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    Hey guys, can you give me a helping hand with this question?

    Angle θ is obtuse
    Sinθ = (sqrt5)/3
    Work out the values of Cosθ

    The part where θ is obtuse is throwing me off a bit (sorry if it's hard to understand the question I don't know how to use latex)
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    (Original post by Retsek)
    Hey guys, can you give me a helping hand with this question?

    Angle θ is obtuse
    Sinθ = (sqrt5)/3
    Work out the values of Cosθ

    The part where θ is obtuse is throwing me off a bit (sorry if it's hard to understand the question I don't know how to use latex)
    You can find the magnitude of cos\theta by comparing sides of a triangle.

    The angle \theta is obtuse, so  90^{\circ}<\theta<180^{\circ}, now you need to consider what quadrant that is in and whether cos is positive or negative in that quadrant.
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    (Original post by Retsek)
    Hey guys, can you give me a helping hand with this question?

    Angle θ is obtuse
    Sinθ = (sqrt5)/3
    Work out the values of Cosθ

    The part where θ is obtuse is throwing me off a bit (sorry if it's hard to understand the question I don't know how to use latex)
    This is always a hard question for Level 2 FM students. You can use a few methods but you might like to use the identity:

    \sin^2 \theta + \cos^2 \theta = 1

    If you plug in the value for \sin \theta and rearrange then square root, you'll be left with two possible values, one positive and one negative (remember there are positive and negative square roots).

    So then how do you know which one to choose? You are told that \theta is obtuse so this tells you whether \cos \theta is always negative or always positive. Try drawing the \cos graph and see if you can work out which one it is. Alternatively, let me know if you've heard of the CAST diagram (most Level 2 FM students haven't) and I'll explain it using that.
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    Draw a right angle triangle. Now you know what 2 of the sides are since you know sinθ so you can easily calculate length of the third side (adjacent side).
    From there you can calculate cosθ and then the fact that the angle θ is o tide tells you that cosθ is negative since for π/2<θ<π we have cosθ<0.
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    (Original post by notnek)
    This is always a hard question for Level 2 FM students. You can use a few methods but you might like to use the identity:

    \sin^2 \theta + \cos^2 \theta = 1

    If you plug in the value for \sin \theta and rearrange then square root, you'll be left with two possible values, one positive and one negative (remember there are positive and negative square roots).

    So then how do you know which one to choose? You are told that \theta is obtuse so this tells you whether \cos \theta is always negative or always positive. Try drawing the \cos graph and see if you can work out which one it is. Alternatively, let me know if you've heard of the CAST diagram (most Level 2 FM students haven't) and I'll explain it using that.
    So looking at the Cosine graph, between 90 and 180 (obtuse) it is negative, but I must've gone wrong somewhere because one of my answers is positive and the other is less than -1 (which can't be on the graph?)

    I did:
    1 - (sqrt5)/3
    and
    -1 - (sqrt5/3)

    are equal to Cos^2θ
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    (Original post by Retsek)
    So looking at the Cosine graph, between 90 and 180 (obtuse) it is negative, but I must've gone wrong somewhere because one of my answers is positive and the other is less than -1 (which can't be on the graph?)

    I did:
    1 - (sqrt5)/3
    and
    -1 - (sqrt5/3)

    are equal to Cos^2θ
    You've made a mistake in your working. The identity you need is

    \cos^2 \theta = 1 - \sin^2 \theta

    You have \sin \theta = \frac{\sqrt{5}}{3} \ \ so  \ \ \sin^2 \theta = \frac{5}{9}.

    The place you need to think about positive/negative answers is when you go from \cos^2 \theta to \cos \theta in your working.

    Have another go and post your working if you get stuck.
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    (Original post by notnek)
    You've made a mistake in your working. The identity you need is

    \cos^2 \theta = 1 - \sin^2 \theta

    You have \sin \theta = \frac{\sqrt{5}}{3} \ \ so  \ \ \sin^2 \theta = \frac{5}{9}.

    The place you need to think about positive/negative answers is when you go from \cos^2 \theta to \cos \theta in your working.

    Have another go and post your working if you get stuck.
    Excuse the handwriting is this right?
    Name:  IMG_0699.jpg
Views: 140
Size:  525.2 KB
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    (Original post by Retsek)
    Excuse the handwriting is this right?
    Name:  IMG_0699.jpg
Views: 140
Size:  525.2 KB
    Perfect
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    (Original post by notnek)
    Perfect
    Thank-you so much!
 
 
 
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