Trigonometric identities
An identity is an equation that holds for every value of the variable for which both sides are defined. The trigonometric identities are the tools that let you simplify any trigonometric expression , by rewriting one form into another with the same meaning. There are only a handful of fundamental identities; everything else is derived.
The fundamental identities
Reciprocal :
csc θ = 1 sin θ , sec θ = 1 cos θ , cot θ = 1 tan θ . \csc\theta = \frac{1}{\sin\theta},\quad \sec\theta = \frac{1}{\cos\theta},\quad \cot\theta = \frac{1}{\tan\theta}. csc θ = s i n θ 1 , sec θ = c o s θ 1 , cot θ = t a n θ 1 .
Quotient :
tan θ = sin θ cos θ , cot θ = cos θ sin θ . \tan\theta = \frac{\sin\theta}{\cos\theta},\quad \cot\theta = \frac{\cos\theta}{\sin\theta}. tan θ = c o s θ s i n θ , cot θ = s i n θ c o s θ .
Pythagorean :
sin 2 θ + cos 2 θ = 1. \boxed{\sin^2\theta + \cos^2\theta = 1.} sin 2 θ + cos 2 θ = 1. cos 2 θ \cos^2\theta cos 2 θ 1 + tan 2 θ = sec 2 θ 1 + \tan^2\theta = \sec^2\theta 1 + tan 2 θ = sec 2 θ sin 2 θ \sin^2\theta sin 2 θ 1 + cot 2 θ = csc 2 θ 1 + \cot^2\theta = \csc^2\theta 1 + cot 2 θ = csc 2 θ
Proof of sin 2 θ + cos 2 θ = 1 \sin^2\theta + \cos^2\theta = 1 sin 2 θ + cos 2 θ = 1
On the unit circle, the point at angle θ \theta θ ( cos θ , sin θ ) (\cos\theta, \sin\theta) ( cos θ , sin θ ) 1 1 1 1 = cos 2 θ + sin 2 θ ⟹ cos 2 θ + sin 2 θ = 1. \qed 1 = \sqrt{\cos^2\theta + \sin^2\theta} \implies \cos^2\theta + \sin^2\theta = 1. \qed 1 = cos 2 θ + sin 2 θ ⟹ cos 2 θ + sin 2 θ = 1. \qed
This is the defining property of points on the unit circle. Every other Pythagorean identity descends from it.
Strategies for proving identities
To prove an identity LHS = RHS \text{LHS} = \text{RHS} LHS = RHS
Start with the more complex side .
Convert everything to sin \sin sin cos \cos cos
Use Pythagorean substitutions: e.g. 1 − sin 2 θ = cos 2 θ 1 - \sin^2\theta = \cos^2\theta 1 − sin 2 θ = cos 2 θ
Factor when possible.
Find common denominators.
Show LHS = RHS after simplification.
Sometimes the cleanest proof works on both sides until they meet in the middle.
Worked examples
Example 1. Prove sin θ 1 − cos θ = csc θ + cot θ \dfrac{\sin\theta}{1 - \cos\theta} = \csc\theta + \cot\theta 1 − cos θ sin θ = csc θ + cot θ
Multiply numerator and denominator on the left by 1 + cos θ 1 + \cos\theta 1 + cos θ sin θ ( 1 + cos θ ) ( 1 − cos θ ) ( 1 + cos θ ) = sin θ ( 1 + cos θ ) 1 − cos 2 θ = sin θ ( 1 + cos θ ) sin 2 θ = 1 + cos θ sin θ = csc θ + cot θ . \qed \frac{\sin\theta (1 + \cos\theta)}{(1 - \cos\theta)(1 + \cos\theta)} = \frac{\sin\theta(1 + \cos\theta)}{1 - \cos^2\theta} = \frac{\sin\theta(1 + \cos\theta)}{\sin^2\theta} = \frac{1 + \cos\theta}{\sin\theta} = \csc\theta + \cot\theta. \qed ( 1 − c o s θ ) ( 1 + c o s θ ) s i n θ ( 1 + c o s θ ) = 1 − c o s 2 θ s i n θ ( 1 + c o s θ ) = s i n 2 θ s i n θ ( 1 + c o s θ ) = s i n θ 1 + c o s θ = csc θ + cot θ . \qed
Example 2. Prove tan θ + cot θ = sec θ csc θ \tan\theta + \cot\theta = \sec\theta \csc\theta tan θ + cot θ = sec θ csc θ
LHS = sin θ cos θ + cos θ sin θ = sin 2 θ + cos 2 θ sin θ cos θ = 1 sin θ cos θ = sec θ csc θ = \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta} = \sec\theta \csc\theta = cos θ sin θ + sin θ cos θ = sin θ cos θ sin 2 θ + cos 2 θ = sin θ cos θ 1 = sec θ csc θ \qed \qed \qed
Example 3. Simplify ( sec θ − tan θ ) ( sec θ + tan θ ) (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) ( sec θ − tan θ ) ( sec θ + tan θ )
sec 2 θ − tan 2 θ = 1 \sec^2\theta - \tan^2\theta = 1 sec 2 θ − tan 2 θ = 1
Example 4. Prove 1 + tan 2 θ 1 + cot 2 θ = tan 2 θ \dfrac{1 + \tan^2\theta}{1 + \cot^2\theta} = \tan^2\theta 1 + cot 2 θ 1 + tan 2 θ = tan 2 θ
