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Trig graphs help please :)

Hey guys,

Please could someone be kind enough to help me with the question in the attachment. Any help will be appreciated.
Thanks :smile:
my head hurts now..
Reply 2
Original post by Aty100
Hey guys,

Please could someone be kind enough to help me with the question in the attachment. Any help will be appreciated.
Thanks :smile:


Surely you've learnt function transformations? This is just of cosx\cos x shift by π2\frac{\pi}{2} radians to the right. It will have the same minimum and maximum values as cosx\cos x but the x coordinates of these extrema will be shift to the right as well. As will the roots.
Reply 3
Just derive it and set y' = 0
Edit: rip answer
Reply 4
Original post by GUMI
Just derive it and set y' = 0
Edit: rip answer


Heads up: you mean differentiate, not derive. :-)
f(x + a) --> translates graph horizontally by -a

apply that to your situation?
Reply 6
Original post by Zacken
Heads up: you mean differentiate, not derive. :-)

Thanks, and I didn't know we were talking about transformations, woops x)
Reply 7
Original post by GUMI
Thanks, and I didn't know we were talking about transformations, woops x)


You can do it your way as well. :smile:
Reply 8
Original post by Zacken
Surely you've learnt function transformations? This is just of cosx\cos x shift by π2\frac{\pi}{2} radians to the right. It will have the same minimum and maximum values as cosx\cos x but the x coordinates of these extrema will be shift to the right as well. As will the roots.


I actually haven't learnt function transformations :s-smilie: I did have a lesson the graphs but not any transformations. I'm not sure why the teachers skipped that :/
Reply 9
Original post by Aty100
I actually haven't learnt function transformations :s-smilie: I did have a lesson the graphs but not any transformations. I'm not sure why the teachers skipped that :/


That's very weird. Anyhow, cos(xπ2)\cos(x - \frac{\pi}{2}), why don't you let y=xπ2y = x - \frac{\pi}{2} and then all you have is cosy\cos y which you know how to find the roots of, the maximum and minimum, etc... once you've got all of those, then just back-substitute.
Reply 10
Original post by Zacken
That's very weird. Anyhow, cos(xπ2)\cos(x - \frac{\pi}{2}), why don't you let y=xπ2y = x - \frac{\pi}{2} and then all you have is cosy\cos y which you know how to find the roots of, the maximum and minimum, etc... once you've got all of those, then just back-substitute.


Thank you, I understand that bit now. However do I change the limits for part iii) ?
And if we do then does it become 0 + pi/2 < X < 2pi + pi/2 ??
Reply 11
Original post by Aty100
Thank you, I understand that bit now. However do I change the limits for part iii) ?
And if we do then does it become 0 + pi/2 < X < 2pi + pi/2 ??


Yes. :yes:
Reply 12
Original post by Zacken
Yes. :yes:


Yay, thank you :h:
Reply 13
Original post by Aty100
Yay, thank you :h:


Don't thank me, you did it all by yourself, well done! :biggrin:
Reply 14
Original post by Zacken
You can do it your way as well. :smile:


hehe :biggrin:

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