# Sum of Squares of Sine and Cosine

## Theorem

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

where $\sin$ and $\cos$ are sine and cosine.

### Corollary 1

$\sec^2 x - \tan^2 x = 1 \quad \text {(when$\cos x \ne 0$)}$

### Corollary 2

$\csc^2 x - \cot^2 x = 1 \quad \text {(when$\sin x \ne 0$)}$

## Algebraic Proof

 $\ds 1$ $=$ $\ds \cos 0$ Cosine of Zero is One $\ds$ $=$ $\ds \map \cos {x - x}$ $\ds$ $=$ $\ds \cos x \, \map \cos {-x} - \sin x \, \map \sin {-x}$ Cosine of Sum‎ $\ds$ $=$ $\ds \cos x \cos x - \paren {-\sin x \sin x}$ Cosine Function is Even and Sine Function is Odd $\ds$ $=$ $\ds \cos^2 x + \sin^2 x$

$\blacksquare$

### Warning

Note that we need to start from the algebraic definitions of sine and cosine:

$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$

and then use the proofs of the Cosine of Sum that derive directly from these.

Otherwise these proofs are circular.

## Geometric 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:

 $\ds \sin^2 x + \cos^2 x$ $=$ $\ds \frac{\text{opposite}^2 + \text{adjacent}^2} {\text{hypotenuse}^2}$ $\ds$ $=$ $\ds \frac{\text{hypotenuse}^2} {\text{hypotenuse}^2}$ Pythagoras's Theorem $\ds$ $=$ $\ds 1$

$\blacksquare$

## Unit Circle Proof

Let $P = \tuple {x, y}$ be a point on the circumference of a unit circle whose center is at the origin of a cartesian plane.

$P = \tuple {\cos \theta, \sin \theta}$

The graph of the unit circle is the locus of:

$x^2 + y^2 = 1$

as given by Equation of Circle.

Substituting $x = \cos \theta$ and $y = \sin \theta$ yields:

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

$\blacksquare$

## Euler's Formula Proof

 $\ds \cos^2 x + \sin^2 x$ $=$ $\ds \left({\cos x + i \, \sin x}\right) \, \left({\cos x - i \, \sin x}\right)$ factoring over the complex numbers $\ds$ $=$ $\ds \left({\cos x + i \, \sin x}\right) \, \left({\cos \left({-x}\right) + i \, \sin \left({-x}\right)}\right)$ Cosine Function is Even and Sine Function is Odd $\ds$ $=$ $\ds e^{ix} \, e^{-ix}$ Euler's Formula $\ds$ $=$ $\ds 1$

$\blacksquare$

## Exponential Formulation Proof

 $\ds \cos^2 x + \sin^2 x$ $=$ $\ds \paren {\frac {e^{i x} + e^{-i x} } 2}^2 + \sin^2 x$ Cosine Exponential Formulation $\ds$ $=$ $\ds \paren {\frac {e^{i x} + e^{-i x} } 2}^2 + \paren {\frac {e^{i x} - e^{-i x} } {2 i} }^2$ Sine Exponential Formulation $\ds$ $=$ $\ds \frac {\paren {e^{i x} }^2 + 2 e^{-i x} e^{i x} + \paren {e^{-i x} }^2} 4 + \paren {\frac {e^{i x} - e^{-i x} } {2 i} }^2$ Square of Sum $\ds$ $=$ $\ds \frac {\paren {e^{i x} }^2 + 2 e^{-i x} e^{i x} + \paren {e^{-i x} }^2} 4 + \frac {\paren {e^{i x} }^2 - e^{-i x} e^{i x} + \paren {e^{-i x} }^2} {-4}$ Square of Difference and $i^2 = -1$ $\ds$ $=$ $\ds \frac {e^{2 i x} + 2 + e^{-2 i x} } 4 + \frac {e^{2 i x} - 2 + e^{-2 i x} } {-4}$ Exponential of Sum $\ds$ $=$ $\ds \frac {e^{2 i x} + 2 + e^{-2 i x} - e^{2 i x} + 2 - e^{-2 i x} } 4$ simplifying $\ds$ $=$ $\ds \frac 4 4$ simplifying $\ds$ $=$ $\ds 1$

$\blacksquare$

## Historical Note

The identity:

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

was discovered and documented by Varahamihira in the 6th century CE.