We will prove the fundamental trigonometric identity: sine squared theta plus cosine squared theta equals one. We'll also show that tangent theta equals sine theta divided by cosine theta. Let's start with a right triangle where theta is our angle, h is the hypotenuse, a is the adjacent side, and o is the opposite side.
Let's establish the basic trigonometric definitions. Sine theta equals the opposite side divided by the hypotenuse. Cosine theta equals the adjacent side divided by the hypotenuse. And tangent theta equals the opposite side divided by the adjacent side. These definitions form the foundation for our proof.
Now let's square both sine and cosine definitions. Sine squared theta equals o squared over h squared. Cosine squared theta equals a squared over h squared. When we add them together, we get o squared plus a squared, all divided by h squared.
Now we apply the Pythagorean theorem. In any right triangle, the sum of squares of the two shorter sides equals the square of the hypotenuse: o squared plus a squared equals h squared. Substituting this into our expression, we get h squared over h squared, which equals one. Therefore, sine squared theta plus cosine squared theta equals one. This completes our proof of the fundamental trigonometric identity.
Finally, let's prove that tangent theta equals sine theta divided by cosine theta. We know tangent theta equals o over a. When we divide sine by cosine, we get o over h divided by a over h. This equals o over h times h over a, which simplifies to o over a. Therefore, tangent theta equals sine theta divided by cosine theta. This completes our proof of the fundamental trigonometric relationships.