Can I prove Pythagoras' Theorem using that $\sin^2(\theta)+\cos^2(\theta)=1$? In any right-angled triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).


The theorem can be written as an equation relating the lengths of the sides $a$, $b$, and $c$, often called the Pythagorean equation:
$$a^2 + b^2 = c^2$$

Can I prove Pythagoras' Theorem by the following way?

Actually, my question is: does it violate any rules of mathematics, or is it alright?
Sorry, it may not be a valid question for this site. But I want to know. Thanks.
The usual proof of the identity $\cos^2 t+\sin^2 t=1$ uses the Pythagorean Theorem. So a proof of the Pythagorean Theorem by using the identity is not correct.

True, we can define cosine and sine purely "analytically," by power series, or as the solutions of a certain differential equation. Then we can prove $\cos^2 t+\sin^2 t=1$ without any appeal to geometry. 

But we still need geometry to link these "analytically" defined functions to sides of right-angled triangles. 

Remark: The question is very reasonable. The logical interdependencies between various branches of mathematics are usually not clearly described. This is not necessarily always a bad thing. The underlying ideas of the calculus were for a while quite fuzzy, but calculus was still used effectively to solve problems, Similar remarks can be made about complex numbers.
It is not true that you must use the Pythagorean theorem to prove that $sin^2(x)+ cos^2(x)= 1$. It depends upon how you have defined sine and cosine.  It is, for example, perfectly proper to define $sin(x)= \sum_{n= 0}^\infty \frac{(-1)^n}{(2n+ 1)!}x^{2n+ 1}$ and $cos(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$ or, equivalently, to define $sin(x)$ as the function satisfying the differential equation y''+ y= 0 with initial conditions y(0)= 0, y'(0)= 1 and $cos(x)$ as the function, y, satisfying the differential equation y''+ y= 0 with initial conditions y(0)= 1, y'(0)= 0.  In either case, you can then prove that $sin^2(x)+ cos^2(x)= 1$ without reference to the Pythagorean theorem.