Part 3/4:
First, we express the opposite and adjacent sides of the triangle using sine and cosine:
We can derive the opposite side ( O ) as ( O = H \cdot \sin(x) )
The adjacent side ( A ) can be derived as ( A = H \cdot \cos(x) )
By substituting these expressions back into the Pythagorean theorem:
[ H^2 = (H \cdot \sin(x))^2 + (H \cdot \cos(x))^2 ]
Simplifying the Identity
Now, squaring both sides gives us:
[ H^2 = H^2 \cdot \sin^2(x) + H^2 \cdot \cos^2(x) ]
Next, to simplify the equation, we can divide both sides by ( H^2 ) (assuming ( H \neq 0 )):
[ 1 = \sin^2(x) + \cos^2(x) ]
This leads us to the conclusion:
[ \sin^2(x) + \cos^2(x) = 1 ]