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tags:
Amplitude Cosine Period Phase Shift Sine Trigonometry Trigonometry Series

Transformations of Trig Graphs Exercise 3

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Write the sine and cosine functions for the graph.

A graph from -2 pi to 2 pi of a transformed sine curve with max values of 5 at -pi over 6, pi over 2, 7 pi over 6, etc. and min values of 3 at pi over 6, 5 pi over 6, 9pi over 6, etc.
 

Answer: \begin{align*}
    f(x) &= -\sin(3x)+4 \\
    f(x) &= \cos\left(3\left(x+\frac{\pi}{6}\right)\right)+4
\end{align*}

Solution Method:
\(f(x)=A\sin(B(x- C))+ D\)
            \(f(x)=A\cos(B(x- C))+ D\)
            A = amplitude
            B = \(\frac{2\pi}{\text{period}}\)
            C = phase(horizontal) shift
            D = vertical shift

Sine and cosine graphs only differ by a phase shift, so any sinusoidal graph can be written as a sine and a cosine function. Furthermore, amplitude, period, and vertical shift will be the same for both functions. The only difference will be phase shift and possibly a reflection. 

The first thing we need to do is find the center axis. We can see that on this graph the center axis is \(y=4,\) because the maximums and minimums are equidistant from it. The location of this center axis tells us our vertical shift. Our axis is up 4 from the x-axis so \(D=4.\) We can also use this center axis to tell us amplitude. The maximum and minimums are up one and down 1 from the center axis, so our amplitude \(A=1.\) 

Next, we need to find the phase shift, C, for both sine and cosine.  We know that cosine intercepts the y-axis at a maximum. So I want to find the nearest maximum to the y-axis and that will tell me how far cosine is horizontally shifted. There is a maximum left \(\frac{\pi}{6}\) from the y-axis. So my phase shift for cosine is \(C=+\frac{\pi}{6}\).

 We know that sine intercepts the y-axis at the origin and then increases. Luckily for us, this graph intersects the y-axis at the center axis, but it decreases instead of increases. So I have two options: I can find the nearest intersection of the center axis that increases to the right, or I can reflect my sine graph and make the function negative, so it decreases first. I'm going to do the latter, so then my sine function is negative, and \(C=0.\)

Lastly, we need to find the period and calculate B. A period of sine starts at the center axis and has a minimum and maximum, so if I start here at the y-intercept, a minimum and maximum away puts the end of my cycle at \(\frac{2\pi}{3}.\) So the period of this graph is \(\frac{2\pi}{3}.\) Then to find B, we have \(B = \frac{2\pi}{\frac{2\pi}{3}} =3.\) 

So now we put it all together and get the equations
\begin{align*}
    f(x) &= -\sin(3x)+4 \\
    f(x) &= \cos\left(3\left(x+\frac{\pi}{6}\right)\right)+4
\end{align*}

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