Trigonometric Identities and Equations

In short: A trig identity is true for every angle, like $sin^{2} θ + cos^{2} θ = 1$ and $tan θ = \frac{sin θ}{cos θ}$ . 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 $sin θ = 0.6$ for an acute angle, use the inverse:

$θ = sin^{- 1} (0.6) \approx 36. 9^{\circ}$

Rearrange first if needed: $4 cos θ = 3$ gives $cos θ = 0.75$ , so $θ \approx 41. 4^{\circ}$ .

Two identities from one triangle

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

Opposite over adjacent gives

$tan θ = \frac{sin θ}{cos θ}$

Pythagoras gives

$sin^{2} θ + cos^{2} θ = 1$

Here $sin^{2} θ$ means $(sin θ)^{2}$ .

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 11Reciprocal ratios and more identities

$cosec θ = \frac{1}{sin θ} sec θ = \frac{1}{cos θ} cot θ = \frac{1}{tan θ} = \frac{cos θ}{sin θ}$

Divide $sin^{2} θ + cos^{2} θ = 1$ by $cos^{2} θ$ and by $sin^{2} θ$ to get:

$1 + tan^{2} θ = sec^{2} θ 1 + cot^{2} θ = cosec^{2} θ$

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 $tan θ < 0$
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 $\frac{1 - cos ^{2} θ}{sin θ cos θ} = tan θ$ .

LHS $= \frac{sin ^{2} θ}{sin θ cos θ} = \frac{sin θ}{cos θ} = tan θ =$ RHS.

2. Using a sign condition

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

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

Example. $sin θ = \frac{3}{5}$ and $tan θ < 0$ . Find $cos θ$ , $tan θ$ and $sec θ$ .

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 $θ$ is in quadrant 2.
Find the sizes. Forget the signs for a moment. $sin θ = \frac{3}{5}$ means opposite 3 and hypotenuse 5. Pythagoras gives the third side: $5^{2} - 3^{2} = 4$ . So the sizes are $cos θ = \frac{4}{5}$ and $tan θ = \frac{3}{4}$ .
Put in the signs. In quadrant 2 only sin is positive, so $cos θ = - \frac{4}{5}$ and $tan θ = - \frac{3}{4}$ .
Reciprocals keep the sign of the ratio they come from: $sec θ = \frac{1}{cos θ} = - \frac{5}{4}$ . Likewise $cosec θ = \frac{5}{3}$ and $cot θ = - \frac{4}{3}$ .

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: $cos^{2} θ = 1 - \frac{9}{25} = \frac{16}{25}$ , so $cos θ = \pm \frac{4}{5}$ . 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: $\frac{1}{1 - sin x} + \frac{1}{1 + sin x} = \frac{2}{1 - sin ^{2} x} = \frac{2}{cos ^{2} x}$ .

Year 11Solving 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 $0^{\circ} \leq θ \leq 36 0^{\circ}$ there are usually two answers, and you must find both.

The method

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

Quadrant	1	2	3	4
Degrees	$α$	$18 0^{\circ} - α$	$18 0^{\circ} + α$	$36 0^{\circ} - α$
Radians	$α$	$π - α$	$π + α$	$2 π - α$

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°.

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

Example. Solve $2 cos θ + 1 = 0$ for $0^{\circ} \leq θ \leq 36 0^{\circ}$ .

Isolate: $2 cos θ = - 1$ , so $cos θ = - \frac{1}{2}$ .
Related angle: $cos^{- 1} (\frac{1}{2}) = 6 0^{\circ}$ . (Use $\frac{1}{2}$ , not $- \frac{1}{2}$ .)
Cos is negative in quadrants 2 and 3.
Quadrant 2: $θ = 18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ . Quadrant 3: $θ = 18 0^{\circ} + 6 0^{\circ} = 24 0^{\circ}$ .
Check: $cos 12 0^{\circ} = - 0.5$ and $cos 24 0^{\circ} = - 0.5$ . Both are in the domain.

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 $36 0^{\circ}$ to move an answer into the domain. For $- 18 0^{\circ} \leq θ \leq 18 0^{\circ}$ , the answer $24 0^{\circ}$ becomes $24 0^{\circ} - 36 0^{\circ} = - 12 0^{\circ}$ , so the solutions are $12 0^{\circ}$ and $- 12 0^{\circ}$ .

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 $72 0^{\circ}$ is two full turns, so add $36 0^{\circ}$ to each answer as well: $12 0^{\circ}$ , $24 0^{\circ}$ , $48 0^{\circ}$ and $60 0^{\circ}$ .