LHS = sec 2 θ csc 2 θ = 1 / cos 2 θ 1 / sin 2 θ = sin 2 θ cos 2 θ = tan 2 θ = \dfrac{\sec^2\theta}{\csc^2\theta} = \dfrac{1/\cos^2\theta}{1/\sin^2\theta} = \dfrac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta = csc 2 θ sec 2 θ = 1/ sin 2 θ 1/ cos 2 θ = cos 2 θ sin 2 θ = tan 2 θ \qed \qed \qed
Example 5 (harder). Prove cos θ 1 − tan θ + sin θ 1 − cot θ = sin θ + cos θ \dfrac{\cos\theta}{1 - \tan\theta} + \dfrac{\sin\theta}{1 - \cot\theta} = \sin\theta + \cos\theta 1 − tan θ cos θ + 1 − cot θ sin θ = sin θ + cos θ
Common denominator: 1 − tan θ = cos θ − sin θ cos θ 1 - \tan\theta = \dfrac{\cos\theta - \sin\theta}{\cos\theta} 1 − tan θ = cos θ cos θ − sin θ cos θ 1 − tan θ = cos 2 θ cos θ − sin θ \dfrac{\cos\theta}{1 - \tan\theta} = \dfrac{\cos^2\theta}{\cos\theta - \sin\theta} 1 − tan θ cos θ = cos θ − sin θ cos 2 θ
Similarly 1 − cot θ = sin θ − cos θ sin θ 1 - \cot\theta = \dfrac{\sin\theta - \cos\theta}{\sin\theta} 1 − cot θ = sin θ sin θ − cos θ sin θ 1 − cot θ = sin 2 θ sin θ − cos θ = − sin 2 θ cos θ − sin θ \dfrac{\sin\theta}{1 - \cot\theta} = \dfrac{\sin^2\theta}{\sin\theta - \cos\theta} = -\dfrac{\sin^2\theta}{\cos\theta - \sin\theta} 1 − cot θ sin θ = sin θ − cos θ sin 2 θ = − cos θ − sin θ sin 2 θ
Adding:
cos 2 θ − sin 2 θ cos θ − sin θ = ( cos θ − sin θ ) ( cos θ + sin θ ) cos θ − sin θ = sin θ + cos θ . \qed \frac{\cos^2\theta - \sin^2\theta}{\cos\theta - \sin\theta} = \frac{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)}{\cos\theta - \sin\theta} = \sin\theta + \cos\theta. \qed c o s θ − s i n θ c o s 2 θ − s i n 2 θ = c o s θ − s i n θ ( c o s θ − s i n θ ) ( c o s θ + s i n θ ) = sin θ + cos θ . \qed
Try it yourself
Prove ( 1 − sin θ ) ( 1 + sin θ ) = cos 2 θ (1 - \sin\theta)(1 + \sin\theta) = \cos^2\theta ( 1 − sin θ ) ( 1 + sin θ ) = cos 2 θ
Prove sin 4 θ − cos 4 θ = sin 2 θ − cos 2 θ \sin^4\theta - \cos^4\theta = \sin^2\theta - \cos^2\theta sin 4 θ − cos 4 θ = sin 2 θ − cos 2 θ
Show 1 + cos θ sin θ = sin θ 1 − cos θ \dfrac{1 + \cos\theta}{\sin\theta} = \dfrac{\sin\theta}{1 - \cos\theta} sin θ 1 + cos θ = 1 − cos θ sin θ
Prove tan θ sin θ + cos θ = sec θ \tan\theta \sin\theta + \cos\theta = \sec\theta tan θ sin θ + cos θ = sec θ
Show sec 4 θ − sec 2 θ = tan 4 θ + tan 2 θ \sec^4\theta - \sec^2\theta = \tan^4\theta + \tan^2\theta sec 4 θ − sec 2 θ = tan 4 θ + tan 2 θ
Prove 1 − tan 2 θ 1 + tan 2 θ = cos 2 θ \dfrac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta 1 + tan 2 θ 1 − tan 2 θ = cos 2 θ
Simplify sin θ − cos θ + 1 sin θ + cos θ − 1 \dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} sin θ + cos θ − 1 sin θ − cos θ + 1
If sin θ + cos θ = a \sin\theta + \cos\theta = a sin θ + cos θ = a sin θ cos θ \sin\theta \cos\theta sin θ cos θ a a a
Prove csc 2 θ − cot 2 θ = 1 \csc^2\theta - \cot^2\theta = 1 csc 2 θ − cot 2 θ = 1
Simplify sin θ sec θ + 1 + sin θ sec θ − 1 \dfrac{\sin\theta}{\sec\theta + 1} + \dfrac{\sin\theta}{\sec\theta - 1} sec θ + 1 sin θ + sec θ − 1 sin θ
Show 1 + sec θ sec θ = sin 2 θ 1 − cos θ \dfrac{1 + \sec\theta}{\sec\theta} = \dfrac{\sin^2\theta}{1 - \cos\theta} sec θ 1 + sec θ = 1 − cos θ sin 2 θ
If sec θ + tan θ = a \sec\theta + \tan\theta = a sec θ + tan θ = a sec θ − tan θ \sec\theta - \tan\theta sec θ − tan θ
Pitfalls / Tricks
sin 2 θ \sin^2\theta sin 2 θ ( sin θ ) 2 (\sin\theta)^2 ( sin θ ) 2 sin ( sin θ ) \sin(\sin\theta) sin ( sin θ ) sin ( θ 2 ) \sin(\theta^2) sin ( θ 2 ) When dividing by a trigonometric function, ensure it is not zero. State excluded angles if necessary.
An identity is not an equation to solve for θ \theta θ θ \theta θ
Insight. Almost every identity collapses if you first convert to sin \sin sin cos \cos cos