Sum of Squares of Sine and Cosine

Theorem

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

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

Corollaries

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 1 + \tan^2 x\)	\(=\)	\(\displaystyle \sec^2 x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(when $\cos x \ne 0$)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 1 + \cot^2 x\)	\(=\)	\(\displaystyle \csc^2 x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(when $\sin x \ne 0$)

where:

• $\tan$ and $\cot$ are tangent and cotangent.
• $\sec$ and $\csc$ are secant and cosecant.

Algebraic Proof

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 1\)	\(=\)	\(\displaystyle \cos 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Cosine of Zero is One
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cos \left({x - x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cos x \cos \left({- x}\right) - \sin x \sin \left({- x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sine and Cosine of Sum‎
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cos x \cos x - - \sin x \sin x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Cosine Function is Even, Sine Function is Odd
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cos^2 x + \sin^2 x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Notes

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

• $\displaystyle \sin x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
• $\displaystyle \cos x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$

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

Otherwise these proofs are circular.

Geometric Proof

• Starting with $\sin x$ and $\cos x$:

$\displaystyle \sin x = \frac{\text{opposite}}{\text{hypotenuse}}$

$\displaystyle \cos x = \frac{\text{adjacent}}{\text{hypotenuse}}$

• Squaring both sides and adding them together gives:

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

$\blacksquare$

Notes

This proof requires that we start from the trigonometric definitions of sine and cosine, otherwise the proofs are circular.

Unit Circle Proof

Every point $(x,y)$ on the unit circle is $(\cos \theta,\sin \theta)$.

By definition, the graph of the unit circle is the locus of

$x^2 + y^2 = 1$

$\implies$

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

$\blacksquare$

Notes

This proof depends on the unit circle definitions of sine and cosine, otherwise the proof is too circular.

Proof of Corollaries

• $1 + \tan^2 x = \sec^2 x$ (when $\cos x \ne 0$):

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \cos^2 x + \sin^2 x\)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 1 + \frac {\sin^2 x} {\cos^2 x}\)	\(=\)	\(\displaystyle \frac {1} {\cos^2 x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	dividing both sides by $\cos^2 x$ when $\cos x \ne 0$
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 1 + \tan^2 x\)	\(=\)	\(\displaystyle \sec^2 x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definitions of tangent and secant

$\blacksquare$

• $1 + \cot^2 x = \csc^2 x$(when $\sin x \ne 0$):

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sin^2 x + \cos^2 x\)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 1 + \frac {\cos^2 x} {\sin^2 x}\)	\(=\)	\(\displaystyle \frac {1} {\sin^2 x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	dividing both sides by $\sin^2 x$ when $\sin x \ne 0$
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 1 + \cot^2 x\)	\(=\)	\(\displaystyle \csc^2 x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definitions of cotangent and cosecant

$\blacksquare$

Also see

• Difference of Squares of Hyperbolic Cosine and Sine

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $5.19, \ 5.20, \ 5.21$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 16.3 \ (3) \ \text{(i)}$