A point traveling around a circle creates a repeating wave when we record one of its coordinates. This connects the geometry of the unit circle to the family of functions
By the end of this lesson, you will be able to read sine and cosine from the unit circle, explain why \(\sin^2\theta+\cos^2\theta=1\text{,}\) and use \(A\text{,}\)\(w\text{,}\) and \(T\) to describe and sketch a cosine wave. You need the Pythagorean theorem, coordinates, and basic algebra; no previous trigonometry lesson is required.
Work with pencil and paper as you read. The interactive exercises are selected from the Math Trailhead’s existing trigonometry worksheets. Enter exact answers such as pi/3 and sqrt(3)/2 unless a problem requests decimals. Use radians unless a degree symbol appears.
SubsectionThe unit circle: angles become coordinates
The unit circle has center \((0,0)\) and radius \(1\text{.}\) Start at \((1,0)\) and turn through an angle \(\theta\text{.}\) Positive angles turn counterclockwise; negative angles turn clockwise. The angle determines a point on the circle.
An angle in radians is the signed distance traveled along a circle divided by its radius. On the unit circle, the size of the angle equals the arc length. One full turn is \(2\pi\) radians because the circumference is \(2\pi\text{.}\) Therefore
To convert degrees to radians, multiply by \(\pi/180\text{;}\) to convert radians to degrees, multiply by \(180/\pi\text{.}\) For example, \(120^\circ=2\pi/3\) radians and \(-\pi/4=-45^\circ\text{.}\)
In the first quadrant, this agrees with right-triangle trigonometry: cosine is adjacent side divided by hypotenuse and sine is opposite side divided by hypotenuse. Here the hypotenuse is \(1\text{.}\) The coordinate definition continues to work in every quadrant, including points on the axes.
Signs come from coordinates: in quadrant I both are positive; in II cosine is negative and sine positive; in III both are negative; in IV cosine is positive and sine negative. Also, each coordinate lies between \(-1\) and \(1\text{.}\)
You can reconstruct the familiar exact values. At \(\pi/4\text{,}\) the triangle’s legs have the same length \(a\text{,}\) so \(2a^2=1\) gives \(a=\sqrt{2}/2\text{.}\) Bisect an equilateral triangle of side \(1\text{:}\) the short leg is \(1/2\text{,}\) the hypotenuse is \(1\text{,}\) and the long leg is \(\sqrt{3}/2\text{.}\) This gives the \(\pi/6\) and \(\pi/3\) values.
A reference angle is the acute angle from a point’s radius to the horizontal axis. Reflect a first-quadrant point to obtain other quadrants, keeping coordinate magnitudes and changing their signs. For example, \(5\pi/6\) is in quadrant II with reference angle \(\pi/6\text{,}\) so its point is \((-\sqrt{3}/2,1/2)\text{.}\)
Adding or subtracting \(2\pi\) makes one extra full turn and returns to the same point. Thus \(-\pi/6\) and \(11\pi/6\) have the same sine and cosine. This is why both functions repeat every \(2\pi\text{.}\)
If you’re having trouble seeing that, think of \(\frac{1 \text{ft}}{12 \text{in}}\) or \(\frac{60 \text{min}}{1 \text{hr}}\text{.}\) They’re both equivalent to 1 as well.
If you’re having trouble seeing that, think of \(\frac{1 \text{ft}}{12 \text{in}}\) or \(\frac{60 \text{min}}{1 \text{hr}}\text{.}\) They’re both equivalent to 1 as well.
