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Put angles on the unit circle, convert between degrees and radians, read signs by quadrant, and get amplitude, period and midline from a wave's equation.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson sine and cosine stop being ratios in a right triangle and become coordinates of a point on the unit circle, which lets them keep working for angles bigger than a right angle and for angles past a full turn. You measure angles in radians as well as degrees, read the sign of each function off the quadrant, and then unroll the walk round the circle into a wave whose amplitude, period and midline you can read from its equation — or build from a description of something that repeats.
In a right triangle you already write $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$. You plot points $(x, y)$ and know the circle of radius $1$ about the origin is $x^2 + y^2 = 1$. This lesson glues them together: put the triangle inside that circle and the hypotenuse is $1$, so the ratios become coordinates.
Unit circle: the circle of radius $1$ centred at the origin.
Standard position: an angle measured anticlockwise from the positive $x$-axis, vertex at the origin.
Radian: the angle cutting off an arc of length $1$ on the unit circle. A full turn is $2\pi$ radians, a half turn $\pi$.
Quadrant: one of the four quarters the axes cut the plane into, numbered anticlockwise from the top right.
Quadrantal angle: one whose point lands on an axis.
Reference angle: the acute angle between the radius and the $x$-axis.
Draw the unit circle and measure an angle $\theta$ anticlockwise from the positive $x$-axis. The point where that angle meets the circle is defined to be $(\cos\theta, \sin\theta)$. Because the radius is $1$, the old right-triangle ratios come out unchanged for acute angles — but this definition also works at $150^\circ$, at $270^\circ$ and past a full turn, where no triangle exists.
Angles are measured in radians as often as in degrees. A half turn is $180^\circ$ and also $\pi$ radians, so degrees become radians when multiplied by $\frac{\pi}{180}$: $30^\circ = \frac{\pi}{6}$, $45^\circ = \frac{\pi}{4}$, $90^\circ = \frac{\pi}{2}$.
The quadrantal points come straight off the picture: $(1, 0)$, $(0, 1)$, $(-1, 0)$, $(0, -1)$ at $0^\circ$, $90^\circ$, $180^\circ$, $270^\circ$. Between them the signs follow the quadrant, since $\cos\theta$ is the $x$ and $\sin\theta$ is the $y$. Positive in quadrants 1 to 4: both, only sine, neither, only cosine.
Three angles are worth knowing by heart: $30^\circ$ gives $(\frac{\sqrt{3}}{2}, \frac{1}{2})$, $45^\circ$ gives $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$, $60^\circ$ gives $(\frac{1}{2}, \frac{\sqrt{3}}{2})$. Every exact value here is one of those with a sign attached.
Another way: picture
A circle of radius $1$, a point on it, a vertical drop to the $x$-axis and a horizontal run back to the origin. The run is $\cos\theta$, the drop is $\sin\theta$, and Pythagoras on that triangle reads $\cos^2\theta + \sin^2\theta = 1$.
Another way: story
A point walks anticlockwise round the circle at a steady pace. Its shadow on the $x$-axis is the cosine, its shadow on the $y$-axis the sine. Both slide back and forth forever between $-1$ and $1$, which is why these functions repeat.
Converting the wrong way. To go from degrees to radians multiply by $\frac{\pi}{180}$; to come back multiply by $\frac{180}{\pi}$. Check the size: $90^\circ$ should give about $1.57$, not $5157$.
"Sine is the $x$." It is the $y$. The point is $(\cos\theta, \sin\theta)$, in alphabetical order, and cosine comes first.
Losing the sign outside quadrant 1. $\sin 150^\circ$ and $\sin 30^\circ$ are both $\frac{1}{2}$, but their cosines differ by a minus sign. Size comes from the reference angle, sign from the quadrant, and you need both.
Multiply by $\frac{\pi}{180}$: $210 \times \frac{\pi}{180} = \frac{210}{180}\pi$.
The degrees go on top, $180$ underneath.
$\frac{210}{180} = \frac{7}{6}$, so $210^\circ = \frac{7}{6}\pi$ radians.
A little more than $\pi$, which fits: $210^\circ$ is just past a half turn.
$240^\circ$ is $60^\circ$ past $180^\circ$, so it lies in quadrant 3 with reference angle $60^\circ$.
Its distance to the nearest half of the $x$-axis.
At $60^\circ$ the point is $(\frac{1}{2}, \frac{\sqrt{3}}{2})$; in quadrant 3 both coordinates are negative.
Size from the reference angle, sign from the quadrant.
So $\cos 240^\circ = -\frac{1}{2}$ and $\sin 240^\circ = -\frac{\sqrt{3}}{2}$.
$135^\circ$ is $45^\circ$ short of $180^\circ$, so it is in quadrant 2 with reference angle $45^\circ$.
At $45^\circ$ both coordinates are $\frac{\sqrt{2}}{2}$; in quadrant 2 the $x$ turns negative, so the point is $(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.
Convert $45^\circ$ to radians. Give the coefficient of $\pi$.
Answer:
In which quadrant does the angle $260^\circ$ lie?
Match each quadrant to the signs of $\sin\theta$ and $\cos\theta$ there.
| $\sin\theta$ positive, $\cos\theta$ positive | $\sin\theta$ positive, $\cos\theta$ negative | $\sin\theta$ negative, $\cos\theta$ negative | $\sin\theta$ negative, $\cos\theta$ positive | |
|---|---|---|---|---|
| Quadrant 1, for instance $70^\circ$ | ||||
| Quadrant 2, for instance $160^\circ$ | ||||
| Quadrant 3, for instance $250^\circ$ | ||||
| Quadrant 4, for instance $340^\circ$ |
Where on the unit circle is the point at $90^\circ$? Fill in $x$ and $y$.
