Every trig identity in this course descends from just one Pythagorean fact — learning to trace that descent, rather than memorise the endpoints, is what makes proofs fall into place.
10 min read4 worked examples
The foundations
Two identities underpin almost everything in trigonometry so far:
The second is the Pythagorean identity, following directly from Pythagoras' theorem applied to a right-angled triangle in the unit circle. Dividing it through by $\cos^2\theta$ or $\sin^2\theta$ produces two further identities involving the reciprocal functions from earlier in this topic.
Derivationsec²θ = 1 + tan²θ, by dividing the Pythagorean identity by cos²θ
1
Start from $\sin^2\theta+\cos^2\theta=1$.
The Pythagorean identity.
2
Divide every term by $\cos^2\theta$: $\dfrac{\sin^2\theta}{\cos^2\theta} + \dfrac{\cos^2\theta}{\cos^2\theta} = \dfrac{1}{\cos^2\theta}$.
Dividing both sides by the same non-zero quantity preserves the equation.
3
$\tan^2\theta + 1 = \sec^2\theta$.
sin θ/cos θ = tan θ, and 1/cos θ = sec θ, by definition.
▪ QED
Derivationcsc²θ = 1 + cot²θ, by dividing the Pythagorean identity by sin²θ
1
Start from $\sin^2\theta+\cos^2\theta=1$.
The Pythagorean identity.
2
Divide every term by $\sin^2\theta$: $\dfrac{\sin^2\theta}{\sin^2\theta} + \dfrac{\cos^2\theta}{\sin^2\theta} = \dfrac{1}{\sin^2\theta}$.
Dividing both sides by the same non-zero quantity preserves the equation.
3
$1 + \cot^2\theta = \csc^2\theta$.
cos θ/sin θ = cot θ, and 1/sin θ = csc θ, by definition.
▪ QED
The full identity toolkit
$\sin^2\theta+\cos^2\theta=1$, $\quad\sec^2\theta=1+\tan^2\theta$, $\quad\csc^2\theta=1+\cot^2\theta$, together with $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ and $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$ — plus the addition, double-angle and R cos/sin identities from earlier lessons — form the complete set of tools for proving further trig identities.
Strategy for proving identities
To prove that two expressions are identically equal (true for all valid $\theta$, not just solving for specific values):
A working method
Start from the more complicated side (usually the one with more terms or a fraction), and work towards the simpler side using known identities — never manipulate both sides at once, or the logic becomes circular. Convert everything to sines and cosines if stuck, since these have the fewest identities to juggle. Look for Pythagorean substitutions ($\sin^2\theta+\cos^2\theta=1$ or its variants) whenever a squared term appears. Combine fractions over a common denominator when the target side is a single fraction or whole number.
Small-angle approximations
For an angle $\theta$ measured in radians and close to zero, the graphs of $\sin\theta$, $\cos\theta$ and $\tan\theta$ can be approximated by simple polynomials — because near $\theta=0$, each curve is almost indistinguishable from its tangent line (or a simple quadratic, for cosine):
These approximations only hold when $\theta$ is in radians — the clean relationship $\sin\theta\approx\theta$ is a direct consequence of how the radian itself is defined via arc length. Using a degree value here would give a badly wrong approximation. These results are especially useful for simplifying limits and small-angle physics problems where an exact trig expression is unwieldy.
Worked example 1
Proving a straightforward identity
Prove that $\dfrac{1}{\sec\theta} + \tan\theta\sin\theta \equiv \dfrac{1}{\cos\theta}$.
MethodConvert everything to sines and cosines, then combine into one fraction
$= 1 + 2\sin\theta\cos\theta$, using the Pythagorean identity.
sin²θ + cos²θ = 1.
4
$= 1+\sin2\theta$, using the double angle formula.
sin 2θ = 2 sin θ cos θ.
5
This matches the right-hand side, so the identity is proven.
LHS has been reduced exactly to RHS.
▪ proven
Examiner's eye
Recognising $2\sin\theta\cos\theta$ as "half of $\sin2\theta$'s formula" is a pattern worth watching for — any time an expanded expression contains a $\sin\theta\cos\theta$ term, the double angle identity is very likely to be the next useful step.
Worked example 3
Applying the small-angle approximations
Show that for small $x$ (in radians), $\dfrac{1-\cos3x}{x\sin4x} \approx \dfrac{9}{8}$.
MethodApply the small-angle approximations to each trig term, then simplify
1
For small $x$: $\cos3x \approx 1-\dfrac{(3x)^2}{2} = 1-\dfrac{9x^2}{2}$, so $1-\cos3x \approx \dfrac{9x^2}{2}$.
Apply cos θ ≈ 1 − θ²/2 with θ = 3x, then isolate 1 − cos 3x.
The x² cancels, leaving a constant — confirming the claimed approximation.
▪ shown
Why this matters
Without the small-angle approximations, this limit-style expression would be far harder to evaluate directly as $x\to0$ (both numerator and denominator tend to $0$). The approximations turn an awkward $\frac00$-type expression into simple algebra, which is exactly why they're so useful in calculus and physics.
Worked example 4
A synoptic identity combining multiple techniques · Challenge
Prove that $\dfrac{\cos2\theta}{1-\sin2\theta} \equiv \dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}$, for $\cos\theta\ne\sin\theta$.
MethodRewrite cos 2θ and the "1" using Pythagorean and double angle identities, then factorise
1
Left-hand side numerator: $\cos2\theta = \cos^2\theta-\sin^2\theta = (\cos\theta-\sin\theta)(\cos\theta+\sin\theta)$.
Use the cos²θ − sin²θ form of cos 2θ, then factorise as a difference of two squares.
2
Left-hand side denominator: $1-\sin2\theta = (\sin^2\theta+\cos^2\theta) - 2\sin\theta\cos\theta$.
Replace 1 using the Pythagorean identity, and sin 2θ using the double angle formula, so both terms are quadratic in sin θ, cos θ.
Cancel a common factor of $(\cos\theta-\sin\theta)$ (valid since $\cos\theta\ne\sin\theta$): $= \dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}$.
One factor of (cos θ − sin θ) cancels from top and bottom.
6
This matches the right-hand side, so the identity is proven.
LHS has been reduced exactly to RHS.
▪ proven
Look deeper — three identities, one proof
This single proof leaned on the Pythagorean identity, the double angle formula for $\cos2\theta$, and algebraic factorisation as a difference of two squares and a perfect square — a genuine synthesis of everything covered across this topic. Spotting that both numerator and denominator hide a common factor of $(\cos\theta-\sin\theta)$ is the key structural insight that unlocks the whole proof.
Common mistake
Manipulating both sides of an identity simultaneously ("working from the middle outwards") — this can accidentally use the very fact being proved, making the argument circular and invalid. Always start from just one side (conventionally the more complex one) and manipulate it alone until it matches the other side exactly.
Check yourself
Prove that $\tan\theta+\cot\theta \equiv \dfrac{2}{\sin2\theta}$. (Hint: convert both terms on the left to sines and cosines, combine into one fraction, then apply the Pythagorean and double-angle identities.)