Author Topic: trig problem! (Read 631 times) Tweet Share

Highacheiver · « **on:** September 01, 2012, 02:01:13 pm »

okay i know this may seem easy but can someone pleas tell me the x intercepts and y interecept to this problem and how they got it?!

y=2cos2(x+ π/4)+sqrt(3)

it would be mush appreciated!

Lasercookie · « **Reply #1 on:** September 01, 2012, 02:29:17 pm »

Assuming that you meant $y=2cos(2(x+ \frac{\pi}{4}))+\sqrt{3}$ and not something slightly different. You haven't given a restricted domain, so I'm also assuming you want the general solution over R.

y-int:
$let \ x = 0$

$y=2cos(2(\frac{\pi}{4}))+\sqrt{3}$

$y=2cos(\frac{\pi}{2})+\sqrt{3}$

$y=\sqrt{3}$

x-int:

This process is nowhere near as lengthy on paper, since you don't need to state every single step (I tend to sketch out graphs too, which I'm not going to bother showing in this post here). Maybe you could simplify this a bit by making use of identities or using one of the slightly different methods.

$let \ y = 0$

$0=2cos(2(x + \frac{\pi}{4}))+\sqrt{3}$

$-\frac{\sqrt{3}}{2} = cos(2(x + \frac{\pi}{4})$

$let \ u = 2(x + \frac{\pi}{4})$

$-\frac{\sqrt{3}}{2} = cos(u)$

Solving for u:
cos is -ve in second and third quadrants.

reference angle: $\theta = \frac{\pi}{6}$

Second quadrant $u = \pi - \frac{\pi}{6} = \frac{5\pi}{6}+ 2\pi \cdot n$

Third quadrant: $u = \pi + \frac{\pi}{6} = \frac{7\pi}{6} + 2\pi \cdot n$

where n is an integer (we don't have a restricted domain, so there'll be infinitely many x-intercepts).

Now let's sub back in u.

$\frac{5\pi}{6}+ 2\pi \cdot n = 2(x + \frac{\pi}{4})$

$\frac{7\pi}{6}+ 2\pi \cdot n = 2(x + \frac{\pi}{4})$

Now let's solve for x:

$\frac{5\pi}{6}+ 2\pi \cdot n = 2(x + \frac{\pi}{4})$

$\frac{5\pi}{12}+ \pi \cdot n = x + \frac{\pi}{4}$

$\frac{5\pi}{12} - \frac{\pi}{4} + \pi \cdot n = x$

$x = \frac{\pi}{6}+ \pi \cdot n$

$\frac{7\pi}{6}+ 2\pi \cdot n = 2(x + \frac{\pi}{4})$

$\frac{7\pi}{12} + \pi \cdot n = x + \frac{\pi}{4}$

$x = \frac{7\pi}{12} - \frac{\pi}{4} + \pi \cdot n$

$x = \frac{\pi}{3} + \pi \cdot n$

So our x-intercepts over R are:
$x = \frac{\pi}{3} + \pi \cdot n$ and $x = \frac{\pi}{6}+ \pi \cdot n$ where n is an integer

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Author Topic: trig problem! (Read 631 times) Tweet Share

Highacheiver

trig problem!

Lasercookie

Re: trig problem!

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