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    Hey guys I wonder if you could help with the attached question:

    I'm gonna focus on question 5 here (question 4 I was able to do but only using compound angle formulae though the main answer in the mark scheme used sin(30)=1\2 and then subbing 't' for tanx I think which I didn't understand one bit )

    For question 5 part b I get the answer (63\55) but the mark scheme says 33\65 and gets sin(beta) as -5\13 whereas I got (5\13)?!

    Help would be appreciated... Thanks

    PS sorry for the long winded explanation :awesome:

    EDIT: I meant I got cos(alpha)=5\13 and the ms says -5\13 thus getting part a wrong also
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    Can you show your working for sin(b)?

    Also, what did you get for part a)?
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    (Original post by james22)
    Can you show your working for sin(b)?

    Also, what did you get for part a)?
    Apologies check my edit for changes

    For part a I got 2-√3

    For b its messy but basically I have a right angled triangle of hyp=13 opp=12 and adj=5 for the angle alpha
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    (Original post by Branny101)
    Hey guys I wonder if you could help with the attached question:

    I'm gonna focus on question 5 here (question 4 I was able to do but only using compound angle formulae though the main answer in the mark scheme used sin(30)=1\2 and then subbing 't' for tanx I think which I didn't understand one bit )

    For question 5 part b I get the answer (63\55) but the mark scheme says 33\65 and gets sin(beta) as -5\13 whereas I got (5\13)?!

    Help would be appreciated... Thanks

    PS sorry for the long winded explanation :awesome:

    EDIT: I meant I got cos(alpha)=5\13 and the ms says -5\13
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    As alpha is obtuse angle the cosine is negative
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    (Original post by ztibor)
    As alpha is obtuse angle the cosine is negative
    I understand the angle is obtuse but could you clarify? Why negative?

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    (Original post by Branny101)
    I understand the angle is obtuse but could you clarify? Why negative?

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    Think about the graph
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    (Original post by keromedic)
    Think about the graph
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    Still not getting it bro

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    (Original post by Branny101)
    Still not getting it bro

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    For an obtuse angle \theta, 90\leq \theta \leq 180.
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    (Original post by keromedic)
    For an obtuse angle \theta, 90\leq \theta \leq 180.
    360-(arccos(5\13))?!?!

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    I've not fully understood this either in previous exam questions.

    Having seen the responses, I presume if you take the negative of the acute angle, you end up hitting the cos graph between 90 and 180, which makes it obtuse.

    I presume the same applies for tan and sin? If we're told the angle is obtuse, we find the answer and multiply by -1?

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    Use the All Slappers Take Cash diagram for the sign.

    Divide a square into four and label each quadrant like this

    SA
    TC
    Start from the horizontal line between A and C and got anticlockwise.

    A - between 0 and 90, all sines, cosines and tans are positive
    S - between 90 and 180, only sine is positive
    T - between 180 and 270, only tan is positive
    C - between 270 and 360, only cosine is positive

    This can be verified by considering the graphs of all of these functions
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    (Original post by MarkProbio)
    I've not fully understood this either in previous exam questions.

    Having seen the responses, I presume if you take the negative of the acute angle, you end up hitting the cos graph between 90 and 180, which makes it obtuse.

    I presume the same applies for tan and sin? If we're told the angle is obtuse, we find the answer and multiply by -1?

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    Ahh right so its like some unknown rule... If you find out why this is please let me know


    Oh and thanks

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    (Original post by Angryification)
    Use the All Slappers Take Cash diagram for the sign.

    Divide a square into four and label each quadrant like this

    SA
    TC
    Start from the horizontal line between A and C and got anticlockwise.

    A - between 0 and 90, all sines, cosines and tans are positive
    S - between 90 and 180, only sine is positive
    T - between 180 and 270, only tan is positive
    C - between 270 and 360, only cosine is positive

    This can be verified by considering the graphs of all of these functions
    Thanks buddy! But if I were just to consider the graphs how would this work (for some reason I find the CAST or ASTC a bit more confusing)?!

