Trigonometry and Modelling

What you'll learn

Solve trigonometric equations over a specified interval without missing solutions.
Derive and use double-angle and triple-angle identities.
Rewrite $asin⁡x+bcos⁡xa\sin x+b\cos x$ as one shifted sine or cosine wave.
Use trigonometry to model repeating situations, such as daily temperature.

1. Angle intervals and inverse trig

At A-Level, the interval matters just as much as the equation. You might be asked for solutions in degrees, such as $0∘≤x<360∘0^\circ\leq x<360^\circ$ , or in radians, such as $0≤x<2π0\leq x<2\pi$ .

Always check your calculator mode before using inverse trig, such as $sin^{-1}$ , $cos^{-1}$ or $tan^{-1}$ .

Definition

Period

The period of a trigonometric function is the angle length after which its values repeat. Sine and cosine have period 360° or $2π2\pi$ ; tangent has period 180° or $π\pi$ .

A reference angle is the acute angle your calculator gives before you use the quadrant information to find all required solutions.

Example

Solving a basic cosine equation

Solve $cos⁡θ=0.25\cos\theta=0.25$ for $0≤θ<2π0\leq\theta<2\pi$ . The interval uses $π\pi$ , so set your calculator to radians and find the reference angle:
$α=cos⁡−1(0.25)=1.318…\alpha=\cos^{-1}(0.25)=1.318\ldots$
Cosine is positive in the first and fourth quadrants, so the two solutions are:

A unit-circle diagram showing the reference angle for cos θ = 0.25 and the two solutions in quadrants I and IV.

$$
\theta=\alpha \quad \text{or} \quad \theta=2\pi-\alpha
$$

3. Round to 3 d.p.:

$$
\theta=1.318,\ 4.965
$$

2. Identities from compound-angle formulae

Definition

Trigonometric identity

A trigonometric identity is an equation involving trig functions that is true for every value for which both sides are defined. The symbol $≡\equiv$ means “identically equal to”.

The key compound-angle formulae are:

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bcos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡Btan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\begin{aligned} \sin(A+B)&=\sin A\cos B+\cos A\sin B\\ \cos(A+B)&=\cos A\cos B-\sin A\sin B\\ \tan(A+B)&=\frac{\tan A+\tan B}{1-\tan A\tan B} \end{aligned}

Key Idea

Double angles

To get a double-angle identity, replace $B$ by $A$ in a compound-angle formula.

Example

Deriving double-angle identities

Put $B = A$ into the sine formula:
$sin⁡2A=sin⁡Acos⁡A+cos⁡Asin⁡A=2sin⁡Acos⁡A\sin 2A=\sin A\cos A+\cos A\sin A=2\sin A\cos A$
Put $B = A$ into the cosine formula:
$cos 2A=\cos^2 A-\sin^2 A$
Use $sin⁡2A+cos⁡2A≡1\sin^2 A+\cos^2 A\equiv1$ to create two alternative forms:
$cos 2A=2\cos^2 A-1=1-2\sin^2 A$
Put $B = A$ into the tangent formula:
$tan⁡2A=tan⁡A+tan⁡A1−tan⁡Atan⁡A=2tan⁡A1−tan⁡2A\tan 2A=\frac{\tan A+\tan A}{1-\tan A\tan A}=\frac{2\tan A}{1-\tan^2 A}$

The same approach gives the triple-angle identities:

sin⁡3θ≡3sin⁡θ−4sin⁡3θ,cos⁡3θ≡4cos⁡3θ−3cos⁡θ\sin 3\theta \equiv 3\sin\theta-4\sin^3\theta,\qquad \cos 3\theta \equiv 4\cos^3\theta-3\cos\theta

3. Using identities to solve equations

Many equations look like cubics, but an identity can turn them into a much simpler trig equation.

