# Using unit circle to explain $\cos(0) = 1$ and $\sin(90) = 1$

We have been taught $\cos(0) = 1$ and $\sin(90) = 1$.

But, how do I visualize these angles on the unit circle?

Suppose you have an angle $\theta$ in the unit circle. Then, the functions $\cos\theta$ and $\sin\theta$ represent the $x$ and $y$ components respectively. See the image below.

Plugging the values $0$ and $90$ you will get it.

• If the brown point is (a,b), cos(0) = adjacent/hypotenuse = a/1 and sin(90) will be opposite/hypotenuse = b/1, so how do cos(0) becomes 0 and sin(90) = 1? Jul 24, 2017 at 14:48
• what should the values for a & b be? Jul 24, 2017 at 14:49
• @youcanlearnanything You want to see why when $\theta=0$ then $\cos(\theta)=0$ and when $\theta=\pi/2$ ($90^\circ$) then $\sin(\theta)=1$. In those cases, the brown point in the image would be in the $x$-axis (because, again, $\theta=0$) and in the $y$-axis (for $\theta=\pi/2$). Then, in the first case the cosine (the $x$ value) must be $0$, and the sine (the $y$ value) must be $1$.
– Edu
Jul 24, 2017 at 15:00
• I have updated the 2nd picture. In the illustration, when cos(0) = 0, x and y should be (0,0) since the angle is 0, and for sin(90) = 1, the x is 0 and y = 1, but where is the hypotenuse? Jul 26, 2017 at 14:10
• the green line represents the opposite, and blue line represents the adjacent. Jul 26, 2017 at 14:17

Here is one way to do it:

Here is another one regarding the sine function:

• I like this one the best because it is the least cluttered, and because it contains a huge fraction of the trig you need for the rest of your life packed in it. Jul 24, 2017 at 14:33

$\theta$ is the angle measured from the $x$-axis in a counter clockwise direction.

We know that

$$\sin\theta=\frac{OPPOSITE}{HYPOTENUSE}$$

$$\cos\theta=\frac{ADJACENT}{HYPOTENUSE}$$

For a unit circle $r=1$

$$x=\cos \theta$$

$$y=\sin \theta$$

When $\theta=0$

The length is of $ADJACENT=HYPOTENUSE$

Adjacent means horizontal distance measure from the origin to the point.

Meaning if we ratio them

$$\cos 0=1$$

Opposite is the vertical distance. We don't have any vertical distance when $\theta=0$

$$\sin 0=0$$

Let's us see for the opposite length

As $\theta=\frac{\pi}{2}$

We see that the horizontal distance is disappearing for $\theta=\frac{\pi}{2}$ and the vertical distance is increasing.At $\theta=\frac{\pi}{2}$ the vertical distance equals to the hypotenuse.

Taking

$$\sin\frac{\pi}{2}=1$$

$$\cos\frac{\pi}{2}=0$$

The standard way is with the right angle leg along the positive $x$-axis, so that e.g. acute angles are exactly the ones with the other angle leg within the first quadrant.

Then look at the point where the second angle leg intersects the unit circle. It will have coordinates $(\cos x, \sin x)$.

For instance, $\cos 0^\circ = 1$ because the angle $0$ has its second leg along the positive $x$-axis, and it intersects the unit circle in $(\cos0^\circ, \sin 0^\circ)$. But that point is also $(1,0)$, since it's along the positive $x$-axis. Thus the two sets of coordinates are equal, and we get $\cos 0^\circ = 1$ and $\sin 0^\circ = 0$. The case for $\cos90^\circ = 0$ and $\sin 90^\circ$ is sone similarily, but this time the second angle leg goes straight up along the positive $y$-axis.