An Interesting Trig Circle

We are going to go through a (light) development of some trig identities by using geometry. We will not get into specifics or prove these claims in this post. There might be a post later to prove these statements. So without further ado let’s start with a circle.


Choose two points on the circle that are not antipodes. We shall label them A and B. 


We shall now form the triangle with the center, A, and B.


Now drop the perpendicular from the center to the line segment AB.


The word “sine” comes from a Latin mistranslation of the Arabic word  jayb which is a translation of the Sanskrit word jy\overline{a} which means “bowstring.” Cosine means “complement of sine.”, so named because it is a phase shift of sine by \pi/2 radians.


One might be able to see that if we were to rotate this picture by \pi/2 radians, the current length of \sin\theta would become the cosine length and vice versa. Now if we draw a line segment parallel to AB with the center and upper point on the circle as endpoints, we see that there is a length leftover. Also there is a length left over from the horizontal line. We call these lengths \mathrm{cvs}\theta and \mathrm{vsin}\theta respectively.


Now take the line tangent to the circle at point A. From A to the intersection of the tangent line and horizontal line is the length \tan\theta. The other obvious length created by this tangent line is \cot\theta.


Now the length from the center to the other clear point on the horizontal line is exactly \sec\theta. Leaving the complementary case to be \csc\theta.


Finally the two obvious remaining lengths are what we call \mathrm{exsec} and \mathrm{excsc}.


Now from just the sine function we can obtain all of the others. We present the following identities.


\cos\theta = \sin(\theta + \frac{\pi}{2})

\mathrm{vsin}\theta  = 1-\cos\theta = 1-\sin(\theta + \frac{\pi}{2})

\mathrm{cvs}\theta = 1-\sin\theta

\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\sin\theta}{\sin(\theta + \frac{\pi}{2})}

\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{\sin(\theta + \frac{\pi}{2})}{\sin\theta}

\sec\theta = \frac{1}{\cos\theta} = \frac{1}{\sin(\theta + \frac{\pi}{2})}

\csc\theta = \frac{1}{\sin\theta}

\mathrm{exsec}\theta = \sec\theta - 1 = \frac{1}{\sin\theta} - 1

\mathrm{excsc}\theta = \csc\theta - 1 = \frac{1}{\sin\theta} - 1

From this picture, many more identities could be conjectures such as (\cot\theta + \tan\theta)^2 = \sec^2\theta + \csc^2\theta and $\cos\theta + \mathrm{vsin}\theta + \mathrm{exsec}\theta = \sec\theta$.We encourage the reader to further explore this circle!


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s