a. This is our reference angle. \({\frac{\pi }{4}}\) is in the first quadrant, where the \(x\)-coordinate (and therefore cosine) is positive.So, \(\cos({\frac{\pi }{4}}) = {\frac{\sqrt{2}}{2}}\text{.}\)
b.\({\frac{3\pi }{4}}\) is in the second quadrant, where the \(x\)-coordinate (and therefore cosine) is negative.So, \(\cos({\frac{3\pi }{4}}) = {-\frac{\sqrt{2}}{2}}\text{.}\)
c.\({\frac{5\pi }{4}}\) is in the third quadrant, where the \(x\)-coordinate (and therefore cosine) is negative.So, \(\cos({\frac{5\pi }{4}}) = {-\frac{\sqrt{2}}{2}}\text{.}\)
d.\({\frac{7\pi }{4}}\) is in the fourth quadrant, where the \(x\)-coordinate (and therefore cosine) is positive.So, \(\cos({\frac{7\pi }{4}}) = {\frac{\sqrt{2}}{2}}\text{.}\)
e.\({\frac{9\pi }{4}}\) is coterminal with \({\frac{\pi }{4}}\text{,}\) and both are in the first quadrant, where the \(x\)-coordinate (and therefore cosine) is positive.So, \(\cos({\frac{9\pi }{4}}) = {\frac{\sqrt{2}}{2}}\text{.}\)
f.\({\frac{-3\pi }{4}}\) is coterminal with \({\frac{5\pi }{4}}\text{,}\) and both are in the third quadrant, where the \(x\)-coordinate (and therefore cosine) is negative.So, \(\cos({\frac{-3\pi }{4}}) = {-\frac{\sqrt{2}}{2}}\text{.}\)
g.\({\frac{-5\pi }{4}}\) is coterminal with \({\frac{3\pi }{4}}\text{,}\) and both are in the second quadrant, where the \(x\)-coordinate (and therefore cosine) is negative.So, \(\cos({\frac{-5\pi }{4}}) = {-\frac{\sqrt{2}}{2}}\text{.}\)
h.\({\frac{-7\pi }{4}}\) is coterminal with \({\frac{\pi }{4}}\text{,}\) and both are in the first quadrant, where the \(x\)-coordinate (and therefore cosine) is positive.So, \(\cos({\frac{-7\pi }{4}}) = {\frac{\sqrt{2}}{2}}\text{.}\)
a. This is our reference angle. \(\displaystyle{\frac{\pi }{4}}\) is in the first quadrant, where the \(y\)-coordinate (and therefore sine) is positive.So, \(\sin\left(\displaystyle{\frac{\pi }{4}}\right) = \displaystyle{\frac{\sqrt{2}}{2}}\text{.}\)
b.\(\displaystyle{\frac{3\pi }{4}}\) is in the second quadrant, where the \(y\)-coordinate (and therefore sine) is positive.So, \(\sin\left(\displaystyle{\frac{3\pi }{4}}\right) = \displaystyle{\frac{\sqrt{2}}{2}}\text{.}\)
c.\(\displaystyle{\frac{5\pi }{4}}\) is in the third quadrant, where the \(y\)-coordinate (and therefore sine) is negative.So, \(\sin\left(\displaystyle{\frac{5\pi }{4}}\right) = \displaystyle{-\frac{\sqrt{2}}{2}}\text{.}\)
d.\(\displaystyle{\frac{7\pi }{4}}\) is in the fourth quadrant, where the \(y\)-coordinate (and therefore sine) is negative.So, \(\sin\left(\displaystyle{\frac{7\pi }{4}}\right) = \displaystyle{-\frac{\sqrt{2}}{2}}\text{.}\)
e.\(\displaystyle{\frac{9\pi }{4}}\) is coterminal with \(\displaystyle{\frac{\pi }{4}}\text{,}\) and both are in the first quadrant, where the \(y\)-coordinate (and therefore sine) is positive.So, \(\sin\left(\displaystyle{\frac{9\pi }{4}}\right) = \displaystyle{\frac{\sqrt{2}}{2}}\text{.}\)
f.\(\displaystyle{\frac{-3\pi }{4}}\) is coterminal with \({\frac{5\pi }{4}}\text{,}\) and both are in the third quadrant, where the \(y\)-coordinate (and therefore sine) is negative.So, \(\sin\left(\displaystyle{\frac{-3\pi }{4}}\right) = \displaystyle{-\frac{\sqrt{2}}{2}}\text{.}\)
g.\(\displaystyle{\frac{-5\pi }{4}}\) is coterminal with \(\displaystyle{\frac{3\pi }{4}}\text{,}\) and both are in the second quadrant, where the \(y\)-coordinate (and therefore sine) is positive.So, \(\sin({\frac{-5\pi }{4}}) = {\frac{\sqrt{2}}{2}}\text{.}\)
h.\(\displaystyle{\frac{-7\pi }{4}}\) is coterminal with \({\frac{\pi }{4}}\text{,}\) and both are in the first quadrant, where the \(y\)-coordinate (and therefore sine) is positive.So, \(\sin({\frac{-7\pi }{4}}) = {\frac{\sqrt{2}}{2}}\text{.}\)
Every point \((u,v)\) on the unit circle is one unit from the origin. The Pythagorean theorem says \(u^2+v^2=1\text{.}\) Substitute \(u=\cos\theta\) and \(v=\sin\theta\text{:}\)
That is the whole derivation. The notation \(\sin^2\theta\) means \((\sin\theta)^2\text{,}\) not \(\sin(\theta^2)\text{.}\) In other quadrants, the triangle’s side lengths are the absolute values of the coordinates. Squaring removes the signs, so the same identity holds for every angle, including the axis points.
