Sum of Squares of Sine and Cosine/Proof 2

Theorem

$\cos^2 x + \sin^2 x = 1$

Proof

From the trigonometric definitions of sine and cosine:

\(\ds \sin x\)

\(=\)

\(\ds \frac{\text{opposite} } {\text{hypotenuse} }\)

\(\ds \cos x\)

\(=\)

\(\ds \frac{\text{adjacent} } {\text{hypotenuse} }\)

Then:

\(\ds \sin^2 x + \cos^2 x\)

\(=\)

\(\ds \frac{\text{opposite}^2 + \text{adjacent}^2} {\text{hypotenuse}^2}\)

squaring both sides and adding

\(\ds \)

\(=\)

\(\ds \frac{\text{hypotenuse}^2} {\text{hypotenuse}^2}\)

Pythagoras's Theorem

\(\ds \)

\(=\)

\(\ds 1\)

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Definitions of the ratios