# Trigonometric Identities and Equations

> Key identities, reciprocal ratios, proving identities and solving trig equations over a domain.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/trig-identities-and-equations/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: Two keys, one lock
Ben's grandma has two keys that look completely different. One is old and brass. One is shiny and new. Both open the same front door.

Trig identities are like that. $1 - \cos^2\theta$ and $\sin^2\theta$ look different, but they are always equal, for every angle. Swapping one for the other is often the trick that turns a hard equation into an easy one.

**In short:** A trig identity is true for every angle, like $\sin^2\theta + \cos^2\theta = 1$ and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$. To solve a trig equation, get the ratio alone, find the related angle, then use ASTC to find every answer in the domain.

## Starting point: Years 9 and 10

### Solving a simple trig equation

To solve $\sin\theta = 0.6$ for an acute angle, use the inverse:

$$\theta = \sin^{-1}(0.6) \approx 36.9^\circ$$

Rearrange first if needed: $4\cos\theta = 3$ gives $\cos\theta = 0.75$, so $\theta \approx 41.4^\circ$.

### Calculator tip: The calculator gives only one answer
`SHIFT sin` gives just one solution, between $-90^\circ$ and $90^\circ$. Use the quadrant rule to find the other solutions in the domain, then check each one by working out its sine on the calculator.

### Two identities from one triangle

Take a right triangle with hypotenuse 1 and angle $\theta$. The opposite side is $\sin\theta$ and the adjacent side is $\cos\theta$.

Opposite over adjacent gives

$$\tan\theta = \frac{\sin\theta}{\cos\theta}$$

Pythagoras gives

$$\sin^2\theta + \cos^2\theta = 1$$

Here $\sin^2\theta$ means $(\sin\theta)^2$.

### Worked example: Using the identity
If $\sin\theta = \dfrac{3}{5}$ and $\theta$ is acute, find $\cos\theta$ and $\tan\theta$.

$\cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$, so $\cos\theta = \dfrac{4}{5}$. Then $\tan\theta = \dfrac{3/5}{4/5} = \dfrac{3}{4}$.

Or draw a 3, 4, 5 triangle and read them off.

### Key idea: An identity is true for every angle
An **equation** like $\sin\theta = 0.5$ is true only for some angles. An **identity** like $\sin^2\theta + \cos^2\theta = 1$ is true for all of them.

### Watch out: sin²θ is not sin(θ²)
$\sin^2 30^\circ = (\sin 30^\circ)^2 = 0.25$. It is not $\sin 900^\circ$.

**Mind map**

```text
Identities and equations
  Core identities
    tan = sin / cos
    sin² + cos² = 1
  Reciprocal ratios
    cosec = 1/sin
    sec = 1/cos
    cot = 1/tan
  Proving
    Start with the harder side
    Change to sin and cos
  Solving
    Isolate the trig ratio
    Related angle
    ASTC for every quadrant
    Check the domain
```

### Year 11: Reciprocal ratios and more identities
$$\operatorname{cosec}\theta = \frac{1}{\sin\theta} \qquad \sec\theta = \frac{1}{\cos\theta} \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$$

Divide $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$ and by $\sin^2\theta$ to get:

$$1 + \tan^2\theta = \sec^2\theta \qquad 1 + \cot^2\theta = \operatorname{cosec}^2\theta$$

**Three ways you will use these identities**

| Use case | You will see |
|---|---|
| 1. Proving an identity | "Prove that" or "Show that" |
| 2. Using a sign condition | one ratio, and only the sign of another, such as $\tan\theta < 0$ |
| 3. Factorising inside identities | powers such as $\sin^4 x$, or two fractions added together |

**1. Proving an identity**

*Use it when* a question says "prove" or "show that".

*How:* start from the more complicated side and change it one step at a time until it matches the other side. Changing everything to sin and cos usually helps. Do not move terms across the equals sign: you are showing the two sides are the same, not solving.

*Example.* Prove $\dfrac{1 - \cos^2\theta}{\sin\theta\cos\theta} = \tan\theta$.

LHS $= \dfrac{\sin^2\theta}{\sin\theta\cos\theta} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta =$ RHS.

**2. Using a sign condition**

*Use it when* you are given one ratio and only the **sign** of another, such as $\tan\theta < 0$.

*How:* the sign tells you which quadrant $\theta$ is in, and the quadrant decides the sign of every other ratio. Work in this order: quadrant first, then sizes, then signs.

*Example.* $\sin\theta = \dfrac{3}{5}$ and $\tan\theta < 0$. Find $\cos\theta$, $\tan\theta$ and $\sec\theta$.