At 90 degrees, (x, y) = (px, py)
You already read a parabola's graph off $y = a(x - h)^2 + k$: $a$ stretches it, $h$ slides it sideways, $k$ slides it up. You now know that $\sin\theta$ is the height of a point walking round the unit circle. Unroll that walk along a time axis and you get a wave, and the very same three letters do the very same three jobs to it.
Periodic: a function that repeats the same block of values forever.
Period: the length of one repeat, measured along the horizontal axis.
Amplitude: half the distance from the lowest value to the highest — how far the curve rises above its middle. Always positive.
Midline: the horizontal line halfway between the peaks and the troughs.
Phase shift: a sideways slide of the whole curve.
Frequency: how many full cycles fit in one unit of time; the reciprocal of the period.
Let the angle be time and let the point keep walking round the circle. Its height traces $y = \sin t$: a wave that starts at $0$, peaks at $1$, returns through $0$ to $-1$, and is back where it began after $2\pi$. Its horizontal position traces $y = \cos t$, the same wave started at its peak. Both have period $2\pi$, amplitude $1$ and midline $y = 0$.
Now put the three letters in: $$y = a\sin(bt) + k.$$ Amplitude is $|a|$: the curve rises $|a|$ above the midline and falls $|a|$ below it. A negative $a$ flips the wave upside down but does not change its amplitude, because a distance has no sign.
Period is $\frac{2\pi}{b}$: the inside must grow by $2\pi$ for one full cycle, and it grows $b$ times as fast as $t$ does. So $y = \sin(3t)$ has period $\frac{2\pi}{3}$, and $y = \sin\!\left(\frac{t}{2}\right)$ has period $4\pi$.
Midline is $y = k$, so the greatest value is $k + |a|$ and the least is $k - |a|$.
To build a model from a description, run this backwards. Given the highest and lowest values, the amplitude is half their difference and the midline is their average. Given how long one repeat takes, solve $\frac{2\pi}{b} = \text{period}$ for $b$. Choose cosine when the quantity is at its highest at $t = 0$ and sine when it is at its midline there.
Another way: picture
One cycle of $y = 3\sin(2t) + 1$. A dashed line at $y = 1$ is the midline; the peak at $y = 4$ and the trough at $y = -2$ are each $3$ from it; the pattern is finished and starting again by $t = \pi$.
Another way: story
A Ferris wheel. Its radius is the amplitude, the height of its axle is the midline, and how long it takes to go round once is the period. Nothing else about your seat's height can change.
"The amplitude is the whole height." For $y = 3\sin x$ the curve runs from $-3$ to $3$, a total height of $6$, but the amplitude is $3$. Amplitude is measured from the midline, not from trough to peak.
"A bigger $b$ means a longer wave." The opposite. The period of $\sin(bx)$ is $\frac{2\pi}{b}$, so $b = 4$ packs four cycles into the space $\sin x$ needs for one.
Forgetting the midline in a maximum. The greatest value of $2\sin x - 5$ is not $2$; it is $-5 + 2 = -3$. Find the middle first, then go one amplitude up.
Amplitude $= |-4| = 4$; the minus flips the curve, so it goes down first.
A distance is never negative.
Period $= \frac{2\pi}{3}$, so three whole cycles fit where $\sin t$ fits one.
Divide $2\pi$ by the coefficient inside.
Midline $y = 2$, so the values run from $2 - 4 = -2$ up to $2 + 4 = 6$.
Amplitude $= \frac{18 - 2}{2} = 8$; midline $= \frac{18 + 2}{2} = 10$.
Half the range, then the average.
Period $40$ gives $\frac{2\pi}{b} = 40$, so $b = \frac{\pi}{20}$.
Highest at $t = 0$ means cosine: $h = 8\cos\!\left(\frac{\pi}{20}t\right) + 10$.
Check: at $t = 0$ this gives $18$.
Period $= \frac{2\pi}{4} = \frac{\pi}{2}$.
Midline $-3$, amplitude $5$, so the least value is $-3 - 5 = -8$.
$y = -\sin(6x) - 7$. What is the amplitude?
Answer:
$y = \sin(8x)$ has period $k\pi$. What is $k$?
Answer:
$y = 4\sin x + 1$. What is the greatest value $y$ takes?
Answer:
$y = -6\sin(5x) + 2$. What is the amplitude?
Answer:
The tide at a harbour runs between $4$ m and $10$ m, is at its highest at $t = 0$, and repeats every $8$ hours. Model it as $h = a\cos(bt) + k$ with $b$ a multiple of $\pi$: fill in $a$, $k$, and the coefficient of $\pi$ in $b$.
h = amp cos(coef pi t) + mid
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Convert $60^\circ$ to radians. Give the coefficient of $\pi$.
Answer:
$y = \sin(2x)$ has period $k\pi$. What is $k$?
Answer:
You can place an angle on the unit circle and read a wave from its equation. Convert $150^\circ$ to radians, say which quadrant $200^\circ$ is in and what sign $\cos 200^\circ$ has, and give the amplitude, period and midline of $y = 6\sin(2t) - 1$.
8. Your turn: where is the point at $135^\circ$?, step 2
19. Your turn: the period and least value of $y = 5\sin(4t) - 3$, step 2