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C3 trig question help

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-maths-papers/worked-solution/worked-solution.php?paper_id=33&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.
(edited 7 years ago)
Jan07Q8.png21.0KB
Is it to do with the range of x and y values?
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.


Please attach the question :h:
Original post by SeanFM
Please attach the question :h:



Sorry! I forgot. I've added it now.
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Sorry but I think siny =/= cos y+90, siny = cos90-y and that is why
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Original post by Rafoleeno
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That makes sense to me now. Thank you for putting so much effort into this. It's really helped. :smile:
Original post by etherealinsanity
That makes sense to me now. Thank you for putting so much effort into this. It's really helped. :smile:


No problem good luck!

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