1. **Find the quadrant.** Sin is positive in quadrants 1 and 2 (ASTC). Tan is negative in quadrants 2 and 4. Only quadrant 2 is on both lists, so $\theta$ is in quadrant 2.
2. **Find the sizes.** Forget the signs for a moment. $\sin\theta = \dfrac{3}{5}$ means opposite 3 and hypotenuse 5. Pythagoras gives the third side: $\sqrt{5^2 - 3^2} = 4$. So the sizes are $\cos\theta = \dfrac{4}{5}$ and $\tan\theta = \dfrac{3}{4}$.
3. **Put in the signs.** In quadrant 2 only sin is positive, so $\cos\theta = -\dfrac{4}{5}$ and $\tan\theta = -\dfrac{3}{4}$.
4. **Reciprocals keep the sign** of the ratio they come from: $\sec\theta = \dfrac{1}{\cos\theta} = -\dfrac{5}{4}$. Likewise $\operatorname{cosec}\theta = \dfrac{5}{3}$ and $\cot\theta = -\dfrac{4}{3}$.

*Diagram: Angle theta in quadrant 2. The point (minus 4, 3) makes a right triangle with sides 3 up, 4 to the left of the origin, and hypotenuse 5*
*In quadrant 2 the point is (−4, 3) and the hypotenuse is 5. So sin θ = 3/5, cos θ = −4/5 and tan θ = 3/(−4) = −3/4.*

You can also get the size from the identity: $\cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$, so $\cos\theta = \pm\dfrac{4}{5}$. The identity cannot choose between $+$ and $-$. Only the quadrant can.

**3. Factorising inside identities**

*Use it when* you see higher powers such as $\sin^4 x$, or two fractions added together.

*How:* use ordinary algebra (difference of two squares, a common denominator), then look for $\sin^2 x + \cos^2 x$ to replace with 1.

*Examples.*
- Difference of two squares: $\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) = \sin^2 x - \cos^2 x$.
- Common denominator: $\dfrac{1}{1 - \sin x} + \dfrac{1}{1 + \sin x} = \dfrac{2}{1 - \sin^2 x} = \dfrac{2}{\cos^2 x}$.

### Year 11: Solving over a domain
The calculator gives only **one** angle. But two different angles can have the same sine, cosine or tangent, one in each of two quadrants. So for $0^\circ \leq \theta \leq 360^\circ$ there are usually **two** answers, and you must find both.

**The method**

1. **Isolate** the ratio: get $\sin\theta$, $\cos\theta$ or $\tan\theta$ alone.
2. **Find the related angle** $\alpha$: use the inverse on the value **without** its minus sign. It is always acute.
3. **Choose the quadrants** with ASTC: the two quadrants where the ratio has the sign you need.
4. **Turn $\alpha$ into an angle in each of those quadrants:**

| Quadrant | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Degrees | $\alpha$ | $180^\circ - \alpha$ | $180^\circ + \alpha$ | $360^\circ - \alpha$ |
| Radians | $\alpha$ | $\pi - \alpha$ | $\pi + \alpha$ | $2\pi - \alpha$ |

*Diagram: The same acute angle alpha drawn from the x-axis in each quadrant. Quadrant 1 is alpha, quadrant 2 is 180 degrees minus alpha, quadrant 3 is 180 degrees plus alpha, and quadrant 4 is 360 degrees minus alpha*
*The same acute angle α sits against the x-axis in every quadrant. Quadrant 2 stops α short of 180°, quadrant 3 goes α past 180°, and quadrant 4 stops α short of 360°.*

5. **Check** every answer is inside the domain, and test it on the calculator.

**Example.** Solve $2\cos\theta + 1 = 0$ for $0^\circ \leq \theta \leq 360^\circ$.

1. Isolate: $2\cos\theta = -1$, so $\cos\theta = -\dfrac{1}{2}$.
2. Related angle: $\cos^{-1}\left(\dfrac{1}{2}\right) = 60^\circ$. (Use $\dfrac{1}{2}$, not $-\dfrac{1}{2}$.)
3. Cos is negative in quadrants 2 and 3.
4. Quadrant 2: $\theta = 180^\circ - 60^\circ = 120^\circ$. Quadrant 3: $\theta = 180^\circ + 60^\circ = 240^\circ$.
5. Check: $\cos 120^\circ = -0.5$ and $\cos 240^\circ = -0.5$. Both are in the domain.

*Diagram: Unit circle. The vertical line x equals minus one half meets the circle at 120 degrees and 240 degrees, each 60 degrees from the negative x-axis*
*cos θ is the x value on the unit circle. The line x = −1/2 meets the circle twice: at 120° and 240°, each 60° from the negative x-axis.*

**A different domain.** Add or subtract $360^\circ$ to move an answer into the domain. For $-180^\circ \leq \theta \leq 180^\circ$, the answer $240^\circ$ becomes $240^\circ - 360^\circ = -120^\circ$, so the solutions are $120^\circ$ and $-120^\circ$.