So \(\sin\theta=\pm4/5\text{.}\) Quadrant II has positive vertical coordinate, hence \(\sin\theta=4/5\text{.}\) The identity determines the magnitude; the quadrant determines the sign.
Draw a unit circle and explain the identity in one sentence. Then find \(\cos\theta\) when \(\sin\theta=-5/13\) and \(\theta\) is in quadrant III. Finally, simplify \(1-\sin^2\theta\text{.}\)
The squared coordinates of a point on a circle of radius \(1\) sum to \(1\text{.}\) Here \(\cos^2\theta=1-25/169=144/169\text{;}\) quadrant III makes cosine negative, so \(\cos\theta=-12/13\text{.}\) For every angle, \(1-\sin^2\theta=\cos^2\theta\text{.}\)
Now let the angle vary and graph its horizontal coordinate against the angle. This produces \(y=\cos x\text{.}\) As the point goes around the circle, its horizontal coordinate follows the pattern \(1,0,-1,0,1\) at the five landmarks above. Plotting the vertical coordinate instead gives \(y=\sin x\text{,}\) with pattern \(0,1,0,-1,0\text{.}\)
The midline is the line halfway between the maximum and minimum; here it is \(y=0\text{.}\) The amplitude is the distance from the midline to a peak; here it is \(1\text{.}\) The period is the length of the shortest positive horizontal shift that repeats the entire graph; here it is \(2\pi\text{.}\) A maximum and the next minimum are only half a period apart.
To see this on the circle, rotate \((u,v)\) clockwise by a quarter turn. Its coordinates become \((v,-u)\text{,}\) so the new horizontal coordinate is the old vertical coordinate. This also lets us describe sine waves using the cosine family.
In \(f(x)=A\cos(wx+T)\text{,}\) call the angle inside cosine \(\theta=wx+T\text{.}\) Starting with an input \(x\text{,}\) first compute this angle, then read its horizontal coordinate on the unit circle, then multiply that coordinate by \(A\text{.}\) The parameter \(w\) is often written as the Greek letter \(\omega\text{.}\) We use \(P\) for the period: \(T\) here is an angle, not the period.
Multiplying by \(A\) multiplies every height by \(A\text{.}\) Thus \(3\cos x\) ranges from \(-3\) to \(3\text{,}\) and \(\tfrac12\cos x\) ranges from \(-1/2\) to \(1/2\text{.}\) A negative \(A\) also reflects the graph across the horizontal axis: \(-3\cos x\) begins at \(-3\) rather than \(3\text{.}\) Amplitude is a distance, so it is \(|A|\text{,}\) never negative.
\(y = {-2\sin\mathopen{}\left(x\right)}\) is the graph of sine with negative coefficient (so it is the vertically-reflected, or ’upside-down’, version of sine) and amplitude 2; this corresponds to graph C.
