To form an equation of a (simple) sine or cosine curve from a graph or from given information, we need to identify the key features of the cyclic function. We can begin by writing the general form of the equations. Namely:
$f\left(x\right)=a\sin\left(bx-c\right)+d$f(x)=asin(bx−c)+d or $f\left(x\right)=a\cos\left(bx-c\right)+d$f(x)=acos(bx−c)+d
From this point, we need to determine the values of the constants $a$a, $b$b, $c$c and $d$d using the information at hand. Recall from previous chapters that:
Find the equation of the cosine curve that has undergone the following transformations:
a vertical translation of $4$4 units upwards
a vertical dilation by a multiple of $2$2
a horizontal translation resulting in a phase shift of $22\frac{1}{2}^\circ$2212° to the left
a horizontal dilation by a multiple of $\frac{1}{2}$12
The general form of the cosine equation is $f\left(x\right)=a\cos\left(bx-c\right)+d$f(x)=acos(bx−c)+d, for which we need values of $a$a, $b$b, $c$c and $d$d.
We are told that the vertical translation is $4$4 units up. So, $d=4$d=4
We are told that there is a vertical dilation by the multiple $2$2. This means the amplitude is $a=2$a=2.
Given that there is a horizontal dilation by the multiple $\frac{1}{2}$12, the period is adjusted from being $360^\circ$360° to half of that amount, which is $180^\circ$180°. Since $\frac{360^\circ}{b}=180^\circ$360°b=180°, we have $b=2$b=2.
There is a phase shift of $22\frac{1}{2}^\circ$2212° to the left. This is $\frac{c}{b}$cb. (Remember that shifting left is positive.) So, $\frac{c}{2}=22\frac{1}{2}^\circ$c2=2212° and therefore $c=45^\circ$c=45°.
There is no reflection. So, the value of $a$a will be positive.
Putting all this together into the general form we determine the equation:
$f\left(x\right)=2\cos\left(2x+45^\circ\right)+4$f(x)=2cos(2x+45°)+4.
Find the equation of the curve in the graph below.
We mark in the visible details.
The maximum is at $y=0.1$y=0.1 and the minimum is at $y=-1.1$y=−1.1. The average of these locates the centre line, $y=-0.5$y=−0.5. The amplitude is $0.1-(-0.5)=0.6$0.1−(−0.5)=0.6.
Crests of the waveform occur at $-180^\circ$−180° and $540^\circ$540° so that the period is $540-(-180)=720^\circ$540−(−180)=720°. This is twice the standard period of $360^\circ$360°. Therefore, the period constant is $\frac{1}{2}$12.
The curve looks like a cosine curve shifted $180^\circ$180° to the left. It could also be interpreted as a sine curve reflected in the $x$x-axis and this would make the amplitude constant negative with no phase shift.
Considering this as a sine function with no phase shift and inserting the constant values into $y(x)=a\sin\left(bx-c\right)+d$y(x)=asin(bx−c)+d, we have
$y(x)=-0.6\sin\frac{x}{2}-0.5$y(x)=−0.6sinx2−0.5
If we think of it as a sine function without a reflection but with a shift of $360^\circ$360° to the right, we would write the formula as
$y(x)=0.6\sin\left(\frac{x}{2}-180^\circ\right)-0.5$y(x)=0.6sin(x2−180°)−0.5
Or, if we think of it as a cosine curve shifted to the left by $180^\circ$180°, we could write the formula as
$y(x)=0.6\cos\left(\frac{x}{2}+90^\circ\right)-0.5$y(x)=0.6cos(x2+90°)−0.5
You should verify that these forms are equivalent.
Determine the equation of the graphed function given that it is of the form $y=a\sin x$y=asinx or $y=a\cos x$y=acosx, where $x$x is in degrees.
Determine the equation of the graphed function given that it is of the form $y=\sin bx$y=sinbx or $y=\cos bx$y=cosbx, where $b$b is positive and $x$x is in degrees.
Determine the equation of the graphed function given that it is of the form $y=\cos\left(x-c\right)$y=cos(x−c), where $c$c is the least positive value and $x$x is in degrees.
What are the different ways that we can transform the graph of the sine or cosine function? Recall that a transformation can be described algebraically using the variables $a$a, $b$b, $c$c, and $d$d, where
$\sin\left(x\right)\text{ }\rightarrow\text{ }a\sin\left(bx-c\right)+d$sin(x) → asin(bx−c)+d
We read this as "$\sin\left(x\right)$sin(x) transforms into $a\sin\left(bx-c\right)+d$asin(bx−c)+d". In this lesson we will focus on the role of the variable $c$c in horizontal translations or phase shifts and how it works in conjunction with other transformations.
The graph of $y=\cos x$y=cosx has been shifted $30^\circ$30° to the right. State the equation of the new graph.
