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    a)Find the vale of arc sin 1/2 + arccos 1/2, giving your answer in radians in terms of \pi
    b) Given that angles A and B are defines as A=arcsin x and B= arccos x, -1\leqx\leq1 show that sin A=sinB
    c) Hence deduce that arcsin x + arccos x= \frac{\pi}{2}

    Okay so I have part a and b down, I think;

    a)arcsin (1/2) +arccos(1/2)=30+60=90

    b) A=arcsin x
    Thus x=sin A

    B=arccos x
    Thus x=cos B

    therefore sin A=cos B
    So, how do I proceed with part c?
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    (Original post by mintmocha)
    a)Find the vale of arc sin 1/2 + arccos 1/2, giving your answer in radians in terms of \pi
    b) Given that angles A and B are defines as A=arcsin x and B= arccos x, -1\leqx\leq1
    c) Hence deduce that arcsin x + arccos x= \frac{\pi}{2}



    So, how do I proceed with part c?
    Part (a) asked you to give your answer in terms of radians, whereas you gave your answer in degrees.

    You don't seem to have written out part (b) fully, what you've posted there is not a question.

    For part (c), maybe take the cosine of both sides first. Although I can't be sure what the question is looking for unless you post part (b) properly.
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    (Original post by mintmocha)
    a)Find the vale of arc sin 1/2 + arccos 1/2, giving your answer in radians in terms of \pi
    b) Given that angles A and B are defines as A=arcsin x and B= arccos x, -1\leqx\leq1
    c) Hence deduce that arcsin x + arccos x= \frac{\pi}{2}



    So, how do I proceed with part c?
    a) Needs to be in terms of  \pi

    Ok for c) you've said

    x=sinA
    x=cosA

    How would you convert cos A into a form of sin A? Think: the angles add up to 90 degrees, as already stated in part c.
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    draw a right angled triangle and label the sides - think of \frac{a}{c} as x in your equation

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    It is very similar to a question I just did:

    Edexcel C3, Jan 2007, Question 8.
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    anyone else ?
 
 
 
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