is what I got....

First note $2cos^2\left ( \frac{\theta}{2}\right )=cos(\theta)+1$

$cos^2(\theta)-cos(\theta)-1=0$

let $cos(\theta)=x$

$x^2-x-1=0$

$\left (x-\frac{1+\sqrt{5}}{2}\right )\left (x-\frac{1-\sqrt{5}}{2}\right )=0$

$\theta=cos^{-1}\left (\frac{1-\sqrt{5}}{2}\right )$ and $\theta =360-cos^{-1}\left (\frac{1-\sqrt{5}}{2}\right )$

$\theta=128.2^{\circ}$ and $\theta=231.8^{\circ}$

Hopefully that's correct...

m@tty · « **Reply #21 on:** April 09, 2010, 06:29:37 pm »

$\cos^2(\theta)=2\cos^2\left(\frac{\theta}{2}\right) , \theta\in [0,360]^\circ$

$\left(\cos(\theta)\right)^2=2\cos^2\left(\frac{\theta}{2}\right)$

$\left(2\cos^2\left(\frac{\theta}{2}\right)-1\right)^2 =2\cos^2\left(\frac{\theta}{2}\right)$

$4\cos^4\left(\frac{\theta}{2}\right)-4\cos^2\left(\frac{\theta}{2}\right)+1=2\cos^2\left(\frac{\theta}{2}\right)$

Let $\cos^2\left(\frac{\theta}{2}\right)=x$

$4x^2-4x+1=2x$

$4x^2-6x+1=0$

$x=\frac{6\pm\sqrt{36-16}}{8}$

$x=\frac{3}{4} \pm \frac{\sqrt{5}}{4}$

Sub $x=\cos^2\left(\frac{\theta}{2}\right)$

$\cos^2\left(\frac{\theta}{2}\right)=\frac{1}{4}(3 \pm \sqrt{5})$

$\cos\left({\frac{\theta}{2}\right)=\pm\sqrt{\frac{1}{4}(3\pm \sqrt{5})}$

$\frac{\theta}{2}=\cos^{-1}\left(\pm\frac{1}{2} \sqrt{3 -\sqrt{5}}\right)$ (as $\frac{1}{2}\sqrt{3+\sqrt{5}}>1$ and obviously has no solutions.)

$\frac{1}{2}\sqrt{3-\sqrt{5}}\approx 0.44$

So,
$\frac{\theta}{2} \approx \cos^{-1}(0.44)$ , $\frac{\theta}{2}\approx\cos^{-1}(-0.44)$ .

There is only one solution for each of the above for $\frac{\theta}{2} \in [0,180]^\circ$ .

So, $\theta = 2\cdot \cos^{-1}\left(\pm\frac{1}{2} \sqrt{3-\sqrt{5}}\right) \approx 128.17, 231.83$

EDIT:
I'm going to come back later and fix this up.... The other way was muuuch better...

EDIT EDIT: Fixed, same as Stroodle's answers. Just discovered by an infinitely more arduous path

Stroodle · « **Reply #22 on:** April 09, 2010, 06:37:42 pm »

Woohoo. We got one of the same answers!!

m@tty · « **Reply #23 on:** April 09, 2010, 06:41:22 pm »

I made a mistake...

Stroodle · « **Reply #24 on:** April 09, 2010, 06:46:50 pm »

Was a hard question...

Our trig sac was all analysis, and full of questions like this. I got rank one with only 63%...

m@tty · « **Reply #25 on:** April 09, 2010, 07:11:06 pm »

That's pretty low for rank 1.. Must have been quite hard.

Fixed now. Your way was much better, Stroodle.

I forgot to take both the positive and negative square root at one point, hence only receiving one solution.

Stroodle · « **Reply #26 on:** April 09, 2010, 08:40:47 pm »

Easy to do. That's what I hate about trig - seems like half the time picking the long way or the short way is just a matter of chance. Just hope we don't pick the long way on exam questions..

darkphoenix · « **Reply #27 on:** April 10, 2010, 01:45:20 pm »

Quote from: Stroodle on April 09, 2010, 05:23:20 pm

This is what I got....

First note $2cos^2\left ( \frac{\theta}{2}\right )=cos(\theta)+1$

Thanks guys.

So you got this from the orginal equation of $2cos^2(\theta) -1 = cos(2\theta)$ ?

By diving by 2?

brightsky · « **Reply #28 on:** April 10, 2010, 02:02:33 pm »

$2 \cos^2 \left(\frac{\theta}{2}\right)$

$= \cos^2 \left(\frac{\theta}{2}\right) + \cos^2 \left(\frac{\theta}{2}\right)$

$= \cos^2 \left(\frac{\theta}{2}\right) + \sin^2 \left(\frac{\theta}{2}\right) + \cos^2 \left(\frac{\theta}{2}\right) - \sin^2 \left(\frac{\theta}{2}\right)$

$= 1 + \left(\cos^2 \left(\frac{\theta}{2}\right) - \sin^2 \left(\frac{\theta}{2}\right)\right)$

By the double angle formula (i.e. $\cos (2x) = \cos^2(x) - \sin^2(x)$ )

$= 1 + \cos(\theta)$

darkphoenix · « **Reply #29 on:** April 10, 2010, 02:45:23 pm »

Ah alrite thanks man

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