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Thread: [SOLVED] Trig. and Sin Wave Question

  1. #1
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    [SOLVED] Trig. and Sin Wave Question

    If anyone can help me with these two questions I would appreciate it. Thank you!

    Trig. question:

    "The expression $\displaystyle [2cos(theta)/sin2(theta)]$ is equivalent to...?
    Choices:
    (1) [csc theta]
    (2) [sec theta]
    (3) [cot theta]
    (4) [sin theta]

    Work done so far:
    $\displaystyle [Cos2 theta/Sin2 theta] = [1-sin^2 theta/1-cos^2 theta] $
    ...where to go from here?

    Sin Wave Question:
    "A student attaches one end of a rope to a wall at a fixed point 3 feet above the ground, as shown in the accompanying diagram, and moves the other end of the rope up and down, producing a wave described by the equation $\displaystyle y = a sin bx + c$. The range of the rope’s height above the ground is between 1 and 5 feet. The period of the wave is $\displaystyle [4(pie symbol)]$. Write the equation that represents this wave."


    Work done so far:
    Amp.= 3
    Freq.= 2
    Range=...5?

    Equation: 3 sin 2x + ...?

    Side Note: I'm new to using "LaTex"...
    Last edited by MrOats; May 16th 2008 at 09:08 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by MrOats View Post
    If anyone can help me with these two questions I would appreciate it. Thank you!

    Trig. question:
    "The expression $\displaystyle [2cos(theta)/sin2(theta)]$ is equivalent to...?
    Choices:
    (1) [csc theta]
    (2) [sec theta]
    (3) [cot theta]
    (4) [sin theta]

    Work done so far:
    $\displaystyle [Cos2 theta/Sin2 theta] = [1-sin^2 theta/1-cos^2 theta] $
    ...where to go from here?

    Sin Wave Question:
    "A student attaches one end of a rope to a wall at a fixed point 3 feet above the ground, as shown in the accompanying diagram, and moves the other end of the rope up and down, producing a wave described by the equation $\displaystyle y = a sin bx + c$. The range of the rope’s height above the ground is between 1 and 5 feet. The period of the wave is $\displaystyle [4(pie symbol)]$. Write the equation that represents this wave."


    Work done so far:
    Amp.= 3
    Freq.= 2
    Range=...5?

    Equation: 3 sin 2x + ...?

    Side Note: I'm new to using "LaTex"...
    For the first one

    $\displaystyle \sin(2\theta)=2\sin(\theta)\cos(\theta)$

    $\displaystyle \therefore\frac{2\cos(\theta)}{\sin(2\theta)}=\fra c{2\cos(\theta)}{2\sin(\theta)\cos(\theta)}$

    I think you can take it from there
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by MrOats View Post
    If anyone can help me with these two questions I would appreciate it. Thank you!

    Trig. question:
    "The expression $\displaystyle [2cos(theta)/sin2(theta)]$ is equivalent to...?
    Choices:
    (1) [csc theta]
    (2) [sec theta]
    (3) [cot theta]
    (4) [sin theta]

    Work done so far:
    $\displaystyle [Cos2 theta/Sin2 theta] = [1-sin^2 theta/1-cos^2 theta] $
    ...where to go from here?

    Sin Wave Question:
    "A student attaches one end of a rope to a wall at a fixed point 3 feet above the ground, as shown in the accompanying diagram, and moves the other end of the rope up and down, producing a wave described by the equation $\displaystyle y = a sin bx + c$. The range of the rope’s height above the ground is between 1 and 5 feet. The period of the wave is $\displaystyle [4(pie symbol)]$. Write the equation that represents this wave."


    Work done so far:
    Amp.= 3
    Freq.= 2
    Range=...5?

    Equation: 3 sin 2x + ...?

    Side Note: I'm new to using "LaTex"...
    It moves from 1 to 5 for the second one...so amplitude is 4

    Since in the equation $\displaystyle a\sin(bx+c)$

    The period is $\displaystyle \frac{2\pi}{|b|}$

    So you have $\displaystyle \frac{2\pi}{|b|}=4\pi$

    Solve for b and consider that the lowest point is one and the highest point is 5 to figure out the c value which just shifts the sine wave vertically
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  4. #4
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    Exclamation Please check/help me further.

    Please check, someone, to make sure I did these right? Thank you!

    For this question:
    I got $\displaystyle {cos(theta)/sin(theta)}$, which is $\displaystyle cot(theta)$.

    Am I correct? If not, please help me.

    For the second question:
    I got Amplitude = 4, Period = [2(pie)/b], Frequency = 2/4
    ...Not so sure about this one...

    And for the equation: y= 4 Sin (2/4(x)+4)

    ...I really am unsure about this one. Someone please clarify for me? I need to get "solid" in my head what the amplitude, frequency, period, and range are...And solid, a=, b=, c=...answers.

    Please help me demystify this sin wave question! Thank you.
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  5. #5
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    Quote Originally Posted by MrOats View Post
    Please check, someone, to make sure I did these right? Thank you!

    For this question:
    I got $\displaystyle {cos(theta)/sin(theta)}$, which is $\displaystyle cot(theta)$.

    Am I correct? If not, please help me.
    No you are wrong...

    $\displaystyle \frac{2\cos \theta}{\sin 2\theta} = \frac{2\cos \theta}{2\sin \theta\cos \theta} = \frac1{\sin\theta} = csc\theta$

    So answer is option (1)

    Quote Originally Posted by MrOats View Post
    For the second question:
    I got Amplitude = 4, Period = [2(pie)/b], Frequency = 2/4
    ...Not so sure about this one...

    And for the equation: y= 4 Sin (2/4(x)+4)

    ...I really am unsure about this one. Someone please clarify for me? I need to get "solid" in my head what the amplitude, frequency, period, and range are...And solid, a=, b=, c=...answers.

    Please help me demystify this sin wave question! Thank you.
    I think you are misinterpreting the question ....

    $\displaystyle y = (a\sin bx) + c$
    Is the distance x from the wall or the person?

    Anyway I will assume from the wall,
    Note that when x = 0, then y = 3. So c = 3.

    Since the range of variation for y is c-a to c+a (because $\displaystyle \sin bx$ varies from -1 to 1) and the question tells us the ropes height is from 1 to 5 feet, we have $\displaystyle 3 - a =1$ and $\displaystyle 3 + a = 5$. This gives us $\displaystyle a = 2$

    $\displaystyle 2\sin bx + 3$ has the same period as $\displaystyle \sin bx$. Now since the period of a sin function is $\displaystyle 2\pi$, we are forced to choose $\displaystyle b = \frac12$.

    Thus the final equation is
    $\displaystyle y = 2\sin \frac{x}{2} + 3$

    Side Note:Now since $\displaystyle 4\pi$ is the period, it means $\displaystyle a\sin bx + c = a\sin b(x+4\pi) + c \Rightarrow bx+4\pi b = n\pi+ (-1)^n bx$. Clearly for n=2, the equation reads $\displaystyle 4\pi b = 2\pi$ and thus b=2
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  6. #6
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    Thank you very much Isomorphism!

    I get it now. Finally. =)
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