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OCR C3 Jun 2010 Trig

Hi guys,

My sister's stuck on a particular question on this paper, and after having a look, neither my dad nor I can see how to solve it! We've done the first part, but are having trouble with the second.

8(i) Express 3 cos x + 3 sin x in the form R cos (x-a)...


We've done this and got the right answer:

32cos(xπ4)3\sqrt{2}cos(x-\frac{\pi}{4})

The next part says:

8(ii) The expression T(x) is defined by

T(x)=83cosx+3sinxT(x)=\frac{8}{3cosx+3sinx}

Find the smallest possible value of x satisfying

T(3x)=896T(3x)=\frac{8}{9}\sqrt{6}

giving your answer in an exact form.


I understand that T(3x) isn't the same as 3(T(x)), so I used a "function machine" diagram and concluded that T(3x) = R cos (3x-a), but I get nowhere near the right answer when I try and solve it. I got about 0.61..., whereas it's pi/16. Can anyone shed any light?

Thanks a lot!
Reply 1
Using the previous part you should get T(3x)=a/cos(3x-pi/4) where a=8/(3sqrt(2))

Now just rearange to get cos(3x-pi/4)=some constant (should be easy to do this bit) and then solve how you would normaly solve a trig equation.
Original post by Tortious
Hi guys,

My sister's stuck on a particular question on this paper, and after having a look, neither my dad nor I can see how to solve it! We've done the first part, but are having trouble with the second.



We've done this and got the right answer:

32cos(xπ4)3\sqrt{2}cos(x-\frac{\pi}{4})

The next part says:



I understand that T(3x) isn't the same as 3(T(x)), so I used a "function machine" diagram and concluded that T(3x) = R cos (3x-a), but I get nowhere near the right answer when I try and solve it. I got about 0.61..., whereas it's pi/16. Can anyone shed any light?

Thanks a lot!


From the previous question try writing

T(x)=83cos(x)+3sin(x)=832cos(xπ4) T(x) = \frac{8}{3cos(x) + 3sin(x)} = \frac{8}{3\sqrt{2}cos(x - \frac{\pi}{4})}

Then

T(3x)=832cos(3xπ4)T(3x) = \frac{8}{3\sqrt{2}cos(3x - \frac{\pi}{4})}

Then solve

T(3x)=832cos(3xπ4)=896 T(3x) = \frac{8}{3\sqrt{2}cos(3x - \frac{\pi}{4})} =\frac{8}{9}\sqrt{6}

by rearranging and solving how you would normally for any trig function.

I have never heard of a "function machine" diagram in my entire life :redface:
Reply 3
Original post by Tortious
Hi guys,

My sister's stuck on a particular question on this paper, and after having a look, neither my dad nor I can see how to solve it! We've done the first part, but are having trouble with the second.



We've done this and got the right answer:

32cos(xπ4)3\sqrt{2}cos(x-\frac{\pi}{4})

The next part says:



I understand that T(3x) isn't the same as 3(T(x)), so I used a "function machine" diagram and concluded that T(3x) = R cos (3x-a), but I get nowhere near the right answer when I try and solve it. I got about 0.61..., whereas it's pi/16. Can anyone shed any light?

Thanks a lot!


I'm getting pi/36 and not pi/16. ???
Reply 4
Original post by TheIrrational
I have never heard of a "function machine" diagram in my entire life :redface:


Heh, I'm sure you'll have seen them. It's this:

Function Machine



It's a diagrammatic way of expressing "to make x into the result, subtract alpha, cos everything and then multiply by R".

From the previous question try writing

T(x)=83cos(x)+3sin(x)=832cos(xπ4) T(x) = \frac{8}{3cos(x) + 3sin(x)} = \frac{8}{3\sqrt{2}cos(x - \frac{\pi}{4})}

Then

T(3x)=832cos(3xπ4)T(3x) = \frac{8}{3\sqrt{2}cos(3x - \frac{\pi}{4})}

Then solve

T(3x)=832cos(3xπ4)=896 T(3x) = \frac{8}{3\sqrt{2}cos(3x - \frac{\pi}{4})} =\frac{8}{9}\sqrt{6}

by rearranging and solving how you would normally for any trig function.


I think I see what you've done, but if T(x) can be written in R cos(x-a) format, why is this then put as the denominator of the "8/..." fraction which represents T(x)? Aren't the fraction and "R cos(x-a)" alternative ways of expressing the same thing? :erm:
Reply 5
Original post by steve2005
I'm getting pi/36 and not pi/16. ???


Sorry, just checked the mark scheme and it is pi/36. :colondollar:

Did you use the same method as TheIrrational?
Reply 6
Original post by Tortious
Sorry, just checked the mark scheme and it is pi/36. :colondollar:

Did you use the same method as TheIrrational?


Yes
Reply 7
Just realised I'm an idiot and have looked at this paper for so long I kept misreading the question. I'll work through now, but I think I see what I've done wrong. Thanks for all your help!
Original post by Tortious
Just realised I'm an idiot and have looked at this paper for so long I kept misreading the question. I'll work through now, but I think I see what I've done wrong. Thanks for all your help!


Good luck! Yes, I have seen the function machines before, not for many years though :biggrin:

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