Increasing \(x\) by \(\Delta x\) changes the angle by \(w\Delta x\text{.}\) A complete turn requires an angular change of size \(2\pi\text{.}\) For \(A\ne0\) and \(w\ne0\text{,}\)
For example, \(\cos(2x)\) makes a full turn while \(x\) increases by only \(\pi\text{;}\) its graph is horizontally compressed. The graph of \(\cos(x/2)\) needs \(4\pi\) units for a turn and is horizontally stretched. Larger \(|w|\) means a shorter period.
If \(w<0\text{,}\) the angle decreases as \(x\) increases, so the point travels clockwise. The period still uses \(|w|\text{.}\) Reflecting the circle across the horizontal axis shows \(\cos(-\theta)=\cos\theta\text{,}\) so \(\cos(-2x)=\cos(2x)\text{.}\) With a nonzero phase angle, keep track of that sign too: \(\cos(-2x+T)=\cos(2x-T)\text{.}\)
The graph of \(A\cos(w(x-h))\) is the graph of \(A\cos(wx)\) shifted right by \(h\) if \(h>0\text{,}\) or left by \(|h|\) if \(h<0\text{.}\) Equivalently, solve \(w x+T=0\) to find where the original angle-zero point has moved.
For example, \(\cos(2x+\pi/2)=\cos(2(x+\pi/4))\) shifts \(\cos(2x)\) left by \(\pi/4\text{,}\) not \(\pi/2\text{.}\) The plus sign describes a head start around the circle. The graph reaches a given angle earlier, which explains the shift to the left when \(w>0\text{.}\)
Changing \(T\) does not change amplitude or period. Adding \(2\pi k\) to \(T\text{,}\) for any integer \(k\text{,}\) gives the same function; horizontal shifts that differ by whole periods are equivalent. If you change \(w\) while keeping \(T\) fixed, both the period and the shift \(-T/w\) generally change.
Two exceptions deserve their own check: if \(A=0\text{,}\) then \(f(x)=0\text{;}\) if \(w=0\text{,}\) then \(f(x)=A\cos T\) is constant. Constant functions have no smallest positive period. In particular, do not divide by \(w\) when \(w=0\text{.}\)
For \(g(x)=-2\cos(3x-\pi)\text{,}\) find the amplitude, period, horizontal shift, and \(g(0)\text{.}\) Explain why its amplitude is not negative and its shift is not \(\pi\text{.}\)
The amplitude is \(2\text{,}\) the period is \(2\pi/3\text{,}\) and \(3x-\pi=3(x-\pi/3)\) gives a shift right by \(\pi/3\text{.}\) Also \(g(0)=-2\cos(-\pi)=2\text{.}\) The minus sign reflects the graph; amplitude is a positive distance. The phase angle must be divided by the coefficient of \(x\) to find the shift.
Looking at the grid lines, we see that the interval between 0 and \(\pi\) has been split into thirds, and our graph has been shifted left by one of those thirds.
If we want an equation for the cosine wave that has been shifted left by one-thirds of \(\pi\text{,}\) we need to take \(y = \cos(x)\) and replace \(x\) by \(\left( {x+\frac{\pi }{3}} \right)\text{.}\)
Looking at the grid lines, we see that the interval between 0 and \(\pi\) has been split into quarters, and our graph has been shifted right by one of those quarters.
If we want an equation for the cosine wave that has been shifted right by one-quarters of \(\pi\text{,}\) we need to take \(y = \cos(x)\) and replace \(x\) by \(\left( {x-\frac{\pi }{4}} \right)\text{.}\)
Looking at the grid lines, we see that the interval between 0 and \(\pi\) has been split into sixths, and our graph has been shifted right by one of those sixths.
If we want an equation for the cosine wave that has been shifted right by one-sixths of \(\pi\text{,}\) we need to take \(y = \cos(x)\) and replace \(x\) by \(\left( {x-\frac{\pi }{6}} \right)\text{.}\)
Looking at the grid lines, we see that the interval between 0 and \(\pi\) has been split into halves, and our graph has been shifted left by one of those halves.
