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Thread: please check trig work2

  1. #1
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    please check trig work2

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

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

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

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

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


    Quote Originally Posted by william View Post

    2. in the general equation $\displaystyle y= acos k(x+b)$ 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 $\displaystyle y= 3cos \frac{2}{3} x-4$ 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 $\displaystyle \theta$ <or equal $\displaystyle 2\pi$, if $\displaystyle sin\theta=\frac{3}{5}$ the value of tan $\displaystyle \theta$ is $\displaystyle \frac{4}{3}$
    as $\displaystyle tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$

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


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

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

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




    I agree




    refer to my response to question 1.





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

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




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

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

    this means $\displaystyle k= 4 $ and $\displaystyle y=4\sin(4x)$
    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 $\displaystyle \frac{2\pi}{k}$ therefore this is the general solution.

    Q3

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