## A geometric proof of trigonometric derivatives

As we already know, the first derivatives of and are and respectively. The question is how do we prove it?

Since high school, the only proof that I knew is to use two of the trigonometric identities which are:

- , and
- .

Hence,

.

The derivative of can also be derived using the second identity.

However, there is another proof, which is (I think) more elegant. It is using a geometric interpretation of and . Let us draw a unit circle in .

Thus we have and . We are looking for and, however we are not going to discuss the first derivative of as it can be done in the same way as .

We are now going to estimate , see the second figure. The following relationship applies:

. or .

Thus,

,

where is the chord PQ and is the arc PQ. As goes to zero, the ratio of the arc PQ and the chord PQ tends to 1 and the chord PQ tends to become a tangent line of the unit circle at , therefore . Assuming that the limit of multiplication is the same as the multiplication of limit, we have the following:

.

As mentioned before, the first derivative of can be done the same way.

Also, I should thank Pak Yudi Soeharyadi as I got the idea of this geometric proof from him.

Nice to get to know your all posts and surfing inside!

Thank you.. It is also nice to read yours. Quite motivational.

-salam hangat dari Tangerang-

I do accept as true with all the ideas you’ve presented in your post. They are very convincing and will certainly work. Nonetheless, the posts are too quick for newbies. May just you please extend them a bit from subsequent time? Thanks for the post.

I think your second-listed identity should read $\cos\alpha-\cos\beta = -2 \sin\left(\frac{\alpha-\beta}{2}\right) \sin\left(\frac{\alpha+\beta}{2}\right)$

Thanks for the comment. I edited it.