Example

Solving a cubic-looking sine equation

Solve $3sin⁡θ−4sin⁡3θ=0.63\sin\theta-4\sin^3\theta=0.6$ for $0∘≤θ≤180∘0^\circ\leq\theta\leq180^\circ$ . Recognise the left-hand side:
$3sin⁡θ−4sin⁡3θ=sin⁡3θ3\sin\theta-4\sin^3\theta=\sin 3\theta$
Rewrite the equation and set $u=3θu=3\theta$ . The interval for $u$ is three times as large:

A mapping diagram showing why θ in 0° to 180° becomes u in 0° to 540° when u = 3θ.

$$
\sin u=0.6,\qquad 0^\circ\leq u\leq540^\circ
$$

3. Find the reference angle:

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

4. Sine is positive in the first and second quadrants, and the interval includes one and a half full turns:

A number-line/angle-cycle diagram showing all four u-values for sin u = 0.6 over 0° ≤ u ≤ 540°.

$$
u=36.9^\circ,\ 143.1^\circ,\ 396.9^\circ,\ 503.1^\circ
$$

5. Divide each value by 3:

$$
\theta=12.3^\circ,\ 47.7^\circ,\ 132.3^\circ,\ 167.7^\circ
$$

Common Mistake

Forgetting the enlarged interval

If $u=3θu=3\theta$ , solve for $u$ over three times the original interval first. If you divide too early, you usually miss solutions from later cycles.

4. Reciprocal functions and identity proofs

Definition

Reciprocal trig functions

The secant and cosecant functions are $sec⁡x=1cos⁡x\sec x=\frac{1}{\cos x}$ and $csc⁡x=1sin⁡x\csc x=\frac{1}{\sin x}$ ; they are undefined where their denominators are zero.

When proving an identity, usually start with the more complicated side and simplify towards the other side. Do not try to “move terms across” as if you were solving an equation.

Example

Proving a reciprocal identity

Prove $sec⁡x−cos⁡x≡sin⁡xtan⁡x\sec x-\cos x\equiv\sin x\tan x$ . Start with the left-hand side and replace $sec⁡x\sec x$ :
$sec⁡x−cos⁡x=1cos⁡x−cos⁡x\sec x-\cos x=\frac{1}{\cos x}-\cos x$
Put the terms over a common denominator:
$1cos⁡x−cos⁡x=1−cos⁡2xcos⁡x\frac{1}{\cos x}-\cos x=\frac{1-\cos^2 x}{\cos x}$
Use $1−cos⁡2x≡sin⁡2x1-\cos^2 x\equiv\sin^2 x$ and $tan⁡x=sin⁡xcos⁡x\tan x=\frac{\sin x}{\cos x}$ :
$1−cos⁡2xcos⁡x=sin⁡2xcos⁡x=sin⁡xtan⁡x\frac{1-\cos^2 x}{\cos x}=\frac{\sin^2 x}{\cos x}=\sin x\tan x$

Common Mistake

Domains still matter

A fractional identity is only valid where both sides are defined. Be especially careful when cancelling factors that could be zero.

5. R-form: one wave instead of two

Definition

R-form

R-form rewrites $asin⁡x+bcos⁡xa\sin x+b\cos x$ as a single wave such as $Rsin⁡(x+α)R\sin(x+\alpha)$ or $Rcos⁡(x−α)R\cos(x-\alpha)$ . Here $R > 0$ is the amplitude, and $α\alpha$ is the phase shift.

Expand your chosen form before matching coefficients:

Rsin⁡(x+α)=Rcos⁡αsin⁡x+Rsin⁡αcos⁡xRcos⁡(x+α)=Rcos⁡αcos⁡x−Rsin⁡αsin⁡x\begin{aligned} R\sin(x+\alpha)&=R\cos\alpha\sin x+R\sin\alpha\cos x\\ R\cos(x+\alpha)&=R\cos\alpha\cos x-R\sin\alpha\sin x \end{aligned}

Usually, $R=a2+b2R=\sqrt{a^2+b^2}$ , and $α\alpha$ comes from a tangent ratio.

Tip

Choose signs carefully

The sign in $x+αx+\alpha$ or $x−αx-\alpha$ controls whether the sine or cosine coefficient is positive or negative, so always expand first.

