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Edexcel C2 Trigonometry Help

I'm doing a Silver Tiered C2 Paper
https://a44694f152ee562e38a9454969d2b9a549a03543.googledrive.com/host/0B1ZiqBksUHNYOTR1Zk42aXdIazg/05%20Silver%201%20-%20C2%20Edexcel.pdf
Need help on Question 4. I got 3 out of 7 marks on this question and this is what I did to achieve that:
cos (3x-10) = -0.4
3x-10 = cos^-1(-0.4) = 113.6
3x-10 = 113.6
3x = 113.6+10
3x = 123.6
x = 41.2

I don't understand the steps after that would help me achieve the additional 4 marks. I know what to do after, just don't understand why we need to do that, if that makes sense...?
Reply 1
Avatar for B_9710
B_9710
18
There's more than one answer in the specified range. You only put one answer.
Reply 2
Avatar for Zacken
Zacken
22
Original post by ravioliyears
...


You're asked to find all the solutions. You've only found one.

It's kind of like me asking you to solve x2=9x^2 = 9 and you saying, x=3x=3. I wouldn't give you the marks because you haven't given me all the solutions (x=±3x = \pm 3).

That's the case here.

If cos⁡x=α\cos x = \alpha then x=cos⁡−1αx = \cos^{-1} \alpha is a solution, but you need to find the others.

Here's an image to convince you:



The solutions are the points of intersection and you've only found one of them.

Remember that if 3x−10=cos⁡−1(−0.4)3x - 10= \cos^{-1} (-0.4) is a solution, then so is 3x−10=cos⁡−1(−0.4)+2π3x -10 = \cos^{-1} (-0.4) + 2\pi and +4π+4\pi, etc...

Because cos⁡\cos is a 2pi periodic function, that is, if you add 2pi to its argument, it gives the same thing.
Original post by B_9710
There's more than one answer in the specified range. You only put one answer.


I know that. What I mean is I don't get why we have to do:
3x-10 = 360-113.58
and
3x-10 = 360+113.58
Reply 4
Avatar for B_9710
B_9710
18
Original post by ravioliyears
I know that. What I mean is I don't get why we have to do:
3x-10 = 360-113.58
and
3x-10 = 360+113.58


Well cos⁡(360−x)=cos⁡x \cos (360-x)=\cos x . Draw the graph of y=cosx to try and see why.
Reply 5
Avatar for Zacken
Zacken
22
Original post by ravioliyears
I know that. What I mean is I don't get why we have to do:
3x-10 = 360-113.58
and
3x-10 = 360+113.58


You might want to watch this and then this. :smile:
Original post by ravioliyears
I know that. What I mean is I don't get why we have to do:
3x-10 = 360-113.58
and
3x-10 = 360+113.58


That's how you find the other answers for what x x can be.

You ± 360
Original post by Zacken
You might want to watch this and then this. :smile:


Thanks. Not a fan of using the CAST method. How could I answer it graphically? Sorry - I feel like I'm asking such obvious questions that I should know the answer to.
Reply 8
Avatar for Zacken
Zacken
22
Original post by ravioliyears
Thanks. Not a fan of using the CAST method. How could I answer it graphically? Sorry - I feel like I'm asking such obvious questions that I should know the answer to.


No worries. I avoid the CAST method as well. You should be able to see from the graph that cos⁡\cos (and sin⁡\sin for that matter) are 2π2\pi periodic functions. That is, it repeats itself after every 2π2\pi radians. So if you have cos⁡θ=α\cos \theta = \alpha then not only is θ=arccos⁡α\theta = \arccos \alpha a solution but so is θ=arccos⁡α±2π\theta = \arccos \alpha \pm 2\pi and indeed, arccos⁡α±4π\arccos \alpha \pm 4\pi, ±6π\pm 6\pi... you get the gist.

Also, with cos⁡\cos - the graph should reveal that cos⁡θ=cos⁡(2π−θ)\cos \theta = \cos (2\pi - \theta) so if θ=arccos⁡α\theta = \arccos \alpha is a solution,then so is θ=2π−α\theta = 2\pi - \alpha.
Original post by Zacken
No worries. I avoid the CAST method as well. You should be able to see from the graph that cos⁡\cos (and sin⁡\sin for that matter) are 2π2\pi periodic functions. That is, it repeats itself after every 2π2\pi radians. So if you have cos⁡θ=α\cos \theta = \alpha then not only is θ=arccos⁡α\theta = \arccos \alpha a solution but so is θ=arccos⁡α±2π\theta = \arccos \alpha \pm 2\pi and indeed, arccos⁡α±4π\arccos \alpha \pm 4\pi, ±6π\pm 6\pi... you get the gist.

Also, with cos⁡\cos - the graph should reveal that cos⁡θ=cos⁡(2π−θ)\cos \theta = \cos (2\pi - \theta) so if θ=arccos⁡α\theta = \arccos \alpha is a solution,then so is θ=2π−α\theta = 2\pi - \alpha.

Thanks, Zain! I get it now :woo: Thanks again :biggrin:
Reply 10
Avatar for Zacken
Zacken
22
Original post by ravioliyears
Thanks, Zain! I get it now :woo: Thanks again :biggrin:


No worries. :-)

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