[BlindMath] Interesting way to present Trig to the blind.

[Ray McAllister]

Yes. 
You need a little foundational knowledge, but 

I is the imaginary constant.  I^2 = -1.

Cos(x) = 
1/2 * (e^(I x) + e^( -I x)). 

Sin(x) = -i/2 * (e^(ix) - e^(-I x))


These are derived from euler's Formula 
E^(ix) = cos(x) + I sin(x). 

So, with these formula. You want to show that sin(-x) = - sin(x)  Just run -x instead of X. 

You get -I /2 *(e^(-ix) - e^(ix)) 

Well, just switch the terms, and you have -sin(x).  

You need a little basic complex number work, but not a whole complex analysis course.  What you had in Algebra II is good. 

You want to prove that sin(a+b) = sin(a) cos(b) + sin(b) cos(a).
All you need to do is just run those formulas for sine and cosine with those variables in them and watch how things cancel out.  

Ray. 

-----Original Message-----
From: BlindMath <blindmath-bounces at nfbnet.org> On Behalf Of Pranav Lal via BlindMath
Sent: Tuesday, August 25, 2026 12:20 PM
To: 'Blind Math list for those interested in mathematics' <blindmath at nfbnet.org>
Cc: Pranav Lal <pranav.lal at gmail.com>
Subject: Re: [BlindMath] Interesting way to present Trig to the blind.

Hi Ray,

Tell me more. I have done proofs over 20 years ago. These were proofs like proving the mid point theorem and I think the Pythagoras theorem. They did use trigonometry I think at some points but were mainly focused on the properties of triangles..

What you appear to imply that instead of using geometry, you can use equations to prove these theorems which would be more understandable to a blind person.

Am I understanding you correctly?
Pranav


_______________________________________________
BlindMath mailing list
BlindMath at nfbnet.org
http://nfbnet.org/mailman/listinfo/blindmath_nfbnet.org
To unsubscribe, change your list options or get your account info for BlindMath:
http://nfbnet.org/mailman/options/blindmath_nfbnet.org/raymcal%40att.net
BlindMath Gems can be found at <http://www.blindscience.org/blindmath-gems-home>