Question

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

Answer 1

Consider a right triangle with sides `a, b, c` with `a < b < c`.

By pythagoras,

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

`sqrt(a^2 + b^2) = c ->`because `c` must be positive. We now let the angle opposite side `a` be `theta`. Since `sintheta = a/c` and `costheta = b/c`, we have:

`sqrt((c(sin theta))^2 + (c(costheta)^2) = c`

`sqrt(c^2(sin^2theta + cos^2theta)) = c`

`csqrt(sin^2theta + cos^2theta) = c`

And so `sqrt(sin^2theta + cos^2theta) = 1`, which means that `sin^2theta + cos^2theta = 1`, so our proof is complete.

Hopefully this helps!