Proofs of trigonometric identities

Proofs of trigonometric identities are used to show relations between trigonometric functions. This article will list trigonometric identities and prove them.

Proof
Q.E.D.
 * 1) $$\frac{\sin \theta}{\cos \theta} = \frac{\frac{opposite}{hypotenuse}}{\frac{adjacent}{hypotenuse}} $$(By definition of the sine and cosine)
 * 2) $$\frac{\sin \theta}{\cos \theta} = \frac{\frac{opposite}{hypotenuse}}{\frac{adjacent}{hypotenuse}} (\frac{hypotenuse}{hypotenuse})$$ (By definition of multiplicative identity)
 * 3) $$\frac{\sin \theta}{\cos \theta} = \frac{opposite}{adjacent} = \tan \theta$$ (definition of the tangent)

Proof
Q.E.D.
 * 1) $$a^2 + b^2 = h^2 \!\ $$ (Pythagorean Theorem)
 * 2) $$a = h \sin \theta \!\ $$ and $$b = h \cos \theta \!\ $$ (definition of sine and cosine)
 * 3) $$a^2 = h^2 \sin^2 \theta \!\ $$ and $$b^2 = h^2 \cos^2 \theta \!\ $$ (square both sides of each equation)
 * 4) $$h^2 \sin^2(x) + h^2 \cos^2(x) = h^2 \!\ $$ (substitute expressions for a² and b² into first equation)
 * 5) $$\sin^2(x) + \cos^2(x) = 1 \!$$ (Divide both sides by h²)

Proof
Q.E.D.
 * 1) $$\sin^2(x) + \cos^2(x) = 1 \!\ $$ (Identity 4)
 * 2) $$\tan^2 \theta + 1 = \sec^2 \theta \!\ $$ (multiply both sides by $$\sec^2 \theta \!\ $$)

Proof
Q.E.D.
 * 1) $$\sin^2(x) + \cos^2(x) = 1 \!\ $$ (Identity 4)
 * 2) $$\cot^2 \theta + 1 = \csc^2 \theta \!\ $$ (Multiply both sides by $$\csc^2 \theta \!\ $$)

Proof
Q.E.D.
 * 1) $$ \frac{\tan \theta}{\theta} > 1 $$ (Elementary identity #3)
 * 2) $$ \frac{\sin \theta}{\theta} > \cos \theta $$ (multiply both sides by $$\cos \theta \!\ $$)
 * 3) $$ 1 > \frac{\sin \theta}{\theta} > \cos \theta$$ (Elementary identity #2)
 * 4) $$ \lim_{\theta \to 0} 1 = 1 $$ and $$ \lim_{\theta \to 0} \cos \theta = 1$$ (definition of limit)
 * 5) $$\lim_{\theta \to 0}{\frac{\sin \theta}{\theta}} = 1$$ (Squeeze Theorem)

Proof
Q.E.D.
 * 1) $$\lim_{\theta \to 0}\frac{\cos \theta - 1}{\theta} $$ (Given)
 * 2) $$ = \lim_{\theta \to 0}\frac{\cos^2 \theta - 1}{\theta (\cos \theta + 1)} $$ (multiply by $$ \frac{\cos \theta + 1}{\cos \theta + 1}$$)
 * 3) $$ = \lim_{\theta \to 0}\frac{-\sin^2 \theta}{\theta(\cos \theta + 1)}$$ (Numerator derived from Elementary identity #4)
 * 4) $$ = -\lim_{\theta \to 0} \frac{\sin \theta}{\theta} \lim_{\theta \to 0} \frac{\sin \theta}{\cos \theta + 1} $$ (limit laws)
 * 5) $$-\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = -1 $$ and$$\lim_{\theta \to 0} \frac{\sin \theta}{\cos \theta + 1} = \frac{0}{2} = 0$$ (first limit evaluated using previous identity, second limit is evaluated by substituting θ with 0)
 * 6) $$= -1(0) = 0 \!\ $$

Proof
Q.E.D.
 * 1) $$= \lim_{\phi \to 0}\frac{\sin k\phi}{p\phi} \frac{k}{k} $$ (multiplicative identity)
 * 2) $$= \lim_{\phi \to 0}\frac{k}{p}\frac{\sin k\phi}{k\phi} $$ (reorder variables in denominator)
 * 3) $$= \frac{k}{p} \lim_{\phi \to 0}\frac{\sin k\phi}{k\phi}$$ (limit laws)
 * 4) $$\lim_{\phi \to 0}\frac{\sin k\phi}{k\phi} = \lim_{k\phi \to 0}\frac{\sin k\phi}{k\phi}$$ Substituting θ for kφ, we get Identity 1. Thus,
 * 5) $$= \frac{k}{p} (1) = \frac{k}{p}$$