Think: A shift to the right is a horizontal translation, so we know the new graph will have an equation of the form $y=\cos\left(x-c\right)$y=cos(x−c). We need to determine the magnitude and sign of the variable $c$c.
Do: The graph of $y=\cos x$y=cosx crosses the $y$y-axis at $\left(0,1\right)$(0,1). Since the whole graph is shifted by $30^\circ$30° to the right, this point will move to $\left(30^\circ,1\right)$(30°,1). We can substitute these coordinates into the form of the new equation to obtain the value of $c$c.
$y$y | $=$= | $\cos\left(x-c\right)$cos(x−c) | |
$1$1 | $=$= | $\cos\left(30^\circ-c\right)$cos(30°−c) | (Substituting the point $\left(30^\circ,1\right)$(30°,1)) |
$0$0 | $=$= | $30^\circ-c$30°−c | (Using the fact that $\cos\theta=1$cosθ=1 when $\theta=0$θ=0) |
$c$c | $=$= | $30^\circ$30° | (Adding $c$c to both sides of the equation) |
In the final line we see that $c=30^\circ$c=30°, so that the equation of the new graph is $y=\cos\left(x-30^\circ\right)$y=cos(x−30°).
Reflect: If the horizontal translation had been to the left, we would have found that $c=-30^\circ$c=−30°, so that the new graph would have the equation $y=\cos\left(x-\left(-30^{\circ}\right)\right)=\cos\left(x+30^{\circ}\right)$y=cos(x−(−30∘))=cos(x+30∘).
Consider the graph of the form $f\left(x\right)=\sin\left(b\left(x+c\right)\right)$f(x)=sin(b(x+c)) shown below.
State the equation of the graph. Give your answer in the form where $c$c is the least positive value.
Think: Compared to the graph of $y=\sin x$y=sinx, the graph of $f\left(x\right)$f(x) has been "squashed" horizontally due to the period change caused by $b$b, and translated horizontally due to the phase shift caused by $c$c.
Do: To determine the period change, recall that $y=\sin x$y=sinx has a period of $360^\circ$360°, so that one whole cycle covers $360^\circ$360° along the $x$x-axis. This means that one whole cycle of $\sin bx$sinbx will cover $\frac{360^\circ}{b}$360°b along the $x$x-axis. From the graph we can see that the period of $f\left(x\right)$f(x) is $120^\circ$120°, so $b=\frac{360}{120}=3$b=360120=3.
Next, we see that the point $\left(0^\circ,0\right)$(0°,0) on the graph of $y=\sin x$y=sinx has been translated to the point $\left(-45^\circ,0\right)$(−45°,0) on the graph of $f\left(x\right)$f(x). This corresponds to a phase shift of $\left(-45\right)^\circ$(−45)°, which means the value of the variable $c$c is $-45^\circ$−45°.
Now we can combine the working above and state the equation of the graph as $f\left(x\right)=\sin\left(3\left(x+45^\circ\right)\right)$f(x)=sin(3(x+45°)).
Reflect: How would the value of $b$b and $c$c change if we were told that the equation of the graph had the form $\sin\left(bx+c\right)$sin(bx+c) instead of $\sin\left(b\left(x+c\right)\right)$sin(b(x+c))?
The form of an equation affects how the corresponding graph transforms.
$a\sin\left(x-c\right)$asin(x−c) has a phase shift of $c$c units to the right.
$a\sin\left(b\left(x-c\right)\right)$asin(b(x−c)) has a phase shift of $c$c units to the right.
$a\sin\left(bx-c\right)$asin(bx−c) has a phase shift of $\frac{c}{b}$cb units to the right.
Determine the equation of the graphed function given that it is of the form $y=\sin\left(x-c\right)$y=sin(x−c), where $c$c is the least positive value and $x$x is in degrees.
Determine the equation of the graph given that it is of the form $y=-\cos\left(x+c\right)-d$y=−cos(x+c)−d, where $c$c is the least positive value and $x$x is in degrees.
In Smithers, the temperature is recorded every 6 hours for 48 hours.
Day/Time | Temperature (°C) | Day/Time | Temperature (°C) |
---|---|---|---|
Monday 12am | $11$11 | ||
Monday 6am | $16$16 | Tuesday 6am | $18$18 |
Monday 12pm | $23$23 | Tuesday 12pm | $25$25 |
Monday 6pm | $18$18 | Tuesday 6pm | $17$17 |
Tuesday 12am | $10$10 | Wednesday 12am | $11$11 |
Plot the data over a two-day interval, where $x=0$x=0 corresponds to 12am on Monday.
Graph $f\left(x\right)$f(x), a sine function (with positive sign) that passes through points $P$P, $Q$Q, $R$R and $S$S.
To model the temperature, find $f\left(x\right)$f(x), if it has the form $a\sin\left(b\left(x-c\right)\right)+d$asin(b(x−c))+d, where $a$a, $b$b, $c$c and $d$d are constants, $c$c is the least positive value, and $x$x is in degrees.