Proof:
We want to prove mathematically that sine and cosine functions are orthogonal. Usually, this is stated as:
\[ \int_{0}^{2\pi} \sin(x)\cos(x)\, dx = 0 \]
Step 1: Recall what orthogonal means for functions
Two functions \( f(x) \) and \( g(x) \) are orthogonal over an interval \( [a, b] \) if:
\[ \int_{a}^{b} f(x)g(x)\, dx = 0 \]
Here, we consider:
\[ f(x) = \sin(x), \quad g(x) = \cos(x), \quad a = 0, \quad b = 2\pi \]
Step 2: Set up the integral
The integral we want is:
\[ \int_{0}^{2\pi} \sin(x)\cos(x)\, dx \]
Using the trigonometric identity \( \sin(x)\cos(x) = \tfrac{1}{2}\sin(2x) \), the integral becomes:
\[ \int_{0}^{2\pi} \tfrac{1}{2}\sin(2x)\, dx = \tfrac{1}{2}\int_{0}^{2\pi} \sin(2x)\, dx \]
Step 3: Integrate sin(2x)
\[ \int \sin(2x)\, dx = -\tfrac{1}{2}\cos(2x) \]
So evaluating the definite integral:
\[ \tfrac{1}{2}\left[-\tfrac{1}{2}\cos(2x)\right]_{0}^{2\pi} = -\tfrac{1}{4}\left[\cos(4\pi) - \cos(0)\right] \]
Step 4: Evaluate the cosine
\[ \cos(4\pi) = 1, \qquad \cos(0) = 1 \]
So the integral becomes:
\[ -\tfrac{1}{4}(1 - 1) = 0 \]
Conclusion
\[ \int_{0}^{2\pi} \sin(x)\cos(x)\, dx = 0 \]
This proves that \( \sin(x) \) and \( \cos(x) \) are orthogonal over \( [0, 2\pi] \).
When we multiply \( \sin(x) \) and \( \cos(x) \) pointwise over one period:
- Positive × positive → positive area
- Positive × negative → negative area
- Negative × positive → negative area
- Negative × negative → positive area
The positive and negative areas cancel exactly, so the integral sums to zero. This is why they are orthogonal.