Sum of Squares of Sine and Cosine

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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:

\(\ds \sin^2 x + \cos^2 x\) \(=\) \(\ds \frac{\text{opposite}^2 + \text{adjacent}^2} {\text{hypotenuse}^2}\) squaring both sides and adding
\(\ds \) \(=\) \(\ds \frac{\text{hypotenuse}^2} {\text{hypotenuse}^2}\) 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.


From Sine of Angle in Cartesian Plane and Cosine of Angle in 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
\(\ds \) \(=\) \(\ds 1\)

$\blacksquare$


Proof using Euler's Identities

\(\ds \cos^2 x + \sin^2 x\) \(=\) \(\ds \paren {\frac {e^{i x} + e^{-i x} } 2}^2 + \sin^2 x\) Euler's Cosine Identity
\(\ds \) \(=\) \(\ds \paren {\frac {e^{i x} + e^{-i x} } 2}^2 + \paren {\frac {e^{i x} - e^{-i x} } {2 i} }^2\) Euler's Sine Identity
\(\ds \) \(=\) \(\ds \frac {\paren {e^{i x} }^2 + 2 e^{-i x} e^{i x} + \paren {e^{-i x} }^2} 4 + \paren {\frac {e^{i x} - e^{-i x} } {2 i} }^2\) 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
\(\ds \) \(=\) \(\ds 1\)

$\blacksquare$


Also known as

This result is often referred to as the Pythagorean Identity.


Also see


Historical Note

The identity:

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

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


Sources

in which a mistake occurs
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