*Diagram: On the unit circle, turning 240 degrees anticlockwise and 120 degrees clockwise reach the same point, so 240 degrees and minus 120 degrees are the same answer*
*240° and −120° end at the same point, so subtracting 360° moves an answer into −180° to 180°.*

**Two full turns.** A domain of $0^\circ$ to $720^\circ$ is two full turns, so add $360^\circ$ to each answer as well: $120^\circ$, $240^\circ$, $480^\circ$ and $600^\circ$.

*Diagram: The graph of y equals cos theta from 0 to 720 degrees makes two full waves, and the line y equals minus one half crosses it at 120, 240, 480 and 600 degrees*
*y = cos θ makes a full wave every 360°, so from 0° to 720° the line y = −1/2 crosses it four times: 120° and 240°, then those plus 360°.*

**One answer, or none.** At the edge of the circle there is only one answer: $\sin\theta = 1$ only at $90^\circ$, and $\cos\theta = -1$ only at $180^\circ$. Sine and cosine are never bigger than 1 or smaller than $-1$, so $\sin\theta = 2$ has no solution at all.

**Five kinds of equation**

Every equation below finishes with the method above. The first job is to turn it into "one ratio = a number".

| The equation has | Do this first |
|---|---|
| 1. One ratio squared and on its own | treat it as a quadratic |
| 2. $\sin\theta$ with $\cos^2\theta$ (or $\cos\theta$ with $\sin^2\theta$) | swap the square using $\sin^2\theta + \cos^2\theta = 1$ |
| 3. $\sin x$ and $\cos x$, no squares | divide by $\cos x$ to get $\tan x$ |
| 4. Factors multiplied to give 0 | set each factor equal to 0 |
| 5. $\ln(\ldots) = 0$ | set the inside equal to 1 |

**1. A quadratic in one ratio**

*Use it when* the same ratio appears squared and on its own, like $2\sin^2\theta - \sin\theta - 1 = 0$.

*How:* let $u = \sin\theta$, factorise, then solve each answer as a simple equation.

*Example.* $(2u + 1)(u - 1) = 0$, so $\sin\theta = 1$ or $\sin\theta = -\dfrac{1}{2}$. $\sin\theta = 1$ gives $90^\circ$ only (the top of the circle). $\sin\theta = -\dfrac{1}{2}$ has related angle $30^\circ$ in quadrants 3 and 4, giving $210^\circ$ and $330^\circ$. That is three answers in all.

**2. Two different ratios, one squared**

*Use it when* sin and $\cos^2$ (or cos and $\sin^2$) appear together, like $2\cos^2\theta = 3\sin\theta$.

*How:* swap the square using $\sin^2\theta + \cos^2\theta = 1$, so only one ratio is left. Then it is kind 1.

*Example.* $2(1 - \sin^2\theta) = 3\sin\theta$, that is $2\sin^2\theta + 3\sin\theta - 2 = 0$. Factorise: $(2\sin\theta - 1)(\sin\theta + 2) = 0$. $\sin\theta = -2$ is impossible, so $\sin\theta = \dfrac{1}{2}$, giving $30^\circ$ and $150^\circ$.

**3. Sine and cosine, no squares**

*Use it when* the equation is like $\sin x + \sqrt{3}\cos x = 0$.

*How:* divide every term by $\cos x$, because $\dfrac{\sin x}{\cos x} = \tan x$. First check that $\cos x = 0$ is not a solution: there $\sin x = \pm 1$, and $\pm 1 + 0 \neq 0$.

*Example.* $\tan x = -\sqrt{3}$. The related angle is $\dfrac{\pi}{3}$ and tan is negative in quadrants 2 and 4, so $x = \dfrac{2\pi}{3}$ or $\dfrac{5\pi}{3}$ for $0 \leq x \leq 2\pi$.

**4. A product equal to zero**

*Use it when* factors are multiplied to give 0, like $\cos x(2\sin x - 1) = 0$.

*How:* set each factor equal to 0, solve each one over the domain, then list all the answers together in order. Do not divide by $\cos x$: you would lose its answers.

*Example.* $\cos x = 0$ gives $\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$. $\sin x = \dfrac{1}{2}$ gives $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$. All four answers: $\dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$.

**5. A trig equation hidden inside ln**

*Use it when* you see $\ln(\ldots) = 0$.

*How:* $\ln 1 = 0$, so the inside must equal 1. Then solve the trig equation that is left.

*Example.* $\ln(2\sin x) = 0$ for $0 \leq x \leq 2\pi$. So $2\sin x = 1$, $\sin x = \dfrac{1}{2}$, and $x = \dfrac{\pi}{6}$ or $\dfrac{5\pi}{6}$. Both make $2\sin x$ positive, as ln needs.