Example

R-form, solving and transformations

Express $sin⁡x−3cos⁡x+4\sin x-\sqrt{3}\cos x+4$ in the form $Rsin⁡(x−α)+4R\sin(x-\alpha)+4$ . Expand the target form:
$Rsin⁡(x−α)=Rcos⁡αsin⁡x−Rsin⁡αcos⁡xR\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$
Match coefficients to get $R = 2$ and $α=60∘\alpha=60^\circ$ :

A right-angled coefficient triangle for the R-form expression sin x − √3 cos x = 2 sin(x − 60°).

$$
R\cos\alpha=1,\qquad R\sin\alpha=\sqrt{3}
$$

3. Therefore:

$$
\sin x-\sqrt{3}\cos x+4=2\sin(x-60^\circ)+4
$$

4. To solve $sin⁡x−3cos⁡x+4=5\sin x-\sqrt{3}\cos x+4=5$ for $0∘≤x<360∘0^\circ\leq x<360^\circ$ , set $u=x−60∘u=x-60^\circ$ :

$$
2\sin u+4=5 \Rightarrow \sin u=\frac{1}{2}
$$

5. Since $−60∘≤u<300∘-60^\circ\leq u<300^\circ$ , the solutions are $u=30∘u=30^\circ$ and $u=150∘u=150^\circ$ :

$$
x=90^\circ,\ 210^\circ
$$

6. Since sine lies between -1 and 1, the range is $2≤y≤62\leq y\leq6$ . The transformations from $y=sin⁡xy=\sin x$ are: translate 60° right, stretch vertically by factor 2, then translate 4 units up.

The transformed sine curve y = 2 sin(x − 60°) + 4 showing its midline, amplitude, range and phase shift.

6. Trigonometric modelling

Definition

Trig model

A trig model uses sine or cosine to approximate a repeating real-world situation. The midline is the average value, and the amplitude is the distance from the midline to a maximum.

In a daily model, an angle like $15 t$ means the input increases by 15° each hour. Since $36015=24\frac{360}{15}=24$ , the period is 24 hours.

Example

A daily temperature model

A room temperature, $θ\theta$ °C, is modelled for $0≤t<240\leq t<24$ by:
$θ=10+4sin⁡((15t−120)∘)+3cos⁡((15t−120)∘)\theta=10+4\sin\left((15t-120)^\circ\right)+3\cos\left((15t-120)^\circ\right)$
Let $u=(15t−120)∘u=(15t-120)^\circ$ . Write $4sin⁡u+3cos⁡u4\sin u+3\cos u$ as $5sin⁡(u+α)5\sin(u+\alpha)$ :
$α=tan⁡−1(34)=36.9∘\alpha=\tan^{-1}\left(\frac{3}{4}\right)=36.9^\circ$
The model becomes:

The daily temperature model as a sine curve over 24 hours, showing the midline, amplitude, maximum and the times when θ = 12 °C.

$$
\theta=10+5\sin\left((15t-83.1)^\circ\right)
$$

4. The maximum and minimum temperatures are:

$$
\theta_{\max}=15,\qquad \theta_{\min}=5
$$

5. The maximum occurs when the sine input is 90°:

$$
15t-83.1=90+360n
$$

6. In the interval $0≤t<240\leq t<24$ , use $n = 0$ :

$$
t=\frac{173.1}{15}=11.54\text{ hours}
$$

7. Convert the decimal part to minutes: 0.54 of an hour is about 33 minutes, so the maximum occurs at about 11:33.

To find when the temperature is 12 °C, solve:
$10+5sin⁡((15t−83.1)∘)=1210+5\sin\left((15t-83.1)^\circ\right)=12$
So $sin⁡v=0.4\sin v=0.4$ , where $v = 15 t - 83.1$ and $−83.1∘≤v<276.9∘-83.1^\circ\leq v<276.9^\circ$ :
$156.4∘v=23.6^\circ,\ 156.4^\circ$
Convert back to times:
$15.97t=7.11,\ 15.97$
These are approximately 07:07 and 15:58.

Exam technique

In the exam

Check degrees versus radians before using inverse trig, and make sure your calculator mode matches.
If you substitute $u=kx+αu=kx+\alpha$ , convert the whole interval to an interval for $u$ before solving.
For R-form, expand first, match coefficients carefully, then use the range of sine or cosine for maxima, minima and modelling questions.

Self review

Check yourself

If $0≤x<π0\leq x<\pi$ , what interval should you use for $3 x$ ?
Which version of $cos⁡2x\cos 2x$ is most useful if the equation also contains $sin⁡x\sin x$ ?
How would you find the maximum value of $7+4sin⁡(x−30∘)7+4\sin(x-30^\circ)$ ?

What you'll learn

Solve trigonometric equations over a specified interval without missing solutions.
Derive and use double-angle and triple-angle identities.
Rewrite $asin⁡x+bcos⁡xa\sin x+b\cos x$ as one shifted sine or cosine wave.
Use trigonometry to model repeating situations, such as daily temperature.

