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Math Help - Trig equations

  1. #1
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    Trig equations

    I accidently posted this in the Pre-Calc section, if possible can someone delete that?

    How do I solve these

    Cos^2x + 2sinxcosx - sin^2 x = 0 [0, 360)

    sin(x +1) = cos x [0, 360)

    Cos 4X - Cos 2X = 0 but range is [-180, 180]

    And finally

    Logan has a garden in the shape of a sector of a circle; outer rim of garden is 25 ft long and central angle along of sector is 50 degrees, she wants to add a 3 ft wide walk to the outer rim, how many sq ft of paving does she need to build it?
    Last edited by Alan306090; January 17th 2010 at 08:33 PM.
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    Quote Originally Posted by Alan306090 View Post
    I accidently posted this in the Pre-Calc section, if possible can someone delete that?

    How do I solve these

    Cos^2x + 2sinxcosx - sin^2 x = 0 [0, 360)

    sin(x +1) = cos x [0, 360)

    Cos 4X - Cos 2X but range is [-180, 180]

    And finally

    Logan has a garden in the shape of a sector of a circle; outer rim of garden is 25 ft long and central angle along of sector is 50 degrees, she wants to add a 3 ft wide walk to the outer rim, how many sq ft of paving does she need to build it?
    1. \cos^2{x} + 2\sin{x}\cos{x} - \sin^2{x} = 0

    Let \cos{x} = X and \sin{x} = Y, then

    X^2 + 2XY - Y^2 = 0

    X^2 + 2XY + Y^2 - Y^2 - Y^2 = 0

    (X + Y)^2 - 2Y^2 = 0

    (X + Y)^2 = 2Y^2

    X + Y = \pm \sqrt{2} Y

    X = (-1 \pm \sqrt{2})Y.


    Therefore

    \cos{x} = (-1 - \sqrt{2})\sin{x} or \cos{x} = (-1 + \sqrt{2})\sin{x}.


    Case 1:

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

    \frac{1}{-1 - \sqrt{2}} = \frac{\sin{x}}{\cos{x}}

    -\frac{1}{1 + \sqrt{2}} = \tan{x}

    -\frac{1 - \sqrt{2}}{1 - 2} = \tan{x}

    1 - \sqrt{2} = \tan{x}

    -(\sqrt{2} - 1) = \tan{x}


    Now, using the half angle identity for tangent, we have

    \tan{\frac{\theta}{2}} = \frac{1 - \cos{\theta}}{\sin{\theta}}

    \tan{\frac{45^\circ}{2}} = \frac{1 - \cos{45^\circ}}{\sin{45^\circ}}

     = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}

     = \frac{\frac{\sqrt{2} - 1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}

     = \sqrt{2} - 1


    So this means that the basic angle is \frac{45^\circ}{2} = 22.5^\circ.

    Since tangent is negative in the second and fourth quadrants, this means that x = 157.5^\circ or x = 337.5^\circ.


    Now have a go for Case 2...
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    Quote Originally Posted by Alan306090 View Post
    I accidently posted this in the Pre-Calc section, if possible can someone delete that?

    How do I solve these

    Cos^2x + 2sinxcosx - sin^2 x = 0 [0, 360)

    sin(x +1) = cos x [0, 360)

    Cos 4X - Cos 2X but range is [-180, 180]

    And finally

    Logan has a garden in the shape of a sector of a circle; outer rim of garden is 25 ft long and central angle along of sector is 50 degrees, she wants to add a 3 ft wide walk to the outer rim, how many sq ft of paving does she need to build it?
    2. \sin{(x + 1)} = \cos{x}.


    Use the angle sum identity for sine:

    \sin{(\alpha + \beta)} = \sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}


    we have

    \sin{x}\cos{1} + \cos{x}\sin{1} = \cos{x}

    \cos{x} - \cos{x}\sin{1} = \sin{x}\cos{1}

    \cos{x}(1 - \sin{1}) = \sin{x}\cos{1}

    \frac{1 - \sin{1}}{\cos{1}} = \frac{\sin{x}}{\cos{x}}

    \frac{1 - \sin{1}}{\cos{1}} = \tan{x}

    x = \arctan{\left(\frac{1 - \sin{1}}{\cos{1}}\right)}.
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    Quote Originally Posted by Alan306090 View Post
    I accidently posted this in the Pre-Calc section, if possible can someone delete that?

    How do I solve these

    Cos^2x + 2sinxcosx - sin^2 x = 0 [0, 360)

    sin(x +1) = cos x [0, 360)

    Cos 4X - Cos 2X but range is [-180, 180]

    And finally

    Logan has a garden in the shape of a sector of a circle; outer rim of garden is 25 ft long and central angle along of sector is 50 degrees, she wants to add a 3 ft wide walk to the outer rim, how many sq ft of paving does she need to build it?
    What is \cos{4x} - \cos{2x} equal to? Or are you trying to find another expression for it?
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    Quote Originally Posted by Alan306090 View Post
    I accidently posted this in the Pre-Calc section, if possible can someone delete that?