    Sorry for being so pedantic...

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    (Original post by Branny101)
    I understand the angle is obtuse but could you clarify? Why negative?

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    As a right triangle has no obtuse angle so we define sine, cosine, tangent and cotangent
    of an angle in a unit circle.
    THe sign follows from this definition.
    Do you know the definition of cos x for any X?

    Some rules from the definition

    \cos \alpha=-\cos(\pi - \alpha) e.g \alpha>90^o
    \sin \alpha=\sin(\pi - \alpha)
    \tan \alpha=-\tan(\pi - \alpha)
    \cot \alpha=-cot(\pi - \alpha)
    \cos \alpha=-\cos(\alpha-\pi) e.g \alpha >180^o
    \sin \alpha=-\sin(\alpha-\pi)
    \tan \alpha=\tan(\alpha-\pi) the period of tangent is \pi
    ..... and so on
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    (Original post by Branny101)
    Ahh right so its like some unknown rule... Posted from TSR Mobile
    It's not an "unknown" rule - it's a rule you should have been taught in class or shown in your textbook.

    As ztibor says, when you move from the right-angled triangle to thinking about larger angles (> 90 degrees) you have to extend the definition of sin, cos and tan to deal with larger angles - the CAST diagram and the graphs of the trig functions are just consequences of this wider definition.

    Did you teacher not explain how to think about cos and sin as the x-coordinates of a point moving round the unit circle, and why, for example, this means that cos is negative for obtuse angles while sin remains positive?
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    I've done a visual representation using the graphs, as a maths A2 student we haven't been taught the quadrants method so I find this easier. Am I correct? Maybe this will help you, OP.

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    (Original post by Branny101)
    Thanks buddy! But if I were just to consider the graphs how would this work (for some reason I find the CAST or ASTC a bit more confusing)?!

    Sorry for being so pedantic...

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    Which graphs are positive between 0 and 90?
    Which graph is positive between 90 and 180?
    Which graph is positive between 180 and 270?
    Which graph is positive between 270 and 360?
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    (Original post by davros)
    It's not an "unknown" rule - it's a rule you should have been taught in class or shown in your textbook.

    As ztibor says, when you move from the right-angled triangle to thinking about larger angles (> 90 degrees) you have to extend the definition of sin, cos and tan to deal with larger angles - the CAST diagram and the graphs of the trig functions are just consequences of this wider definition.

    Did you teacher not explain how to think about cos and sin as the x-coordinates of a point moving round the unit circle, and why, for example, this means that cos is negative for obtuse angles while sin remains positive?
    I say unknown as it is definitely unknown to me, I did not learn it as a means of a circle but by graphs - thank you for the insight though I shall look into this.
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    (Original post by ztibor)
    As a right triangle has no obtuse angle so we define sine, cosine, tangent and cotangent
    of an angle in a unit circle.
    THe sign follows from this definition.
    Do you know the definition of cos x for any X?

    Some rules from the definition

    \cos \alpha=-\cos(\pi - \alpha) e.g \alpha>90^o
    \sin \alpha=\sin(\pi - \alpha)
    \tan \alpha=-\tan(\pi - \alpha)
    \cot \alpha=-cot(\pi - \alpha)
    \cos \alpha=-\cos(\alpha-\pi) e.g \alpha >180^o
    \sin \alpha=-\sin(\alpha-\pi)
    \tan \alpha=\tan(\alpha-\pi) the period of tangent is \pi
    ..... and so on
    Wow... These look so unfamiliar. I will try and understand them.

    Thank you
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    (Original post by MarkProbio)
    I've done a visual representation using the graphs, as a maths A2 student we haven't been taught the quadrants method so I find this easier. Am I correct? Maybe this will help you, OP.

    Posted from TSR Mobile
    Hmm if this works it may be a lot easier to remember... Thank you I will try and cross-reference between this and the above given formulae :no:
 
 
 
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