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Another C4 question

Hi.
https://983c9f06eb1f75af6e83364e092b06f1cb5c4277.googledrive.com/host/0B1ZiqBksUHNYZzd2VHZlOXdqZlE/January%202008%20QP%20-%20C4%20OCR.pdf
How is question 7(i) done.
I cant see how to get two equations.
Thanks
Original post by SamuelN98


Expand your brackets to (in gonna use x to replace theta)
Asinx + Acosx+Bcosx-Bsinx=4sinx
So you have 2 equations here, one in terms of cosx and one in terms of sinx
Ainx - Bsinx=4sinx
Acosx+Bcosx=0

Therefore you get A-B=4, A+B=0
Reply 2
Avatar for B_9710
B_9710
18
IF you expand it out you get (A−B)sin⁡θ+(A+b)cos⁡θ (A-B)\sin \theta + (A+b)\cos \theta .
You could also note that
B(cos⁡θ−sin⁡θ)=0B(\cos\theta - \sin\theta) = 0 when θ=45∘\theta = 45^{\circ}

Therefore
A(sin⁡45+cos⁡45)=4sin⁡45A(\sin{45} + \cos{45})=4\sin{45}
As such A=2A = 2

You can then get:
B(cos⁡θ−sin⁡θ)=2sin⁡θ−2cos⁡θB(\cos\theta - \sin\theta)=2\sin\theta - 2\cos\theta

Therefore B=−2B = -2

Thats just another way if you get stuck with the equations method (or expanding is a pain). Usually a longer way though :smile:
Reply 4
Avatar for Zacken
Zacken
22
In the same vein as above, using x=0x=0 immediately gives you A+B=0A+B= 0 and using x=π2x = \frac{\pi}{2} gets you A−B=4A-B=4.
Reply 5
Avatar for SamuelN98
SamuelN98
OP
5
Original post by Zacken
In the same vein as above, using x=0x=0 immediately gives you A+B=0A+B= 0 and using x=π2x = \frac{\pi}{2} gets you A−B=4A-B=4.


How did you chose suitable values of x to substitute in?
Reply 6
Avatar for Zacken
Zacken
22
Original post by SamuelN98
How did you chose suitable values of x to substitute in?


I wanted to choose values that made one of sine and cosine 0 in turn so that I could have A(one thing only) + B(one thing only) instead of A(sqrt(3)/2 + (1/2)) + B(blah) or the likes. So x=0,π2x=0, \frac{\pi}{2} fit perfectly.

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