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    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.
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    Is it to do with the range of x and y values?
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    (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
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    (Original post by SeanFM)
    Please attach the question

    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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    That makes sense to me now. Thank you for putting so much effort into this. It's really helped.
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    (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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