Trigonometric Graphs

NZ Level 7 (NZC) Level 2 (NCEA)

Finding equations of sine and cosine curves

Lesson

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`)=`a``s``i``n`(`b``x`−`c`)+`d` or $f\left(x\right)=a\cos\left(bx-c\right)+d$`f`(`x`)=`a``c``o``s`(`b``x`−`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:

- The vertical translation is the value of $d$
`d`in the general equations. It is the central value of the function and it allows us to construct an imaginary line that the function appears to oscillate around. - The amplitude is the maximum deviation of the function from the central line. This is the value of $a$
`a`in the general form of the equation. - The period is the length of one complete cycle. The value of the period is equal to $\frac{360^\circ}{b}$360°
`b`. If we know the period, we can deduce the value of $b$`b`. - The phase shift is the horizontal translation of the graph compared with an unmodified sine or cosine graph. The phase shift is equal to $\frac{c}{b}$
`c``b`. If the shift is to the right, then $c$`c`is negative. - If the sine curve starts at the central line but decreases, or the cosine curve starts at the minimum and increases, this can be interpreted as a reflection. If there is a reflection, then the value of $a$
`a`has to be negative. (A reflection is equivalent to a phase shift of a suitable size and either approach will lead to a valid formula.)

EXAMPLE 1

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`)=`a``c``o``s`(`b``x`−`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 $\frac{c}{b}$`c``b`. (Remember that shifting left is positive.) So, $\frac{c}{2}=22\frac{1}{2}^\circ$`c`2=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`)=2`c``o``s`(2`x`+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`)=`a``s``i``n`(`b``x`−`c`)+`d`, we have

$y(x)=-0.6\sin\frac{x}{2}-0.5$`y`(`x`)=−0.6`s``i``n``x`2−0.5

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`)=0.6`s``i``n`(`x`2−180°)−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`)=0.6`c``o``s`(`x`2+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`=`a``s``i``n``x` or $y=a\cos x$`y`=`a``c``o``s``x`, where $x$`x` is in degrees.

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Determine the equation of the graphed function given that it is of the form $y=\sin bx$`y`=`s``i``n``b``x` or $y=\cos bx$`y`=`c``o``s``b``x`, where $b$`b` is positive and $x$`x` is in degrees.

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Determine the equation of the graphed function given that it is of the form $y=\cos\left(x-c\right)$`y`=`c``o``s`(`x`−`c`), where $c$`c` is the least positive value and $x$`x` is in degrees.

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Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

Apply graphical methods in solving problems