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# C3 trig question help

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1. Hi,

I was doing the January 2007 C3 paper and got stuck on the last question. I looked on exam solutions
(http://www.examsolutions.net/a-level...3&solution=8.2) and he did it a different way to how I went about trying to answer the question. I've attached the question and answer below.

I started off with:
y=arccos(x) so x=cosy
I then thought about how cosy=sin(90-y), so x=sin(90-y) and arcsinx=90-y
but then I also thought about how cosy=sin(y+90) which would give arcsinx=y+90 which would be wrong. If I didn't do it the examsolutions way, how would I know whether to use sin(y+90) or sin(90-y)?

Also, I know I've used degrees here. Typing pi/2 made it look confusing.
Attached Images

2. Is it to do with the range of x and y values?
3. (Original post by etherealinsanity)
Hi,

I was doing the January 2007 C3 paper and got stuck on the last question. I looked on exam solutions and he did it a different way to how I went about trying to answer the question. I've attached the question and answer below.

I started off with:
y=arccos(x) so x=cosy
I then thought about how cosy=sin(90-y), so x=sin(90-y) and arcsinx=90-y
but then I also thought about how cosy=sin(y+90) which would give arcsinx=y+90 which would be wrong. If I didn't do it the examsolutions way, how would I know whether to use sin(y+90) or sin(90-y)?

Also, I know I've used degrees here. Typing pi/2 made it look confusing.
4. (Original post by SeanFM)

Sorry! I forgot. I've added it now.
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11. Sorry but I think siny =/= cos y+90, siny = cos90-y and that is why
12. Attachment 552976552978Attachment 552976552978552960
Attached Images

13. That makes sense to me now. Thank you for putting so much effort into this. It's really helped.
14. (Original post by etherealinsanity)
That makes sense to me now. Thank you for putting so much effort into this. It's really helped.
No problem good luck!

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