Trigonometric identities

A trigonometric identity is an equation involving trig ratios that holds for all (valid) values of the angle. Three identities , all springing from the Pythagorean theorem , are the foundation.

The three Pythagorean identities

For any acute $\theta$: $\sin^2\theta + \cos^2\theta = 1,$ $1 + \tan^2\theta = \sec^2\theta,$ $1 + \cot^2\theta = \csc^2\theta.$

Derivation

In a right triangle with $\theta$, write opp $= a$, adj $= b$, hyp $= c$. By Pythagoras, $a^2 + b^2 = c^2$.

Divide by $c^2$: $\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1 \implies \sin^2\theta + \cos^2\theta = 1.$

Divide by $b^2$: $\frac{a^2}{b^2} + 1 = \frac{c^2}{b^2} \implies \tan^2\theta + 1 = \sec^2\theta.$

Divide by $a^2$: $1 + \frac{b^2}{a^2} = \frac{c^2}{a^2} \implies 1 + \cot^2\theta = \csc^2\theta.$

So all three are the same theorem , Pythagoras , viewed through different "lenses" of the right triangle.

Strategies for proving identities

A trigonometric identity proof looks like algebra:

1. Pick one side (usually the messier one) and transform it step by step.
2. Replace using known identities , $\sin^2 + \cos^2 = 1$ is the most useful.
3. Convert everything to $\sin$ and $\cos$ if you get stuck. This is the "nuclear option" , it always works but can be slow.
4. Factor and cancel , many identities boil down to common-factor extraction.
5. Cross-multiply if both sides are fractions.

A few useful rewrites:

• $\sec^2\theta - 1 = \tan^2\theta$.
• $\csc^2\theta - 1 = \cot^2\theta$.
• $\sin\theta \cdot \csc\theta = 1$, $\cos\theta \cdot \sec\theta = 1$, $\tan\theta \cdot \cot\theta = 1$.
• $1 - \sin^2\theta = \cos^2\theta$, $1 - \cos^2\theta = \sin^2\theta$.

Worked examples

Example 1. Prove $\dfrac{\sin\theta}{1 + \cos\theta} = \dfrac{1 - \cos\theta}{\sin\theta}$.

Cross-multiply: $\sin^2\theta = (1 - \cos\theta)(1 + \cos\theta) = 1 - \cos^2\theta$. ✓ (Pythagorean identity.) $\blacksquare$

Example 2. Prove $\tan\theta + \cot\theta = \sec\theta \cdot \csc\theta$.

LHS $= \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta \cos\theta} = \dfrac{1}{\sin\theta \cos\theta} = \csc\theta \cdot \sec\theta$. $\blacksquare$

Example 3. Prove $(\sin\theta + \csc\theta)^2 + (\cos\theta + \sec\theta)^2 = 7 + \tan^2\theta + \cot^2\theta$.

Expand LHS: $= \sin^2 + 2 \sin\csc + \csc^2 + \cos^2 + 2 \cos\sec + \sec^2$ $= (\sin^2 + \cos^2) + 2 \cdot 1 + 2 \cdot 1 + \csc^2 + \sec^2$ $= 1 + 4 + \csc^2 + \sec^2$ $= 5 + (1 + \cot^2) + (1 + \tan^2)$ $= 7 + \tan^2 + \cot^2$. $\blacksquare$

Example 4. Prove $\dfrac{1 + \tan^2 A}{1 + \cot^2 A} = \tan^2 A$.

Numerator: $1 + \tan^2 A = \sec^2 A$. Denominator: $1 + \cot^2 A = \csc^2 A$.

Ratio: $\sec^2 A / \csc^2 A = (1/\cos^2 A) / (1/\sin^2 A) = \sin^2 A / \cos^2 A = \tan^2 A$. $\blacksquare$

Example 5. Prove $\dfrac{\cos A}{1 - \sin A} + \dfrac{\cos A}{1 + \sin A} = 2 \sec A$.

LHS $= \cos A \left[\dfrac{1}{1 - \sin A} + \dfrac{1}{1 + \sin A}\right] = \cos A \cdot \dfrac{(1 + \sin A) + (1 - \sin A)}{1 - \sin^2 A} = \cos A \cdot \dfrac{2}{\cos^2 A} = \dfrac{2}{\cos A} = 2 \sec A$. $\blacksquare$

Try it yourself

1. Prove $\sin^2\theta + \cos^2\theta = 1$ using a right triangle.
2. Prove $\sec^2\theta - \tan^2\theta = 1$.
3. Prove $\dfrac{1 - \sin\theta}{\cos\theta} = \dfrac{\cos\theta}{1 + \sin\theta}$.
4. Prove $(1 + \tan^2\theta)(1 - \sin\theta)(1 + \sin\theta) = 1$.
5. Prove $\dfrac{\tan\theta}{1 - \cot\theta} + \dfrac{\cot\theta}{1 - \tan\theta} = 1 + \tan\theta + \cot\theta$.
6. Prove $\dfrac{\sin\theta - 2\sin^3\theta}{2 \cos^3\theta - \cos\theta} = \tan\theta$.
7. Prove $\sec^4\theta - \sec^2\theta = \tan^2\theta + \tan^4\theta$.
8. Prove $(\sin A + \cos A)(\sec A + \csc A) = 2 + \sec A \csc A$.
9. Prove $\dfrac{\cos\theta - \sin\theta + 1}{\cos\theta + \sin\theta - 1} = \csc\theta + \cot\theta$. (Hint: multiply numerator and denominator by $\cos\theta + \sin\theta + 1$.)
10. Prove $(1 + \cot\theta - \csc\theta)(1 + \tan\theta + \sec\theta) = 2$.

Pitfalls / Insight

• $\sin^2\theta$ means $(\sin\theta)^2$, not $\sin(\theta^2)$.
• Don't divide by something that might be zero. Be wary of $\sin\theta - 1, \cos\theta$ in denominators.
• One side at a time. Pick LHS or RHS, transform, end at the other side. Don't manipulate both at once.

Insight. Every classical identity is, at heart, Pythagoras divided by something. Recognise this and proofs become routine.