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Math Help - More expressions

  1. #1
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    More expressions

    Based on the following information from a past paper:

    Since p = sin40 and q = cos110.

    and therefore sin140 = p and cos70 = - q

    One of the questions was:find an expression for cos140

    In the mark scheme the answer was:

    cos^2(X) + sin^2(X) = 1 (I understand the importance of using the trig identity)

    followed by: cos(140) = √(1 - sin^2(X)) (I got this far)

    this is the bit which I do not understand which was in the mark scheme:

    cos(140) = √(1 - p^2)

    How does p = sin x ?

    all the information that it gives is that p = sin40, so how can that be applied in general?
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  2. #2
    Super Member Quacky's Avatar
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    cos^2(x)+sin^2(x)=1.This is the identity, it is not being applied to the example. The mark scheme doesn't actually say this, it would seem, but you should say that x=140 to start.

    cos^2(140)+sin^2(140)=1

    cos^2(140)=1-sin^2(140)

    cos(140)=\sqrt{1-sin^2(140)}

    Now, you know that sin(140)=p

     <br />
cos(140)=\sqrt{1-(p)^2}<br />
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  3. #3
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    Quote Originally Posted by Quacky View Post
    cos^2(x)+sin^2(x)=1<--this is the identity, it is not being applied to the example. The mark scheme doesn't actually say this, it wouls seem, but you should say that x=140 to start.

    cos^2(140)+sin^2(140)=1

    cos^2(140)=1-sin^2(140)

    cos(140)=\sqrt{1-sin^2(140)}

    Now, you know that sin(140)=p

     <br />
cos(140)=\sqrt{1-(p)^2}<br />
    Of course it is! Thank you so much.
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