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    Question from C3 edexcel Jan 07:
    Given that y=arccosx, express arcsinx in terms of y
    -1<x<1, 0<y<pi

    The answer is Pi/2 -y = arcsinx, but I'm not sure how to get there.
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    (Original post by PuffyPenguin)
    Question from C3 edexcel Jan 07:
    Given that y=arccosx, express arcsinx in terms of y
    -1<x<1, 0<y<pi

    The answer is Pi/2 -y = arcsinx, but I'm not sure how to get there.
    Hint : cos(x) = sin(pi/2 - x)

    You're expected to know this identity (and the equivalent sin(x) identity) but you'll rarely need it in an exam. This was one of the worst answered questions in any A Level maths exam.
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    (Original post by Notnek)
    Hint : cos(x) = sin(pi/2 - x)

    You're expected to know this identity (and the equivalent sin(x) identity) but you'll rarely need it in an exam. This was one of the worst answered questions in any A Level maths exam.
    Thanks
    I haven't actually heard of that identity before - just to check, the sin identity is sin(x)= cos(x-pi/2)
    This is because sin(x) is cos(x) shifted Pi/2 to the right, and it's minus Pi/2 because it's in the bracket so signs are reversed, right?
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    (Original post by PuffyPenguin)
    Thanks
    I haven't actually heard of that identity before - just to check, the sin identity is sin(x)= cos(x-pi/2)
    This is because sin(x) is cos(x) shifted Pi/2 to the right, and it's minus Pi/2 because it's in the bracket so signs are reversed, right?
    That’s correct. You may not have seen the identity written out but you’ve probably been aware of it since you knew that cos was a translation of sin. Also things like sin(30)=cos(60) are applications of the identity.
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    I remember using this in my early mechanics classes!

    Note we can see this effect with a right-angled triangle.

    Also can be seen using trigonometric identities:

     \displaystyle \cos((\pi / 2) -x) = \cos(\pi / 2) \cos(x) + \sin(\pi / 2) \sin(x) \, = \, \sin(x).

    So the identity is valid for the given interval of y in the question.
 
 
 
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