Pythagorean Trigonometry a quick proof

[Sin(x)]^2+[Cos(x)]^2=1. This is one of the first proofs that we learn in trigonometry. It is based on the pythagorean formula.

Okay lets have a few definitions:

On a right triangle:

Sin(x)=opposite leg/hypotenuse or =opp/hyp
Cos(x)=adjacent/hypotenuse or =adj/hyp

The pythagorean theorem (on the same triangle)

(opp)^2 + (adj)^2=(hyp)^2

Everything is defined so lets divide both sides of the pythagorean equation by (hyp)^2. We now have:

(opp)^2/(hyp)^2 + (adj)^2/(hyp)^2= 1

Again we know that Sin(x)=opp/hyp and Cos(x)=adj/hyp.

Therfore: [Sin(x)]^2+[Cos(x)]^2=1

We can also do this with other trignometric functions. Lets look at tan(x) and sec(x)

tan(x)=opp/adj and sec(x)=hyp/opp

Back again to the pythagorean formula but we are going to change it a little bit:

adj^2=hyp^2 -opp^2 (This can be done by subtracting opp^2 from both sides)

Lets divide both sides of the pythagorean formula by adj^2 and we have:

1=hyp^2/adj^2-opp^2/adj^2

Now remember: tan(x)=opp/adj and sec(x)=hyp/opp so:

1=[sec(x)]^2-[tan(x)]^2 or [tan(x)]^2 + 1=[sec(x)]^2

Can you do the proof of this trigonometry identity:

[cot(x)]^2 +1=[csc(x)]^2?

Comments 4 comments

billyaustindillon profile image

billyaustindillon 6 years ago

This brings me right back to high school :)


Ben Evans profile image

Ben Evans 6 years ago Author

Hey thank you for reading. I think it is nice to be able to see how these are derived.


quicksand profile image

quicksand 5 years ago

Wow! I too used to write 'soh cah toa' on my palm and memorize it on my way to school ... all I needed to remember about cot, sec, and cosec was that these were the inverse of sin cos and tan!

Cheers!


Ben Evans profile image

Ben Evans 5 years ago Author

I appreciate you reading. I remember that too. The only thing that mixed me up was csc was 1/sin and sec was 1/cos. It kind of seemed it should have been the other way the other way around.

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