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Using the given information to complete the equation $y= A \\cos{ \\left( \\frac{2 \\pi}{P} x  \\right) }+V $ that describes an electromagnetic wave and calculating the smallest angle, $x$, for which $y=y_0$.

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The shape of an electromagnetic wave is modelled by the equation:
$y=\\var{a}\\cos{\\left(\\frac{\\pi x}{\\simplify{{p}/2}}\\right)}+\\var{v}$

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where $x$ is measured in radians.

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The equation

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\\[ \\begin{split} y=\\var{a}\\cos{\\left(\\frac{\\pi x}{\\simplify{{p}/2}}\\right)}+\\var{v} \\end{split} \\]

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is a wave function.

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The general form of a wave function can be written as:

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\\[ \\begin{split} y=A\\cos{\\left(\\frac{2 \\pi}{P}\\left(x-H\\right)\\right)}+V \\end{split} \\]

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Where $V$ is the average value, $A$ is the amplitute, $H$ the phase and $P$ the wavelenght (also called period). 

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By comparying the equation with the general form we can notice that:

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a) The amplitude of the wave is $A=\\var{a}$

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b)The period can be found by solving the equation 

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\\[ \\begin{split} \\frac{\\pi}{\\simplify{{p}/2}} &= \\frac{2 \\pi}{P} \\\\ \\pi P&= \\simplify{{p}/2}\\times 2\\pi \\\\ \\pi P&= {p}\\pi \\\\ P&=\\frac{{p} \\pi}{\\pi} \\\\ P &=\\var{p} \\end{split} \\]

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So, the wavelenght is $P=\\var{p}$.

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To find the smallest value of $x$ for which $y=\\var{y1}$. We need to substitute $y=\\var{y1}$ in the equation and solve for $x$. 

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\\[ \\begin{split} \\var{y1}&=\\var{a}\\cos{\\left(\\frac{\\pi x}{\\simplify{{p}/2}}\\right)}+\\var{v} \\\\ \\var{y1}-\\var{v} &=\\var{a}\\cos{\\left(\\frac{\\pi x}{\\simplify{{p}/2}}\\right)} \\\\ \\var{y1-v} &=\\var{a}\\cos{\\left(\\frac{\\pi x}{\\simplify{{p}/2}}\\right)} \\\\ \\frac{\\var{y1-v}}{\\var{a}} &=\\cos{\\left(\\frac{\\pi x}{\\simplify{{p}/2}}\\right)}\\end{split} \\]

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We take $\\cos^{-1}$ from both sides:

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\\[ \\begin{split} \\cos^{-1} \\left(\\frac{\\var{y1-v}}{\\var{a}} \\right)&=\\frac{\\pi x}{\\simplify{{p}/2}} \\\\ \\simplify{{p}/2} \\cos^{-1} \\left(\\frac{\\var{y1-v}}{\\var{a}} \\right)&=\\pi x \\\\ \\frac{\\simplify{{p}/2} \\cos^{-1} \\left(\\frac{\\var{y1-v}}{\\var{a}} \\right)}{\\pi}&=x \\end{split} \\]

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We can use the calculator to find that $x= \\frac{\\simplify{{p}/2} \\cos^{-1} \\left(\\frac{\\var{y1-v}}{\\var{a}} \\right)}{\\pi}=\\var{ansC}$ radians (rounded to 2 decimal places).

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We can use the calculator to find that $x= \\frac{\\simplify{{p}/2} \\cos^{-1} \\left(\\frac{\\var{y1-v}}{\\var{a}} \\right)}{\\pi}=\\var{ansC}$ radians.

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Write down the amplitude of this wave.

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$A=$[[0]]

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Calculate the wavelength of this wave.

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The wavelength is [[0]].

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Find the smallest value of $x$ for which $y=\\var{y1}$.

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$x=$[[0]] radians.

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Give your answer rounded to 2 decimal places, if needed.

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