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.