### Year 12: Equations with a changed angle
In $\sin 2x$, $\cos 3x$ or $\cos(x - 30^\circ)$, the angle inside is not just $x$. Solve for the **whole inside angle** first, then work back to $x$.

**Why the domain must change.** If $x$ goes from $0$ to $2\pi$, then $2x$ goes from $0$ to $4\pi$: two full turns of the circle. Each turn gives its own answers, so there are twice as many. If you solve only from $0$ to $2\pi$ for $2x$, you lose half of them.

**The method**

1. Isolate the ratio.
2. Call the inside angle $u$ and work out **its** domain: do the same thing to both ends of the domain of $x$.
3. Solve for $u$ in the first turn as usual (related angle, then ASTC).
4. Add $2\pi$ (or $360^\circ$) to each answer for every extra turn, keeping only those still inside the domain of $u$.
5. Change every $u$ back into $x$.

**Example 1.** Solve $\sqrt{2}\sin 2x = 1$ for $0 \leq x \leq 2\pi$.

1. $\sin 2x = \dfrac{1}{\sqrt{2}}$.
2. Let $u = 2x$. Double both ends: $0 \leq u \leq 4\pi$.
3. Related angle $\dfrac{\pi}{4}$. Sin is positive in quadrants 1 and 2: $u = \dfrac{\pi}{4}$ and $u = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$.
4. Second turn: add $2\pi$ to each, giving $u = \dfrac{9\pi}{4}$ and $\dfrac{11\pi}{4}$. Both are less than $4\pi$.
5. $x = \dfrac{u}{2}$: $x = \dfrac{\pi}{8}, \dfrac{3\pi}{8}, \dfrac{9\pi}{8}, \dfrac{11\pi}{8}$.

Check one: $x = \dfrac{9\pi}{8}$ gives $2x = \dfrac{9\pi}{4}$, which is one full turn past $\dfrac{\pi}{4}$, so its sine is $\dfrac{1}{\sqrt{2}}$.

*Diagram: The graph of y equals sin 2x from 0 to 2 pi makes two full waves. The line y equals 1 over root 2 crosses it four times, at pi over 8, 3 pi over 8, 9 pi over 8 and 11 pi over 8*
*y = sin 2x fits two full waves between 0 and 2π, so the line y = 1/√2 crosses it four times, twice in each wave.*

**Example 2: a shifted angle.** Solve $\cos(x - 30^\circ) = \dfrac{1}{2}$ for $0^\circ \leq x \leq 360^\circ$.

1. Let $u = x - 30^\circ$. Subtract $30^\circ$ from both ends: $-30^\circ \leq u \leq 330^\circ$.
2. Related angle $60^\circ$. Cos is positive in quadrants 1 and 4: $u = 60^\circ$ and $u = 360^\circ - 60^\circ = 300^\circ$. Both are inside $-30^\circ$ to $330^\circ$. ($u = -60^\circ$ is not.)
3. Add the $30^\circ$ back: $x = 90^\circ$ and $x = 330^\circ$.

**Halving the angle** works the other way. For $\sin\dfrac{x}{2}$ with $0 \leq x \leq 2\pi$, the inside angle only runs from $0$ to $\pi$, half a turn, so you may get fewer answers.

In Year 12 these equations appear when you set a derivative to zero to find where a wave peaks.

## Quick summary

- $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta = 1$.
- Reciprocal ratios: $\operatorname{cosec}$, $\sec$, $\cot$. Also $1 + \tan^2\theta = \sec^2\theta$.
- Prove identities by working on one side, usually changing to sin and cos.
- To solve: isolate the ratio, find the related angle, use ASTC for all quadrants, check the domain.
- For $\sin bx$, widen the domain by a factor of $b$ before solving.

## Quiz

Try these questions from the chapter quiz.

1. Solve $\sin\theta = 0.5$ for an acute angle $\theta$.
   - A. $30^\circ$
   - B. $45^\circ$
   - C. $60^\circ$
   - D. $0.5^\circ$
   - Answer: A. $30^\circ$. $\sin^{-1}(0.5) = 30^\circ$.

2. Solve $3\tan\theta = 3$ for an acute angle.
   - A. $30^\circ$
   - B. $45^\circ$
   - C. $60^\circ$
   - D. $90^\circ$
   - Answer: B. $45^\circ$. $\tan\theta = 1$ at $45^\circ$.

3. $\dfrac{\sin\theta}{\cos\theta}$ equals:
   - A. $\cot\theta$
   - B. $\tan\theta$
   - C. $\sec\theta$
   - D. 1
   - Answer: B. $\tan\theta$. This is the definition of $\tan\theta$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/trig-identities-and-equations/#quiz