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

[Ray McAllister]

Yes, and here's how to do the Pythagorean Theorem. 
A^2 + b^2 = c^2.  

Now, we rewrite that as 
A^2 / c^2 + b^2 / c^2 = 1.

That's cos^2 (x) + sin^2 (x) = 1.

Now, if you multiply cos^2 + sin^2 from the E formulas, you get 1.  

I^2 = -1.  I^-I = 1. 

E^I * e^I = e^(2i), and e^I * e^(-i) = e^0 or 1.  

So if you say 
(1/2 (e^(ix) + e^(-ix)))^2 (cosine) 
+
(-i/2 (e^(ix) - e^(-ix))) ^2 (sine) 
You get 1. 

No diagrams. 

The idea would be to work up a lesson or 2 explaining the foundation of this.  Part of this probably would be the Taylor series for E, sine, and cosine. Those are infinite series formulas that we can work together into each other and prove all kinds of things.  

Email me privately if you want to look into this more. 

Ray. 



-----Original Message-----
From: BlindMath <blindmath-bounces at nfbnet.org> On Behalf Of Pranav Lal via BlindMath
Sent: Tuesday, August 25, 2026 7:52 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,

This is certainly promising. I was not educated in the USA so do not when algebra 2 is taught. However, as a blind student, this approach would be easier for me to understand vs. the geometry properties approach.

Pranav


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