1. Angle intervals and inverse trig

Always check your calculator mode before using inverse trig, such as $sin^{-1}$ , $cos^{-1}$ or $tan^{-1}$ .

Definition

Period

The period of a trigonometric function is the angle length after which its values repeat. Sine and cosine have period 360° or $2π2\pi$ ; tangent has period 180° or $π\pi$ .

A reference angle is the acute angle your calculator gives before you use the quadrant information to find all required solutions.

Example

Solving a basic cosine equation

Solve $cos⁡θ=0.25\cos\theta=0.25$ for $0≤θ<2π0\leq\theta<2\pi$ . The interval uses $π\pi$ , so set your calculator to radians and find the reference angle:
$α=cos⁡−1(0.25)=1.318…\alpha=\cos^{-1}(0.25)=1.318\ldots$
Cosine is positive in the first and fourth quadrants, so the two solutions are:

A unit-circle diagram showing the reference angle for cos θ = 0.25 and the two solutions in quadrants I and IV.

$$
\theta=\alpha \quad \text{or} \quad \theta=2\pi-\alpha
$$

3. Round to 3 d.p.:

$$
\theta=1.318,\ 4.965
$$

2. Identities from compound-angle formulae

Definition

Trigonometric identity

A trigonometric identity is an equation involving trig functions that is true for every value for which both sides are defined. The symbol $≡\equiv$ means “identically equal to”.

The key compound-angle formulae are:

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bcos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡Btan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\begin{aligned} \sin(A+B)&=\sin A\cos B+\cos A\sin B\\ \cos(A+B)&=\cos A\cos B-\sin A\sin B\\ \tan(A+B)&=\frac{\tan A+\tan B}{1-\tan A\tan B} \end{aligned}

Key Idea

Double angles

To get a double-angle identity, replace $B$ by $A$ in a compound-angle formula.

Example

Deriving double-angle identities

Put $B = A$ into the sine formula:
$sin⁡2A=sin⁡Acos⁡A+cos⁡Asin⁡A=2sin⁡Acos⁡A\sin 2A=\sin A\cos A+\cos A\sin A=2\sin A\cos A$
Put $B = A$ into the cosine formula:
$cos 2A=\cos^2 A-\sin^2 A$
Use $sin⁡2A+cos⁡2A≡1\sin^2 A+\cos^2 A\equiv1$ to create two alternative forms:
$cos 2A=2\cos^2 A-1=1-2\sin^2 A$
Put $B = A$ into the tangent formula:
$tan⁡2A=tan⁡A+tan⁡A1−tan⁡Atan⁡A=2tan⁡A1−tan⁡2A\tan 2A=\frac{\tan A+\tan A}{1-\tan A\tan A}=\frac{2\tan A}{1-\tan^2 A}$

The same approach gives the triple-angle identities:

sin⁡3θ≡3sin⁡θ−4sin⁡3θ,cos⁡3θ≡4cos⁡3θ−3cos⁡θ\sin 3\theta \equiv 3\sin\theta-4\sin^3\theta,\qquad \cos 3\theta \equiv 4\cos^3\theta-3\cos\theta

3. Using identities to solve equations

Many equations look like cubics, but an identity can turn them into a much simpler trig equation.

Example

Solving a cubic-looking sine equation

Solve $3sin⁡θ−4sin⁡3θ=0.63\sin\theta-4\sin^3\theta=0.6$ for $0∘≤θ≤180∘0^\circ\leq\theta\leq180^\circ$ . Recognise the left-hand side:
$3sin⁡θ−4sin⁡3θ=sin⁡3θ3\sin\theta-4\sin^3\theta=\sin 3\theta$
Rewrite the equation and set $u=3θu=3\theta$ . The interval for $u$ is three times as large:

A mapping diagram showing why θ in 0° to 180° becomes u in 0° to 540° when u = 3θ.

$$
\sin u=0.6,\qquad 0^\circ\leq u\leq540^\circ
$$

3. Find the reference angle:

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

4. Sine is positive in the first and second quadrants, and the interval includes one and a half full turns:

A number-line/angle-cycle diagram showing all four u-values for sin u = 0.6 over 0° ≤ u ≤ 540°.

$$
u=36.9^\circ,\ 143.1^\circ,\ 396.9^\circ,\ 503.1^\circ
$$

5. Divide each value by 3:

$$
\theta=12.3^\circ,\ 47.7^\circ,\ 132.3^\circ,\ 167.7^\circ
$$

Common Mistake

Forgetting the enlarged interval

If $u=3θu=3\theta$ , solve for $u$ over three times the original interval first. If you divide too early, you usually miss solutions from later cycles.

