使用单位圆求角的正弦值

Consider the following diagram:

Here, we can see that having picked the point $(a,b)$ in the first quadrant of the plane, the line from the origin to $(a,b)$ forms a hypotenuse of a right triangle. This line has length $r$.

The length of the adjacent is just the distance of the right angle from the origin, so its value is just $a$. Similarly, the length of the opposite is $b$.

Therefore, from our definitions of sine and cosine, $$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{r}$$ and also $$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{b}{r}.$$

Now, consider the following diagram. In this one, $r=1$.

It turns out that now the value of the coordinate on the circle is just $(\cos\theta, \sin\theta)$!

Recall that in our first graph, $\sin\theta = \frac{a}{r}$ and $\cos\theta = \frac{b}{r}$. When $r = 1$, $a = \sin\theta$ and $b = \cos\theta$. Therefore, the $x$ coordinate is just $\cos\theta$ and the $y$coordinate $\sin\theta$.

Given any unit circle and an angle $\theta$, you can just draw that angle in. At the point where the angle meets the unit circle, the values of cosine and sine can be read straight from the $x$ and $y$ coordinates!

Thinking about sines and cosines in a visual manner like this can make your life much easier, especially when you're dealing with special angles like $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$, etc. This is because the only possible values for sines and cosines of these special angles are $\pm\frac{1}{2}$, $\pm\frac{\sqrt{2}}{2}$, and $\pm\frac{\sqrt{3}}{2}$.

As an example, if you picture the angle $\frac{\pi}{3}$, the point on the unit circle would be further up than it is to the right, so $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos\frac{\pi}{3} = \frac{1}{2}$.

Now for a few practice questions, and we're done!

Visualising Sine, Cosine, and Tangent in a Unit Circle

Mathematicians came up with a way to neatly represent $\sin$, $\cos$, and $\tan$ of an angle, by inscribing it in a unit circle:

Looking at this triangle, we can observe that the hypotenuse of the triangle is the length of the radius of the circle, which is one unit. To find $\sin$ and $\cos$ in terms of the given angle $\alpha$ (alpha, the angle in pink), we can use SOH and CAH.

So $\sin\alpha = \frac{\text{opposite}}{1}$ and $\cos \alpha = \frac{\text{adjacent}}1$. In other words, the length of the adjacent side is the value of $\cos \alpha$ and the length of the opposite side is the value of $\sin \alpha$.

Now if we bring this circle to a cartesian plane, where the radius of the circle is the origin, we will get something like this:

With this, we can now say that the coordinates of the point where the triangle in the first quadrant touches the circle is $(\cos \alpha , \sin \alpha)$. With this representation, we can now proceed to find the values of angles that are larger than $90$ degrees or even negative!

So if we want to find the values of $\sin$ and $\cos$ of the angle $180^{\circ} - \alpha$, we can consider the triangle in Quadrant $\text{II}$. The coordinates of the point of intersection between the circle and the triangle in Quadrant $\text{II}$ will give us the values of $\sin$ and $\cos (180 - \alpha)$.

Since the angle $\alpha$ is the same, the values of $\sin$ and $\cos$ remain the same but since they are on the negative $x$ side, the value of $\cos (180- \alpha)$ will be $-\cos \alpha$ instead of $\cos \alpha$ and since the point is still on the positive $y$-axis, $\sin (180-\alpha) = \sin \alpha$.

Now let's think about reflex angles. Try your hand at finding the values of $\sin$ and $\cos$ of angles in the third and fourth quadrant! Be careful with the signs of the values, and have fun!

Once you've mastered this topic, you can extend it to solving trig equations and inverses!

The unit circle is a circle with radius 1 centered at the origin of the Cartesian Plane.

In a pair of coordinates $(x,y)$ on the unit circle $x^2+y^2=1$, coordinate $x$ is the cosine of the angle formed by the point, the origin, and the x-axis.

Coordinate $y$ is the sine of the angle.

The tangent of the angle is $\frac{y}{x}$.

Now you try:

(Video) Find Cos and Sin Values Using the Unit Circle Hand Trick!

by Paris Mielke

In this video by Paris Mielke, you will learn the hand trick to use the unit circle!