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Math Help - AS.AQA.Page.220.Revision Exercise.Q8.Q19.Trigonometry.

  1. #1
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    AS.AQA.Page.220.Revision Exercise.Q8.Q19.Trigonometry.

    Q8.
    a)i)Express in sin^2 x in terms of cosX.
    ii) By writing cosX=Y show that the equation 7cosX + 2 - 4 sin^2 x = 0 is equivalent to 4y^2 + 7y -2 = 0.

    b) Solve the equation 4y^2 + 7y -2 = 0.

    c)Hence solve the equation 7 cosX + 2 - 4 sin^2X = 0, giving all solutions to the nearest 0.1(degree) in the interval 0(degree)< x < 360(degree).


    Q19.A student models the evening lighting-up time by the equation L = 6.125 - 2.25cos(πt/6) where the time, L hours pm, is always in GMT ( Greenwich Mean Time) and t is in months, starting in mid-December. The model assumes that all months are equally long.

    a)Calculate the value of L for mid-January and for mid-May

    b)Find, by solving an appropriate equation, the two months in the year when the lighting up time will be 5pm (GMT).

    c)Write down an equation for L if t were to be in months starting in mid-March.
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    Quote Originally Posted by ansonbound View Post
    Q8.
    a)i)Express in sin^2 x in terms of cosX.
    ii) By writing cosX=Y show that the equation 7cosX + 2 - 4 sin^2 x = 0 is equivalent to 4y^2 + 7y -2 = 0.

    b) Solve the equation 4y^2 + 7y -2 = 0.

    c)Hence solve the equation 7 cosX + 2 - 4 sin^2X = 0, giving all solutions to the nearest 0.1(degree) in the interval 0(degree)< x < 360(degree).


    you should know sin^2 x + cos^2 x = 1
    then 1-sin^2 x=cos^2 x

    same for part ii)

    can you do it now?
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  3. #3
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    Quote Originally Posted by ansonbound View Post
    Q8.
    a)i)Express in sin^2 x in terms of cosX.
    ii) By writing cosX=Y show that the equation 7cosX + 2 - 4 sin^2 x = 0 is equivalent to 4y^2 + 7y -2 = 0.

    b) Solve the equation 4y^2 + 7y -2 = 0.

    c)Hence solve the equation 7 cosX + 2 - 4 sin^2X = 0, giving all solutions to the nearest 0.1(degree) in the interval 0(degree)< x < 360(degree).


    Q19.A student models the evening lighting-up time by the equation L = 6.125 - 2.25cos(πt/6) where the time, L hours pm, is always in GMT ( Greenwich Mean Time) and t is in months, starting in mid-December. The model assumes that all months are equally long.

    a)Calculate the value of L for mid-January and for mid-May

    b)Find, by solving an appropriate equation, the two months in the year when the lighting up time will be 5pm (GMT).

    c)Write down an equation for L if t were to be in months starting in mid-March.
    Hi ansonbound,

    [Q8.a.i] \sin^2 x = 1 - \cos^2 x

    [Q8.a.ii] \cos x = y

    7 \cos x + 2 - 4 \sin^2 x = 0 \equiv 4y^2+7y-2=0

    7y+2-4(1-\cos^2 x)=0

    7y+2-4+4 \cos^2x=0

    7y-2+4y^2=0

    4y^2+7y-2=0

    [Q8.b] 4y^2+7y-2=0

    (4y-1)(y+2)=0

    y=\frac{1}{4} \ \ or \ \ y=-2

    [Q8.c] \boxed{\cos x =\frac{1}{4}}

    \cos x = -2 exceeds the range of cosine

    x=\{75.5, 284.5\}
    Last edited by masters; March 16th 2010 at 12:19 PM.
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  4. #4
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    Thanks for Q.8, can someone help me out for Q19.?
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  5. #5
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    (πt/6)
    π is a "pie"
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