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Math Help - need description on how to do this

  1. #1
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    Unhappy need description on how to do this

    Consider: y=sin(x+pi/9)

    (a) the graph intersects the x-axis at point A. Find the x co-ordinate of A where 0 < x < pi

    (b) Solve the equation sin(x+pi/9)=-1/2 for 0 < x < 2pi

    the "<" are less than or equal to in both cases.

    Thank you
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  2. #2
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    Quote Originally Posted by mamann View Post
    Consider: y=sin(x+pi/9)

    (a) the graph intersects the x-axis at point A. Find the x co-ordinate of A where 0 < x < pi

    (b) Solve the equation sin(x+pi/9)=-1/2 for 0 < x < 2pi

    the "<" are less than or equal to in both cases.

    Thank you
    For part a), when an equation crosses the x axis, the y coordinate is zero, hence the equation becomes:

     0 = sin(x+\frac{\pi}{9}

    Now, when you take the inverse sine of 0 you get an infinite number of solutions, but only one of them lies in the interval [0, \pi] . Find that solution, then it's just a case of solving algebraically for x.

    For part b it is the exact same process except that you need to find take the inverse sine of  \frac{-1}{2} and find the solution to it which lies in the given interval.
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  3. #3
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    yea but how do i solve for x?
    Apparently you cant just apply sin to x and pi/9.
    you have to do something where sin(a+b) = (sin a)(cos b) + (sin b)(cos a)
    and that gives me:
    (sin x)(cos pi/9) + (sin pi/9)(cos x) = 0
    Note: pi/9 is in radians which turns out to be 20 degrees.

    and i got to here but i dont know what to do after this:

    0.94sinx + 0.34cosx = 0
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  4. #4
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    Quote Originally Posted by mamann View Post
    yea but how do i solve for x?
    Apparently you cant just apply sin to x and pi/9.
    you have to do something where sin(a+b) = (sin a)(cos b) + (sin b)(cos a)
    and that gives me:
    (sin x)(cos pi/9) + (sin pi/9)(cos x) = 0
    Note: pi/9 is in radians which turns out to be 20 degrees.

    and i got to here but i dont know what to do after this:

    0.94sinx + 0.34cosx = 0
    Ah, yes, you could do that.

    From there, you divide the equation by  \cos(x) which gives:

     0.94 \frac{\sin(x)}{\cos(x)} + 0.34 = 0

    You should know that  \frac{\sin(x)}{\cos(x)} = \tan(x) , and you can solve from there.
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  5. #5
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    Quote Originally Posted by mamann View Post
    Consider: y=sin(x+pi/9)

    (a) the graph intersects the x-axis at point A. Find the x co-ordinate of A where 0 < x < pi

    (b) Solve the equation sin(x+pi/9)=-1/2 for 0 < x < 2pi

    the "<" are less than or equal to in both cases.

    Thank you
    (a) the graph of y = \sin\left(x + \frac{\pi}{9}\right) is the graph of y = \sin{x} shifted left \frac{\pi}{9} units.

    since y = \sin{x} crosses the x-axis at x = \pi , the graph of y = \sin\left(x + \frac{\pi}{9}\right) will cross the x-axis at x = \pi - \frac{\pi}{9} = \frac{8\pi}{9}

    (b) from the unit circle ...

    \sin\left(\frac{7\pi}{6}\right) = \sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}

    so, solve the equations to get exact solutions ...

     <br />
x + \frac{\pi}{9} = \frac{7\pi}{6}<br />

     <br />
x + \frac{\pi}{9} = \frac{11\pi}{6}<br />
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