If we want an equation for the cosine wave that has been shifted left by one-halves of \(\pi\text{,}\) we need to take \(y = \cos(x)\) and replace \(x\) by \(\left( {x+\frac{\pi }{2}} \right)\text{.}\)
Looking at the grid lines, we see that the interval between 0 and \(\pi\) has been split into thirds, and our graph has been shifted right by one of those thirds.
If we want an equation for the cosine wave that has been shifted right by one-thirds of \(\pi\text{,}\) we need to take \(y = \cos(x)\) and replace \(x\) by \(\left( {x-\frac{\pi }{3}} \right)\text{.}\)
Looking at the grid lines, we see that the interval between 0 and \(\pi\) has been split into sixths, and our graph has been shifted left by five of those sixths.
If we want an equation for the cosine wave that has been shifted left by five-sixths of \(\pi\text{,}\) we need to take \(y = \cos(x)\) and replace \(x\) by \(\left( {x+\frac{5\pi }{6}} \right)\text{.}\)
SubsectionBuild a graph, then read an equation from a graph
For \(w>0\text{,}\) the five quarter-turn landmarks occur at \(x=h,h+P/4,h+P/2,h+3P/4,h+P\text{.}\) Their heights are \(A,0,-A,0,A\text{.}\) This works even when \(A<0\text{:}\) the first point is then a minimum. Join the landmarks with a smooth wave and repeat.
Here \(A=3\text{,}\)\(w=2\text{,}\) and \(T=\pi/2\text{.}\) The amplitude is \(3\text{,}\)\(P=\pi\text{,}\) and \(h=-\pi/4\text{.}\) Each quarter period has length \(\pi/4\text{.}\)
The graph starts this cycle at a maximum, falls through the midline, reaches a minimum, and returns to the maximum. As a check, \(f(0)=3\cos(\pi/2)=0\text{.}\)
The midline is \(0\) and the amplitude is \((2-(-2))/2=2\text{.}\) Choose \(A=2\text{.}\) The period is \(5-1=4\text{,}\) so choose \(w=2\pi/4=\pi/2\text{.}\) Place a peak at \(h=1\text{:}\)
Then, remember that from a function \(\sin( B x )\) or \(\cos( B x )\text{,}\) the period \(p\) is given by the equation \(\displaystyle{ p = \frac{ 2 \pi }{ B } }\text{.}\)
Some Trailhead problems include a vertical translation: \(f(x)=D+A\cos(wx+T)\text{.}\) The added \(D\) moves the midline to \(y=D\text{;}\) the maximum and minimum become \(D+|A|\) and \(D-|A|\text{.}\) Amplitude, period, and horizontal shift stay the same. For instance, \(4+2\cos(3x)\) has midline \(4\text{,}\) range \([2,6]\text{,}\) and period \(2\pi/3\text{.}\)
The same transformation rules apply to \(D+A\sin(w(x-h))\text{.}\) For \(w>0\text{,}\) its quarter-cycle heights, starting at \(x=h\text{,}\) are \(D,D+A,D,D-A,D\text{.}\) If you convert it to cosine using the quarter-turn identity, its phase angle is \(T=-wh-\pi/2\text{.}\)
A wave with midline \(0\) has amplitude \(4\text{,}\) period \(6\text{,}\) and a maximum at \(x=-1\text{.}\) Write one equation in the form \(A\cos(wx+T)\text{.}\)
Amplitude \(2\text{,}\) period \(4\pi\text{,}\) shift right \(\pi\text{.}\) The points are \((\pi,2)\text{,}\)\((2\pi,0)\text{,}\)\((3\pi,-2)\text{,}\)\((4\pi,0)\text{,}\) and \((5\pi,2)\text{.}\)
Choose \(A=4\text{,}\)\(w=\pi/3\text{,}\) and \(T=\pi/3\text{,}\) giving \(4\cos(\pi x/3+\pi/3)\text{.}\) At \(x=-1\) the angle is zero; six units later it is \(2\pi\text{.}\)
The angle \(2x\) still locates a point on a circle of radius \(1\text{.}\) Its squared coordinates sum to \(1\text{.}\) Changing the angle changes the point, not the radius.
