How do you prove `sin^2x + cos^2x = 1`?

Consider a right angled triangle with an internal angle `theta`:

Then:

`sin theta = a/c`

`cos theta = b/c`

So:

`sin^2 theta + cos^2 theta = a^2/c^2+b^2/c^2 = (a^2+b^2)/c^2`

By Pythagoras `a^2+b^2 = c^2`, so `(a^2+b^2)/c^2 = 1`

So given Pythagoras, that proves the identity for `theta in (0, pi/2)`

For angles outside that range we can use:

`sin (theta + pi) = -sin (theta)`

`cos (theta + pi) = -cos (theta)`

`sin (- theta) = - sin(theta)`

`cos (- theta) = cos(theta)`

So for example:

`sin^2 (theta + pi) + cos^2 (theta + pi) = (-sin theta)^2 + (-cos theta)^2 = sin^2 theta + cos^2 theta = 1`

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Pythagoras theorem

Given a right angled triangle with sides `a`, `b` and `c` consider the following diagram:

The area of the large square is `(a+b)^2`

The area of the small, tilted square is `c^2`

The area of each triangle is `1/2ab`

So we have:

`(a+b)^2 = c^2 + 4 * 1/2ab`

That is:

`a^2+2ab+b^2 = c^2+2ab`

Subtract `2ab` from both sides to get:

`a^2+b^2 = c^2`

The formula for a circle centred at the origin is

`x^2+y^2=r^2`

That is, the distance from the origin to any point `(x,y)` on the circle is the radius `r` of the circle.

Picture a circle of radius `r` centred at the origin, and pick a point `(x,y)` on the circle:

graph{(x^2+y^2-1)((x-sqrt(3)/2)^2+(y-0.5)^2-0.003)=0 [-2.5, 2.5, -1.25, 1.25]}

If we draw a line from that point to the origin, its length is `r`. We can also draw a triangle for that point as follows:

graph{(x^2+y^2-1)(y-sqrt(3)x/3)((y-0.25)^4/0.18+(x-sqrt(3)/2)^4/0.000001-0.02)(y^4/0.00001+(x-sqrt(3)/4)^4/2.7-0.01)=0 [-2.5, 2.5, -1.25, 1.25]}

Let the angle at the origin be theta (`theta`).

Now for the trigonometry.

For an angle `theta` in a right triangle, the trig function `sin theta` is the ratio `"opposite side"/"hypotenuse"`. In our case, the length of the side opposite of `theta` is the `y`-coordinate of our point `(x,y)`, and the hypotenuse is our radius `r`. So:

`sin theta = "opp"/"hyp" = y/r" "<=>" "y=rsintheta`

Similarly, `cos theta` is the ratio of the `x`-coordinate in `(x,y)` to the radius `r`:

`cos theta = "adj"/"hyp"=x/r" "<=>" "x = rcostheta`

So we have `x=rcostheta` and `y=rsintheta`. Substituting these into the circle formula gives

`"      "x^2"     "+"      "y^2"     "=r^2`
`(rcostheta)^2+(rsintheta) ^2 = r^2`
`r^2cos^2theta + r^2 sin^2 theta = r^2`

The `r^2`'s all cancel, leaving us with

`cos^2 theta + sin^2 theta = 1`

This is often rewritten with the `sin^2` term in front, like this:

`sin^2 theta + cos^2 theta = 1`

And that's it. That's really all there is to it. Just as the distance between the origin and any point `(x,y)` on a circle must be the circle's radius, the sum of the squared values for `sin theta` and `cos theta` must be 1 for any angle `theta`.