4. Reciprocal functions and identity proofs

Definition

Reciprocal trig functions

The secant and cosecant functions are $sec⁡x=1cos⁡x\sec x=\frac{1}{\cos x}$ and $csc⁡x=1sin⁡x\csc x=\frac{1}{\sin x}$ ; they are undefined where their denominators are zero.

When proving an identity, usually start with the more complicated side and simplify towards the other side. Do not try to “move terms across” as if you were solving an equation.

Example

Proving a reciprocal identity

Prove $sec⁡x−cos⁡x≡sin⁡xtan⁡x\sec x-\cos x\equiv\sin x\tan x$ . Start with the left-hand side and replace $sec⁡x\sec x$ :
$sec⁡x−cos⁡x=1cos⁡x−cos⁡x\sec x-\cos x=\frac{1}{\cos x}-\cos x$
Put the terms over a common denominator:
$1cos⁡x−cos⁡x=1−cos⁡2xcos⁡x\frac{1}{\cos x}-\cos x=\frac{1-\cos^2 x}{\cos x}$
Use $1−cos⁡2x≡sin⁡2x1-\cos^2 x\equiv\sin^2 x$ and $tan⁡x=sin⁡xcos⁡x\tan x=\frac{\sin x}{\cos x}$ :
$1−cos⁡2xcos⁡x=sin⁡2xcos⁡x=sin⁡xtan⁡x\frac{1-\cos^2 x}{\cos x}=\frac{\sin^2 x}{\cos x}=\sin x\tan x$

Common Mistake

Domains still matter

A fractional identity is only valid where both sides are defined. Be especially careful when cancelling factors that could be zero.

5. R-form: one wave instead of two

Definition

R-form

Expand your chosen form before matching coefficients:

Rsin⁡(x+α)=Rcos⁡αsin⁡x+Rsin⁡αcos⁡xRcos⁡(x+α)=Rcos⁡αcos⁡x−Rsin⁡αsin⁡x\begin{aligned} R\sin(x+\alpha)&=R\cos\alpha\sin x+R\sin\alpha\cos x\\ R\cos(x+\alpha)&=R\cos\alpha\cos x-R\sin\alpha\sin x \end{aligned}

Usually, $R=a2+b2R=\sqrt{a^2+b^2}$ , and $α\alpha$ comes from a tangent ratio.

Tip

Choose signs carefully

The sign in $x+αx+\alpha$ or $x−αx-\alpha$ controls whether the sine or cosine coefficient is positive or negative, so always expand first.

Example

R-form, solving and transformations

Express $sin⁡x−3cos⁡x+4\sin x-\sqrt{3}\cos x+4$ in the form $Rsin⁡(x−α)+4R\sin(x-\alpha)+4$ . Expand the target form:
$Rsin⁡(x−α)=Rcos⁡αsin⁡x−Rsin⁡αcos⁡xR\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$
Match coefficients to get $R = 2$ and $α=60∘\alpha=60^\circ$ :

A right-angled coefficient triangle for the R-form expression sin x − √3 cos x = 2 sin(x − 60°).

$$
R\cos\alpha=1,\qquad R\sin\alpha=\sqrt{3}
$$

3. Therefore:

$$
\sin x-\sqrt{3}\cos x+4=2\sin(x-60^\circ)+4
$$

4. To solve $sin⁡x−3cos⁡x+4=5\sin x-\sqrt{3}\cos x+4=5$ for $0∘≤x<360∘0^\circ\leq x<360^\circ$ , set $u=x−60∘u=x-60^\circ$ :

$$
2\sin u+4=5 \Rightarrow \sin u=\frac{1}{2}
$$

5. Since $−60∘≤u<300∘-60^\circ\leq u<300^\circ$ , the solutions are $u=30∘u=30^\circ$ and $u=150∘u=150^\circ$ :

$$
x=90^\circ,\ 210^\circ
$$

The transformed sine curve y = 2 sin(x − 60°) + 4 showing its midline, amplitude, range and phase shift.

