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    I'm not understanding why  tan \theta  = -\frac {7}{24}

    Find the value Cos \theta and tan \theta, given that Sin \theta  = -\frac {7}{25} and  270^o < \theta < 360^o

    https://gyazo.com/3cdeeda4af99568f00a1f8195d877884
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    (Original post by Naruke)
    I'm not understanding why  tan \theta  = -\frac {7}{24}

    Find the value Cos \theta and tan \theta, given that Sin \theta  = -\frac {7}{25} and  270^o < \theta < 360^o

    https://gyazo.com/3cdeeda4af99568f00a1f8195d877884
    \cos \theta is positive in the fourth quadrant (think CAST diagrams) but tan is negative there.

    In either case, knowing that \cos \theta is positive is enough. Because \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\text{negative}}{\text{pos  itive}} = \text{negative}.
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    Draw a right angle triangle, the sides will be length 7, 24 and 25.  \tan \theta in the range  270^{\circ} \leqslant \theta \leqslant 360^{\circ} is negative, draw a quick sketch of the tan curve if you don't understand this.
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    (Original post by Naruke)
    I'm not understanding why  tan \theta  = -\frac {7}{24}

    Find the value Cos \theta and tan \theta, given that Sin \theta  = -\frac {7}{25} and  270^o < \theta < 360^o

    https://gyazo.com/3cdeeda4af99568f00a1f8195d877884
    sin x = -7/25 -> sin^2x=49/625 -> cos^2x=576/625 -> cosx=+/-24/25. But 270<x<360, so by considering the cosine graph, cos is positive in this region, so cosx=24/25. Now tanx=sinx/cosx = (-7/25)/(24/25)=-7/24.
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    too late
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    (Original post by Zacken)
    \cos \theta is positive in the fourth quadrant (think CAST diagrams) but tan is negative there.

    In either case, knowing that \cos \theta is positive is enough. Because \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\text{negative}}{\text{pos  itive}} = \text{negative}.
    The day we learned about trigo graphs & cast diagrams I was ill.

    The book sucks. Know of anywhere that I can learn this?
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    (Original post by TeeEm)
    too late
    Must you keep posting unneeded and often crudely sarcastic comments on every thread? This really is unacceptable.
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    (Original post by Naruke)
    The day we learned about trigo graphs & cast diagrams I was ill.

    The book sucks. Know of anywhere that I can learn this?
    Here you go.
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    (Original post by constellarknight)
    Must you keep posting unneeded and often crudely sarcastic comments on every thread? This really is unacceptable.
    I am sorry you feel so...
    I beginning to dislike you ...
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    (Original post by TeeEm)
    I am sorry you feel so...
    I beginning to dislike you ...
    The honour... By the way, your IYGB Special Papers are riddled with errors. For example, Paper Z Question 2. The fact that as written, the denominator is sqrt(x+sqrt(x^2-1)) makes it impossible. What was intended was presumably sqrt(x)+sqrt(x^2-1) (note that the two square roots are separate, rather than the second being within the first), which is indeed what you start out with in your solution. I could go on, but I'd rather not. Anyway mate, the point is whatever ego and perceived reputation you may have built up for yourself, you're a nobody. Have a good day!
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    (Original post by constellarknight)
    The honour... By the way, your IYGB Special Papers are riddled with errors. For example, Paper Z Question 2. The fact that as written, the denominator is sqrt(x+sqrt(x^2-1)) makes it impossible. What was intended was presumably sqrt(x)+sqrt(x^2-1) (note that the two square roots are separate, rather than the second being within the first), which is indeed what you start out with in your solution. I could go on, but I'd rather not. Anyway mate, the point is whatever ego and perceived reputation you may have built up for yourself, you're a nobody. Have a good day!
    you clearly have issues
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    (Original post by Zacken)
    Here you go.
    Thanks!

    Another question: Eliminate \theta to give an equation relating x and y:

    c)  x = Sin \theta , y = Cos^2 \theta

    Would this be ok?

    Sin \theta  = \sqrt1-Cos^2\theta

     x = \sqrt1-y

    I don't know how to extend the square root in latex but you catch my drift
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    (Original post by Naruke)
    Thanks!

    Another question: Eliminate \theta to give an equation relating x and y:

    c)  x = Sin \theta , y = Cos^2 \theta

    Would this be ok?

    Sin \theta  = \sqrt1-Cos^2\theta

     x = \sqrt1-y

    I don't know how to extend the square root in latex but you catch my drift
    1. If you want to extend the square root use \sqrt{1-y}, i.e: put the curly braces around it.

    2. That would be okay, I suppose...

    3. It's a very ugly answer, x^2 + y = 1 is the same thing and looks much nicer, but I doubt you'd lose marks for writing down what you have. So yes, it's perfectly correct.
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    (Original post by Zacken)
    1. If you want to extend the square root use \sqrt{1-y}, i.e: put the curly braces around it.

    2. That would be okay, I suppose...

    3. It's a very ugly answer, x^2 + y = 1 is the same thing and looks much nicer, but I doubt you'd lose marks for writing down what you have. So yes, it's perfectly correct.
    Thank you sir
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    (Original post by TeeEm)
    you clearly have issues
    Thanks you narcissistic piece of ****.
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    (Original post by Naruke)
    Thank you sir
    Oi, I'm an A-Level student just like yourself!
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    (Original post by constellarknight)
    Thanks you narcissistic piece of ****.
    Did you have a bad day ?
    I remember you lost it again in another thread the other day ...
    Different username but identical persona...
    How old are you please, if you do not mind me asking?
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    (Original post by Zacken)
    Oi, I'm an A-Level student just like yourself!
    If I consider you an A level student, it will just lower my self - esteem.

    I mean, I've been trying to grow a beard for 6 months but you make it seem light work.

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    (Original post by Naruke)
    If I consider you an A level student, it will just lower my self - esteem.
    Nah, don't be silly. I'd wager your results will be better than mine, for real.

    I mean, I've been trying to grow a beard for 6 months but you make it seem light work.

    :rofl: :rofl: Seriously though, it's great that you're coming on here and asking questions, shows that you want to understand everything and that's going to get you far. :yep:
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    (Original post by TeeEm)
    Did you have a bad day ?
    I remember you lost it again in another thread the other day ...
    Different username but identical persona...
    How old are you please, if you do not mind me asking?
    I do mind you asking. Also what "different username?" I demand that you desist from these libellous accusations at once.
    me postulando maxime vexas, etiam quid dicere conaris verbis "nomen alium"? posco igitur ut statim desistas affirmare famosa haec.
 
 
 
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