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Edexcel A2 C4 Mathematics June 2015 - Official Thread

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Original post by rayquaza17
They didn't say that? They said sec2(y)=2cos(2y)+1\displaystyle sec^2(y) = \frac{2}{cos(2y)+1}


ImageUploadedByStudent Room1434206885.755949.jpg

By implication, this suggests that they are referencing cos^2(y), right?


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Original post by kingaaran
ImageUploadedByStudent Room1434206885.755949.jpg

By implication, this suggests that they are referencing cos^2(y), right?


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I think monii was incorrect when they said about integrating cos^2(y) because you can actually integrate that, but you don't have to do that in this question. (Sorry monii)

Post 993 has the correct answer for this question.
Original post by kingaaran
ImageUploadedByStudent Room1434206885.755949.jpg

By implication, this suggests that they are referencing cos^2(y), right?


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Not sure how well this has come out by hope it's of use.
Original post by rayquaza17
I think monii was incorrect when they said about integrating cos^2(y) because you can actually integrate that, but you don't have to do that in this question. (Sorry monii)

Post 993 has the correct answer for this question.


To rectify, the only way to integrate cos2(y)cos^2(y) is by using the double angle formulae, or is there another 'FP3' method?



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Original post by NikolaT
Not sure how well this has come out by hope it's of use.


Oh, you were talking about the integral of 1/cos^2(y) :redface:


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Original post by kingaaran
To rectify, the only way to integrate cos2(y)cos^2(y) is by using the double angle formulae, or is there another 'FP3' method?

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Yes only sensible way is the double angle formula. :smile:
I can think of only one other way to integrate it, but that involves complex numbers which is FP2. Tbh I think the double angle formula is easier in this case. :tongue:
Hi, can anyone tell me how to do 6(a) from Solomon A? The markscheme isn't very helpful. (not sure why 2sin3xsin2x = cosx-cos5x)


Thanks.
Original post by JamGrip
Hi, can anyone tell me how to do 6(a) from Solomon A? The markscheme isn't very helpful. (not sure why 2sin3xsin2x = cosx-cos5x)


Thanks.


cos(x) = cos(3x - 2x)
cos(5x) = cos(3x + 2x)

Does that help?
So I'm thinking I got 69 + method marks on the C3 paper, so what do people think I'll need roughly for the A*? Aiming for low-mid 60s I think to be safe
Original post by RobHunter97
So I'm thinking I got 69 + method marks on the C3 paper, so what do people think I'll need roughly for the A*? Aiming for low-mid 60s I think to be safe


Depends on the paper bro


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Original post by physicsmaths
Depends on the paper bro


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Obviously, but I mean approximately. 50s, 60s etc.
Original post by RobHunter97
Obviously, but I mean approximately. 50s, 60s etc.


60s
Original post by anonwinner
60s


Thank you :smile:
Original post by kingaaran
To rectify, the only way to integrate cos2(y)cos^2(y) is by using the double angle formulae, or is there another 'FP3' method?



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by parts is another way ...
Reply 1014
Hi, can anyone explain why integrating 6/x is 6lnx and not ln6x? Thanks
Original post by mdc96
Hi, can anyone explain why integrating 6/x is 6lnx and not ln6x? Thanks


check by differentiation ...
Original post by mdc96
Hi, can anyone explain why integrating 6/x is 6lnx and not ln6x? Thanks

If you differentiate ln(6x), it becomes 6/6x which cancels to 1/x. Therefore 6x must become 6ln(x) if you integrate
Original post by rayquaza17
Yes only sensible way is the double angle formula. :smile:
I can think of only one other way to integrate it, but that involves complex numbers which is FP2. Tbh I think the double angle formula is easier in this case. :tongue:


no... you can use integration by parts...in a cunning way
Reply 1018
Original post by RobHunter97
If you differentiate ln(6x), it becomes 6/6x which cancels to 1/x. Therefore 6x must become 6ln(x) if you integrate

Thanks, that makes sense, I had thought dy/dx of lnx was just f'(x)/ x not f'(x)/f(x).
Original post by mdc96
Hi, can anyone explain why integrating 6/x is 6lnx and not ln6x? Thanks


6/x = 6(1/x)

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