    How do I solve these

    Cos^2x + 2sinxcosx - sin^2 x = 0 [0, 360)

    sin(x +1) = cos x [0, 360)

    Cos 4X - Cos 2X but range is [-180, 180]

    And finally

    Logan has a garden in the shape of a sector of a circle; outer rim of garden is 25 ft long and central angle along of sector is 50 degrees, she wants to add a 3 ft wide walk to the outer rim, how many sq ft of paving does she need to build it?
    Worded problem:

    First, you need to find the radius of this sector.

    To do this, remember that C = 2\pi r.

    In your case, since your angle is 50^\circ, this means you have \frac{50}{360} = \frac{5}{36} of the circle.

    So you have

    \frac{5C}{36} = \frac{5\cdot 2\pi r}{36}

    \frac{5C}{36} = \frac{5\pi r}{18}.


    You know that this part of the circumference is 25\,\textrm{ft}.

    So 25\,\textrm{ft} = \frac{5\pi r}{18}

    \frac{25\cdot 18}{5\pi}\,\textrm{ft} = r

    r = \frac{90}{\pi}\,\textrm{ft}.



    Now, you are planning to make a path around the outside of this sector.

    If you draw it, you'll notice that you have a sector inside a sector.

    The first sector has radius \frac{90}{\pi}\,\textrm{ft}, and since the pavers are 3\,\textrm{ft} wide, the second sector has radius \frac{90}{\pi} + 3\,\textrm{ft}.


    A_{\textrm{path}} = A_{\textrm{sector 2}} - A_{\textrm{sector 1}}.


    A_{\textrm{sector 1}} = \pi r^2

     = \pi\left(\frac{90}{\pi}\,\textrm{ft}\right)^2

     = \frac{8100}{\pi} \,\textrm{ft}^2.


    A_{\textrm{sector 2}} = \pi r^2

     = \pi \left(\frac{90}{\pi} + 3\,\textrm{ft}\right)^2

     = \pi \left(\frac{8100}{\pi^2} + \frac{540}{\pi} + 9\right) \,\textrm{ft}^2

     = \left(\frac{8100}{\pi} + 540 + 9\pi\right)\,\textrm{ft}^2.



    So A_{\textrm{path}} = A_{\textrm{sector 2}} - A_{\textrm{sector 1}}

     = \left(\frac{8100}{\pi} + 540 + 9\pi\right)\,\textrm{ft}^2 - \frac{8100}{\pi} \,\textrm{ft}^2

     = \left(540 + 9\pi\right)\,\textrm{ft}^2.
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  6. #6
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    It should be


    Cos 4X - Cos 2X = 0 but range is [-180, 180]

    And prove it, your answers don't seem to correlate to the answers in the back of the book which says:

    3pi\8, 7pi\8, 11pi\8 and 15pi\8 for the first one
    and .285, 3.427 for the second

    And answer to the last should be 121 square ft
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  7. #7
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    Quote Originally Posted by Alan306090 View Post
    It should be


    Cos 4X - Cos 2X = 0 but range is [-180, 180]
    \cos{4x} - \cos{2x} = 0

    \cos^2{2x} - \sin^2{2x} - \cos{2x} = 0

    \cos^2{2x} - (1 - \cos^2{2x}) - \cos{2x} = 0

    2\cos^2{2x} - \cos{2x} - 1 = 0.


    Let \cos{2x} = X so that you have

    2X^2 - X - 1 = 0

    2X^2 - 2X + X - 1 = 0

    2X(X - 1) + 1(X - 1) = 0

    (X - 1)(2X + 1) = 0

    X - 1 = 0 or 2X + 1 = 0

    X = 1 or X = -\frac{1}{2}.


    Case 1:

    \cos{2x} = 1

    2x = \left\{0^\circ, 180^\circ \right\} + 360^\circ n, where n is an integer representing how many times you have gone around the unit circle.

    x = \left\{ 0^\circ, 90^\circ \right\} + 180^\circ n

    x = \left\{-180^\circ , -90^\circ , 0^\circ, 90^\circ, 180^\circ \right\}.


    Case 2:

    \cos{2x} = -\frac{1}{2}.


    \cos{X} = \frac{1}{2} when X = 60^\circ, and cosine is negative in the second and third quadrants.

    So

    2x = \left\{ 120^\circ, 240^\circ \right\} + 360^\circ n

    x = \left\{ 60^\circ, 120^\circ \right\} + 180^\circ n

    x = \left \{-120^\circ , -60^\circ , 60^\circ, 120^\circ \right\}.



    Putting it together you have

    x = \left\{ -180^\circ, -120^\circ, -90^\circ, -60^\circ, 0^\circ, 60^\circ, 90^\circ, 120^\circ, 180^\circ \right \}.
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  8. #8
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    Thank you prove it, though for Case 1,

    The only time Cos2x = 1 [-180, 180] should be 0, no?

    So then 2x = 0 + 360k

    x = 0 + 180k

    Then it should be, -180, 0, and 180? Does this seem right?
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