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: $sin θ = 1$ only at $9 0^{\circ}$ , and $cos θ = - 1$ only at $18 0^{\circ}$ . Sine and cosine are never bigger than 1 or smaller than $- 1$ , so $sin θ = 2$ 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. $sin θ$ with $cos^{2} θ$ (or $cos θ$ with $sin^{2} θ$ )	swap the square using $sin^{2} θ + cos^{2} θ = 1$
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. $ln (\dots) = 0$	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 $2 sin^{2} θ - sin θ - 1 = 0$ .

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

Example. $(2 u + 1) (u - 1) = 0$ , so $sin θ = 1$ or $sin θ = - \frac{1}{2}$ . $sin θ = 1$ gives $9 0^{\circ}$ only (the top of the circle). $sin θ = - \frac{1}{2}$ has related angle $3 0^{\circ}$ in quadrants 3 and 4, giving $21 0^{\circ}$ and $33 0^{\circ}$ . 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 $2 cos^{2} θ = 3 sin θ$ .

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

Example. $2 (1 - sin^{2} θ) = 3 sin θ$ , that is $2 sin^{2} θ + 3 sin θ - 2 = 0$ . Factorise: $(2 sin θ - 1) (sin θ + 2) = 0$ . $sin θ = - 2$ is impossible, so $sin θ = \frac{1}{2}$ , giving $3 0^{\circ}$ and $15 0^{\circ}$ .

3. Sine and cosine, no squares

Use it when the equation is like $sin x + 3 cos x = 0$ .

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

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

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 $\frac{π}{2}$ and $\frac{3 π}{2}$ . $sin x = \frac{1}{2}$ gives $\frac{π}{6}$ and $\frac{5 π}{6}$ . All four answers: $\frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}, \frac{3 π}{2}$ .

5. A trig equation hidden inside ln

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

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 $0 \leq x \leq 2 π$ . So $2 sin x = 1$ , $sin x = \frac{1}{2}$ , and $x = \frac{π}{6}$ or $\frac{5 π}{6}$ . Both make $2 sin x$ positive, as ln needs.

Year 12Equations with a changed angle

In $sin 2 x$ , $cos 3 x$ or $cos (x - 3 0^{\circ})$ , 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 $2 π$ , then $2 x$ goes from $0$ to $4 π$ : 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 $2 π$ for $2 x$ , you lose half of them.

The method

Isolate the ratio.
Call the inside angle $u$ and work out its domain: do the same thing to both ends of the domain of $x$ .
Solve for $u$ in the first turn as usual (related angle, then ASTC).
Add $2 π$ (or $36 0^{\circ}$ ) to each answer for every extra turn, keeping only those still inside the domain of $u$ .
Change every $u$ back into $x$ .

Example 1. Solve $2 sin 2 x = 1$ for $0 \leq x \leq 2 π$ .

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

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

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 $cos (x - 3 0^{\circ}) = \frac{1}{2}$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .

Let $u = x - 3 0^{\circ}$ . Subtract $3 0^{\circ}$ from both ends: $- 3 0^{\circ} \leq u \leq 33 0^{\circ}$ .
Related angle $6 0^{\circ}$ . Cos is positive in quadrants 1 and 4: $u = 6 0^{\circ}$ and $u = 36 0^{\circ} - 6 0^{\circ} = 30 0^{\circ}$ . Both are inside $- 3 0^{\circ}$ to $33 0^{\circ}$ . ( $u = - 6 0^{\circ}$ is not.)
Add the $3 0^{\circ}$ back: $x = 9 0^{\circ}$ and $x = 33 0^{\circ}$ .

Halving the angle works the other way. For $sin \frac{x}{2}$ with $0 \leq x \leq 2 π$ , the inside angle only runs from $0$ to $π$ , 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

$tan θ = \frac{sin θ}{cos θ}$ and $sin^{2} θ + cos^{2} θ = 1$ .
Reciprocal ratios: $cosec$ , $sec$ , $cot$ . Also $1 + tan^{2} θ = sec^{2} θ$ .
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 $sin b x$ , widen the domain by a factor of $b$ before solving.

Quiz

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Solve $sin θ = 0.5$ for an acute angle $θ$ .
1. A $3 0^{\circ}$
2. B $4 5^{\circ}$
3. C $6 0^{\circ}$
4. D $0. 5^{\circ}$
Show answer
A: $3 0^{\circ}$
$sin^{- 1} (0.5) = 3 0^{\circ}$ .
Solve $3 tan θ = 3$ for an acute angle.
1. A $3 0^{\circ}$
2. B $4 5^{\circ}$
3. C $6 0^{\circ}$
4. D $9 0^{\circ}$
Show answer
B: $4 5^{\circ}$
$tan θ = 1$ at $4 5^{\circ}$ .
$\frac{sin θ}{cos θ}$ equals:
1. A $cot θ$
2. B $tan θ$
3. C $sec θ$
4. D 1
Show answer
B: $tan θ$
This is the definition of $tan θ$ .