The Three Fundamental Identities#
These are the building blocks of ALL trig proofs and simplifications:
Identity 1: The Quotient Identity#
$$ \tan\theta = \frac{\sin\theta}{\cos\theta} $$Identity 2: The Squared Identity (Pythagorean)#
$$ \sin^2\theta + \cos^2\theta = 1 $$This can be rearranged:
- $\sin^2\theta = 1 - \cos^2\theta$
- $\cos^2\theta = 1 - \sin^2\theta$
Why $\sin^2\theta + \cos^2\theta = 1$?#
In a right triangle with hypotenuse $r$:
$\sin\theta = \frac{y}{r}$ and $\cos\theta = \frac{x}{r}$
$\sin^2\theta + \cos^2\theta = \frac{y^2}{r^2} + \frac{x^2}{r^2} = \frac{x^2 + y^2}{r^2} = \frac{r^2}{r^2} = 1$ (by Pythagoras)
1. Proving Identities#
The Rules#
- Work with ONE side only — usually the more complicated side.
- Never move things across the = sign (it’s an identity, not an equation).
- Strategy: Convert everything to $\sin$ and $\cos$, then simplify.
Worked Example 1#
Prove: $\frac{\sin\theta}{\tan\theta} = \cos\theta$
LHS = $\frac{\sin\theta}{\frac{\sin\theta}{\cos\theta}} = \sin\theta \times \frac{\cos\theta}{\sin\theta} = \cos\theta$ = RHS ✓
Worked Example 2#
Prove: $\frac{1 - \cos^2\theta}{\sin\theta} = \sin\theta$
LHS = $\frac{\sin^2\theta}{\sin\theta} = \sin\theta$ = RHS ✓
(Used $1 - \cos^2\theta = \sin^2\theta$)
Worked Example 3 (Harder)#
Prove: $\frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta} = \frac{2}{\sin\theta}$
LHS: Common denominator = $\sin\theta(1 + \cos\theta)$
$= \frac{\sin^2\theta + (1 + \cos\theta)^2}{\sin\theta(1 + \cos\theta)}$
$= \frac{\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta}{\sin\theta(1 + \cos\theta)}$
$= \frac{(\sin^2\theta + \cos^2\theta) + 1 + 2\cos\theta}{\sin\theta(1 + \cos\theta)}$
$= \frac{1 + 1 + 2\cos\theta}{\sin\theta(1 + \cos\theta)}$
$= \frac{2 + 2\cos\theta}{\sin\theta(1 + \cos\theta)}$
$= \frac{2(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)}$
$= \frac{2}{\sin\theta}$ = RHS ✓
Worked Example 4 (Factoring)#
Prove: $\cos^2\theta - \sin^2\theta = (1 - \tan^2\theta)\cos^2\theta$
RHS = $(1 - \frac{\sin^2\theta}{\cos^2\theta})\cos^2\theta$
$= \cos^2\theta - \sin^2\theta$ = LHS ✓
2. Identity Proof Strategies#
| If you see… | Try this… |
|---|---|
| $\tan\theta$ | Replace with $\frac{\sin\theta}{\cos\theta}$ |
| $1 - \cos^2\theta$ or $1 - \sin^2\theta$ | Use the Pythagorean identity |
| $\cos^2\theta - \sin^2\theta$ | Factor as $(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)$ |
| Fractions with different denominators | Find a common denominator |
| $\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$ | Recognise as $(\sin\theta + \cos\theta)^2$ |
| A “1” somewhere | Replace with $\sin^2\theta + \cos^2\theta$ |
3. Solving Trigonometric Equations#
The Method#
- Simplify the equation (factor, use identities).
- Isolate the trig ratio (e.g., $\sin\theta = 0.5$).
- Find the reference angle using your calculator or special angles.
- Use the CAST diagram to find ALL solutions in the required interval.
Worked Example 1: Basic#
Solve $2\sin\theta - 1 = 0$ for $\theta \in [0°; 360°]$
$\sin\theta = \frac{1}{2}$
Reference angle: $\theta_{ref} = 30°$
Sin is positive in Q1 and Q2:
$\theta = 30°$ or $\theta = 180° - 30° = 150°$
Worked Example 2: Quadratic#
Solve $2\cos^2\theta - \cos\theta - 1 = 0$ for $\theta \in [0°; 360°]$
Let $k = \cos\theta$:
$2k^2 - k - 1 = 0$
$(2k + 1)(k - 1) = 0$
$k = -\frac{1}{2}$ or $k = 1$
Case 1: $\cos\theta = -\frac{1}{2}$, ref angle = $60°$
Cos negative in Q2 and Q3: $\theta = 120°$ or $\theta = 240°$
Case 2: $\cos\theta = 1$
$\theta = 0°$ or $\theta = 360°$
Worked Example 3: Using identities first#
Solve $\sin^2\theta = 1 - \cos\theta$ for $\theta \in [0°; 360°]$
Replace $\sin^2\theta$ with $1 - \cos^2\theta$:
$1 - \cos^2\theta = 1 - \cos\theta$
$-\cos^2\theta + \cos\theta = 0$
$\cos\theta(-\cos\theta + 1) = 0$
$\cos\theta = 0$ or $\cos\theta = 1$
$\theta = 90°, 270°$ or $\theta = 0°, 360°$
4. General Solutions#
Instead of listing solutions in $[0°; 360°]$, the general solution gives ALL possible angles:
For $\sin\theta = k$:#
$\theta = \theta_{ref} + 360°n$ or $\theta = (180° - \theta_{ref}) + 360°n$, $n \in \mathbb{Z}$
For $\cos\theta = k$:#
$\theta = \pm\theta_{ref} + 360°n$, $n \in \mathbb{Z}$
For $\tan\theta = k$:#
$\theta = \theta_{ref} + 180°n$, $n \in \mathbb{Z}$
Worked Example#
General solution of $\sin\theta = -\frac{\sqrt{3}}{2}$
Reference angle: $60°$
Sin is negative in Q3 and Q4:
$\theta = 180° + 60° + 360°n = 240° + 360°n$
or $\theta = 360° - 60° + 360°n = 300° + 360°n$, $n \in \mathbb{Z}$
🚨 Common Mistakes#
- Moving terms across the = sign in a proof: You’re showing the two sides are equal, not solving. Work with ONE side.
- Missing solutions: Sin and Cos give answers in TWO quadrants. If you only give one, you lose half the marks.
- Dividing by $\sin\theta$ or $\cos\theta$: You lose solutions where $\sin\theta = 0$. Factor instead!
- Not using the identity: If you see $\sin^2\theta$ and $\cos^2\theta$ together, the Pythagorean identity almost always simplifies things.
- General solution — forgetting $n \in \mathbb{Z}$: Always state that $n$ is an integer.
💡 Pro Tip: The “Convert Everything” Strategy#
When stuck on a proof, convert EVERYTHING to $\sin$ and $\cos$. Replace $\tan\theta = \frac{\sin\theta}{\cos\theta}$, and use $\sin^2\theta + \cos^2\theta = 1$ aggressively. This strategy solves 90% of identity proofs.
🔗 Related Grade 11 topics:
- Reduction Formulas & CAST — you MUST reduce angles before using identities
- Sine, Cosine & Area Rules — applies trig to triangles (often combined with identities in exams)
- Quadratic Equations — trig equations often become quadratics (e.g., $2\sin^2\theta - \sin\theta - 1 = 0$)