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Trig. function problem

Hi everyone,

I have been asked to solve the equation in the attachment, between 0 and 2 pi. I have an answer of 5/12 pi (which is right), but I thought 19/12 pi would solve it, but it doesn't.

Why is this - because the cos graph for radians says that it should?
Screen Shot 2016-03-30 at 19.30.44.png9.6KB
Reply 1
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Zacken
22
Original post by Electrogeek
Hi everyone,

I have been asked to solve the equation in the attachment, between 0 and 2 pi. I have an answer of 5/12 pi (which is right), but I thought 19/12 pi would solve it, but it doesn't.

Why is this - because the cos graph for radians says that it should?


We have cos⁡(x−π4)=3/2\cos \left(x - \frac{\pi}{4}\right) = \sqrt{3}/2 so this means that we have since, sqrt(3)/2 is positive:

x−π4=π6 or 2π−π6\displaystyle x - \frac{\pi}{4} = \frac{\pi}{6} \, \text{or} \, 2\pi - \frac{\pi}{6}.
Reply 2
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Electrogeek
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8
Cool. So am I right in saying 5/12 pi is the only answer between 0 and 2 pi?
Reply 3
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Zacken
22
Original post by Electrogeek
Cool. So am I right in saying 5/12 pi is the only answer between 0 and 2 pi?


Well... not quite, remember that if π/6\pi/6 is a solution, then so is π/6±2π\pi/6 \pm 2\pi. This gives x outside 2pi and 0, so we ignore it.

Buuut, x−π/4=2π−π/6−2π=−π/6x- \pi/4 = 2\pi - \pi/6 - 2\pi = -\pi/6 is also a solution...
Reply 4
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Zacken
22
Original post by Electrogeek
Cool. So am I right in saying 5/12 pi is the only answer between 0 and 2 pi?


Whilst I've got a few minutes, I'll write up a thing.

So you know that solutions to trig equations are given by using the CAST method, we take the arctrig function of the absolute value of the RHS, in this case: cos⁡−132=π6\cos^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6}.

Now our solutions are given by the infinite list, we can split this into two infinite lists, one is:

x−π4=π6,π6+2π,π6−2π,π6+4pi,π6−4pi,π6±2npi,…x - \frac{\pi}{4} = \frac{\pi}{6}, \frac{\pi}{6} + 2\pi, \frac{\pi}{6} - 2\pi, \frac{\pi}{6} + 4pi, \frac{\pi}{6} - 4pi, \frac{\pi}{6} \pm 2npi, \ldots

And the other is:

x−π4=2π−π6⏟11π/6,11π6±2π,11π6±4π,11π6 pm6pi,…x - \frac{\pi}{4} = \underbrace{2\pi - \frac{\pi}{6}}_{11\pi/6}, \frac{11\pi}{6} \pm 2\pi, \frac{11\pi}{6} \pm 4\pi, \frac{11\pi}{6}\ pm 6pi, \ldots

Side note: Getting the idea? The reason why this is so is because cos⁡α\cos \alpha is a 2π2\pi-periodic function so we have: cos⁡α=cos⁡(α±2π)\cos \alpha = \cos (\alpha \pm 2\pi) this is true for sin⁡α\sin \alpha as well - tangent is slightly different in the sense that it is a π\pi-periodic function, so tan⁡α=tan⁡(α±π)\tan \alpha = \tan (\alpha \pm \pi) and instead of adding ±2π\pm 2\pi to each solution, you'd add ±π\pm \pi.

Anyways, the first few solutions are x−π4=−11π6,−π6,π6,11π6,13π6,23π6x - \frac{\pi}{4} = -\frac{11\pi}{6}, -\frac{\pi}{6}, \frac{\pi}{6}, \frac{11\pi}{6}, \frac{13\pi}{6}, \frac{23\pi}{6}

So adding π/4\pi/4 to both sides and throwing anything that isn't in our range gives us: x=π12,5π12x = \frac{\pi}{12}, \frac{5\pi}{12}
Reply 5
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Electrogeek
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Original post by Zacken
Well... not quite, remember that if π/6\pi/6 is a solution, then so is π/6±2π\pi/6 \pm 2\pi. This gives x outside 2pi and 0, so we ignore it.

Buuut, x−π/4=2π−π/6−2π=−π/6x- \pi/4 = 2\pi - \pi/6 - 2\pi = -\pi/6 is also a solution...

Cool - I understand now. Thanks. :smile:
Reply 6
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Zacken
22
Original post by Electrogeek
Cool - I understand now. Thanks. :smile:


I typed up something above if you want to have a look. :smile:
Reply 7
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Electrogeek
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8
Original post by Zacken
I typed up something above if you want to have a look. :smile:

So... would these be right for these equations? (answers are in the attachment)
Screen Shot 2016-03-30 at 21.18.09.png17.7KB
Reply 8
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Zacken
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Original post by Electrogeek
So... would these be right for these equations? (answers are in the attachment)


Assuming you mean between 0 and 2pi, I agree with the one for tan⁡\tan but not for sin⁡\sin you should have four solutions for the sin one.
Reply 9
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Electrogeek
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Original post by Zacken
Assuming you mean between 0 and 2pi, I agree with the one for tan⁡\tan but not for sin⁡\sin you should have four solutions for the sin one.

Are the two given for sin right so far?

Or is is -pi/12 (+/- pi), and 7/6 pi (+/- pi)?
(edited 8 years ago)
Reply 10
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Zacken
22
Original post by Electrogeek
Are the two given for sin right so far?

Or is is -pi/12 (+/- pi), and 7/6 pi (+/- pi)?


Should be 2x=2π−π6,2π−π6±2π,π+π6,π+π6±2π2x = 2\pi - \frac{\pi}{6}, 2\pi - \frac{\pi}{6} \pm 2\pi, \pi + \frac{\pi}{6}, \pi + \frac{\pi}{6} \pm 2\pi

and then throw away everything you don't want.
Original post by Zacken
Should be 2x=2π−π6,2π−π6±2π,π+π6,π+π6±2π2x = 2\pi - \frac{\pi}{6}, 2\pi - \frac{\pi}{6} \pm 2\pi, \pi + \frac{\pi}{6}, \pi + \frac{\pi}{6} \pm 2\pi

and then throw away everything you don't want.


Thank you! Makes a lot more sense now. :smile:
Reply 12
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Zacken
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Original post by Electrogeek
Thank you! Makes a lot more sense now. :smile:


No problem.

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