6. Trigonometric modelling

Definition

Trig model

A trig model uses sine or cosine to approximate a repeating real-world situation. The midline is the average value, and the amplitude is the distance from the midline to a maximum.

In a daily model, an angle like $15 t$ means the input increases by 15° each hour. Since $36015=24\frac{360}{15}=24$ , the period is 24 hours.

Example

A daily temperature model

A room temperature, $θ\theta$ °C, is modelled for $0≤t<240\leq t<24$ by:
$θ=10+4sin⁡((15t−120)∘)+3cos⁡((15t−120)∘)\theta=10+4\sin\left((15t-120)^\circ\right)+3\cos\left((15t-120)^\circ\right)$
Let $u=(15t−120)∘u=(15t-120)^\circ$ . Write $4sin⁡u+3cos⁡u4\sin u+3\cos u$ as $5sin⁡(u+α)5\sin(u+\alpha)$ :
$α=tan⁡−1(34)=36.9∘\alpha=\tan^{-1}\left(\frac{3}{4}\right)=36.9^\circ$
The model becomes:

The daily temperature model as a sine curve over 24 hours, showing the midline, amplitude, maximum and the times when θ = 12 °C.

$$
\theta=10+5\sin\left((15t-83.1)^\circ\right)
$$

4. The maximum and minimum temperatures are:

$$
\theta_{\max}=15,\qquad \theta_{\min}=5
$$

5. The maximum occurs when the sine input is 90°:

$$
15t-83.1=90+360n
$$

6. In the interval $0≤t<240\leq t<24$ , use $n = 0$ :

$$
t=\frac{173.1}{15}=11.54\text{ hours}
$$

7. Convert the decimal part to minutes: 0.54 of an hour is about 33 minutes, so the maximum occurs at about 11:33.

To find when the temperature is 12 °C, solve:
$10+5sin⁡((15t−83.1)∘)=1210+5\sin\left((15t-83.1)^\circ\right)=12$
So $sin⁡v=0.4\sin v=0.4$ , where $v = 15 t - 83.1$ and $−83.1∘≤v<276.9∘-83.1^\circ\leq v<276.9^\circ$ :
$156.4∘v=23.6^\circ,\ 156.4^\circ$
Convert back to times:
$15.97t=7.11,\ 15.97$
These are approximately 07:07 and 15:58.

Exam technique

In the exam

Check degrees versus radians before using inverse trig, and make sure your calculator mode matches.
If you substitute $u=kx+αu=kx+\alpha$ , convert the whole interval to an interval for $u$ before solving.
For R-form, expand first, match coefficients carefully, then use the range of sine or cosine for maxima, minima and modelling questions.

Self review

Check yourself

If $0≤x<π0\leq x<\pi$ , what interval should you use for $3 x$ ?
Which version of $cos⁡2x\cos 2x$ is most useful if the equation also contains $sin⁡x\sin x$ ?
How would you find the maximum value of $7+4sin⁡(x−30∘)7+4\sin(x-30^\circ)$ ?

What you'll learn

1. Angle intervals and inverse trig

2. Identities from compound-angle formulae

3. Using identities to solve equations

4. Reciprocal functions and identity proofs

5. R-form: one wave instead of two

6. Trigonometric modelling

Recap questions

Review Trigonometry and Modelling by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Trigonometry and Modelling

What you'll learn

1. Angle intervals and inverse trig

2. Identities from compound-angle formulae

3. Using identities to solve equations

4. Reciprocal functions and identity proofs

5. R-form: one wave instead of two

6. Trigonometric modelling

Recap questions

Review Trigonometry and Modelling by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Trigonometry and Modelling

What you'll learn

1. Angle intervals and inverse trig

2. Identities from compound-angle formulae

3. Using identities to solve equations

4. Reciprocal functions and identity proofs

5. R-form: one wave instead of two

6. Trigonometric modelling

Recap questions

Review Trigonometry and Modelling by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Trigonometry and Modelling Revision Guide

Trigonometry and Modelling

What you'll learn

1. Angle intervals and inverse trig

2. Identities from compound-angle formulae

3. Using identities to solve equations

4. Reciprocal functions and identity proofs

5. R-form: one wave instead of two

6. Trigonometric modelling

Recap questions

Review Trigonometry and Modelling by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Trigonometry and Modelling Revision Guide