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Math Help - please check trig work2

  1. #1
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    please check trig work2

    1. in the general equation y= acos kx the period is determined by the value of x

    2. in the general equation y= acos k(x+b) the phase shift is determined by the value of b

    3. the period of the function y= 3cos \frac{2}{3} x-4 is -4

    4. if \theta is in standard position, and 0 <or equal \theta <or equal 2\pi, if sin\theta=\frac{3}{5} the value of tan \theta is \frac{4}{3}

    5. the equation of a sine function with amplitude 4 and period of 90^o is y=4sin90x
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  2. #2
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    Quote Originally Posted by william View Post
    1. in the general equation y= acos kx the period is determined by the value of x
    no, period is \frac{2\times \pi}{k}


    Quote Originally Posted by william View Post

    2. in the general equation y= acos k(x+b) the phase shift is determined by the value of b
    I agree


    Quote Originally Posted by william View Post

    3. the period of the function y= 3cos \frac{2}{3} x-4 is -4
    refer to my response to question 1.



    Quote Originally Posted by william View Post
    1

    4. if \theta is in standard position, and 0 <or equal \theta <or equal 2\pi, if sin\theta=\frac{3}{5} the value of tan \theta is \frac{4}{3}
    as tan(\theta) = \frac{sin(\theta)}{cos(\theta)}

    then tan(\theta)= \frac{3}{4} \neq \frac{4}{3}


    Quote Originally Posted by william View Post
    5. the equation of a sine function with amplitude 4 and period of 90^o is y=4sin90x
    period is \frac{2\times \pi}{k}

    therefore \frac{\pi}{2} = \frac{2\times \pi}{k}

    this means k= 4 and y=4\sin(4x)
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  3. #3
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    Quote Originally Posted by pickslides View Post
    no, period is \frac{2\times \pi}{k}




    I agree




    refer to my response to question 1.





    as tan(\theta) = \frac{sin(\theta)}{cos(\theta)}

    then tan(\theta)= \frac{3}{4} \neq \frac{4}{3}




    period is \frac{2\times \pi}{k}

    therefore \frac{\pi}{2} = \frac{2\times \pi}{k}

    this means k= 4 and y=4\sin(4x)
    thanks i appreciate the help,my updated answers:

    1. can it be just k? can you just say k?

    3. 2/3 pi

    are they good now?
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  4. #4
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    no, it can't just be k.

    Q1

    The period is found as \frac{2\pi}{k} therefore this is the general solution.

    Q3

    \frac{2\pi}{k} therefore youe answer is \frac{2\pi}{\frac{2}{3}} = 3\pi so the period or length of the cosine wave is 3\pi
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