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    The diagram shows the graphs of y=cos-1(x) and y=tan-1(x), for -1<x<1 in each case. The graphs intersect at the point with coordinates (p,q).

    iii)Deduce that p^4 + p^2 - 1 = 0

    How do I do this?
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    (Original post by *TomTomSatNav*)
    The diagram shows the graphs of y=cos-1(x) and y=tan-1(x), for -1<x<1 in each case. The graphs intersect at the point with coordinates (p,q).

    iii)Deduce that p^4 + p^2 - 1 = 0

    How do I do this?
    What is tan(q) in terms of sin(q) and cos(q)? Hence what is the value of sin(q) in terms of p? Hence use everyone's favourite identity to finish it up.
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    (Original post by RDKGames)
    What is tan(q) in terms of sin(q) and cos(q)? Hence what is the value of sin(q) in terms of p? Hence use everyone's favourite identity to finish it up.
    Thanks for the reply.

    How do i get sin(q) in terms of p? isnt p=sin(q)
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    (Original post by *TomTomSatNav*)
    Thanks for the reply.

    How do i get sin(q) in terms of p? isnt p=sin(q)
    You know that tan(q)=p. You know that cos(q)=p. Again, what is tan(q) in terms of sin(q) and cos(q)? So you can then express sin(q) in terms of p
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    (Original post by RDKGames)
    You know that tan(q)=p. You know that cos(q)=p. Again, what is tan(q) in terms of sin(q) and cos(q)? So you can then express sin(q) in terms of p
    tysm ily (no homo) . Thanks for helping me out, I understand it now!
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    (Original post by *TomTomSatNav*)
    The diagram shows the graphs of y=cos-1(x) and y=tan-1(x), for -1<x<1 in each case. The graphs intersect at the point with coordinates (p,q).

    iii)Deduce that p^4 + p^2 - 1 = 0

    How do I do this?
    You could also use a right angle triangle with hypotenuse length 1 and an angle q, and use tan(q) = opposite/adjacent = p

    and

    cos(q) = adjacent/hypotenuse = p ( = adjacent).

    Then use pythagoras' theorem.
 
 
 
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