Trigonometric transformations

This chapter comprehensively covers essential trigonometric transformation formulas, including compound, double, and triple angle identities, alongside product-sum and sum-product relations. Mastery of these transformations is crucial for simplifying complex expressions, solving trigonometric equations, and proving identities, skills frequently assessed in examinations.

---

Chapter Contents

| Topic |

|---|-------| | 1 | Compound angle formulas | | 2 | Double angle formulas | | 3 | Triple angle formulas | | 4 | Product-sum transformations | | 5 | Sum-product transformations |

---

We begin with Compound angle formulas.

Part 1: Compound angle formulas

Compound Angle Formulas

Overview

Compound angle formulas allow us to break expressions like

\sin(A\pm B)

\cos(A\pm B)

, and

\tan(A\pm B)

into simpler trigonometric parts. They are central in trigonometric simplification, equation solving, exact-value computation, and identity proofs. In exam problems, the challenge is not memorizing them only, but using them with the correct signs and in the right direction. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

State the compound angle formulas correctly.

Compute exact trigonometric values using known angles.

Transform products of standard expressions into simpler forms.

Use compound-angle formulas to solve equations and prove identities.

Avoid sign mistakes in plus/minus formulas.

---

Main Formulas

📐 Sine Compound Angle Formula

$\qquad \sin(A+B)=\sin A\cos B+\cos A\sin B$

$\qquad \sin(A-B)=\sin A\cos B-\cos A\sin B$

📐 Cosine Compound Angle Formula

$\qquad \cos(A+B)=\cos A\cos B-\sin A\sin B$

$\qquad \cos(A-B)=\cos A\cos B+\sin A\sin B$

📐 Tangent Compound Angle Formula

$\qquad \tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$

$\qquad \tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$

These formulas are the foundation for many later trigonometric transformations. ---

Exact-Value Use

💡 Most Common Use

To compute an exact value like
$\qquad \sin 75^\circ$
or
$\qquad \tan 15^\circ$ ,
rewrite the angle as a sum or difference of standard angles such as
$\qquad 45^\circ,\ 30^\circ,\ 60^\circ$

---

Minimal Worked Examples

Example 1 Find

\qquad \sin 75^\circ

Write

\qquad 75^\circ = 45^\circ + 30^\circ

Then

\qquad \sin 75^\circ = \sin(45^\circ+30^\circ)

\qquad = \sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ

\qquad = \dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac12

\qquad = \dfrac{\sqrt6+\sqrt2}{4}

\qquad \boxed{\sin 75^\circ = \dfrac{\sqrt6+\sqrt2}{4}}

--- Example 2 Find

\qquad \cos 15^\circ

Write

\qquad 15^\circ = 45^\circ - 30^\circ

Then

\qquad \cos 15^\circ = \cos(45^\circ-30^\circ)

\qquad = \cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ

\qquad = \dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac12

\qquad = \dfrac{\sqrt6+\sqrt2}{4}

\qquad \boxed{\cos 15^\circ = \dfrac{\sqrt6+\sqrt2}{4}}

---

Sign Discipline

⚠️ Most Common Error

Students often mix the signs in cosine formulas.

Remember:

cosine of sum has a minus

cosine of difference has a plus

That is:

\qquad \cos(A+B)=\cos A\cos B-\sin A\sin B

\qquad \cos(A-B)=\cos A\cos B+\sin A\sin B

::: ---

Standard Derived Results

📐 Useful Special Cases

Setting $B=A$ gives:

$\qquad \sin 2A = 2\sin A\cos A$

$\qquad \cos 2A = \cos^2 A - \sin^2 A$

$\qquad \tan 2A = \dfrac{2\tan A}{1-\tan^2 A}$

These are double-angle formulas derived from compound-angle formulas.

---

Common Patterns

💡 Typical Exam Patterns

exact value of an unusual angle

prove an identity using sum/difference formulas

solve an equation after expanding a compound angle

rewrite an expression into a simpler trigonometric form

derive double-angle identities

---

Common Mistakes

⚠️ Avoid These Errors

❌ mixing the signs in $\cos(A\pm B)$

✅ memorize the pair carefully

❌ applying tangent formula when $1-\tan A\tan B=0$

✅ denominator must be nonzero

❌ using wrong standard angle values

✅ recheck

\sin 30^\circ,\cos 30^\circ,\sin 45^\circ,\cos 60^\circ

, etc.

❌ expanding in the wrong direction

✅ sometimes compressing is better than expanding

---

CMI Strategy

💡 How to Attack Compound-Angle Problems

Rewrite the given angle as a sum or difference of standard angles.

Use the correct formula with signs checked twice.

Simplify exactly before approximating anything.

For identities, expand one side only if possible.

In equation problems, reduce to standard trigonometric forms.

---

Practice Questions

:::question type="MCQ" question="The value of

\sin(45^\circ+30^\circ)

is" options=["

\dfrac{\sqrt6+\sqrt2}{4}

","

\dfrac{\sqrt6-\sqrt2}{4}

","

\dfrac{\sqrt3+1}{2}

","

\dfrac{1}{2}

"] answer="A" hint="Use the sine addition formula." solution="Using

\qquad \sin(A+B)=\sin A\cos B+\cos A\sin B

, we get

\qquad \sin 75^\circ = \sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ

$\qquad = \dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac12 = \dfrac{\sqrt6+\sqrt2}{4}$ Hence the correct option is

\boxed{A}

." ::: :::question type="NAT" question="Find the exact value of

\tan 15^\circ

." answer="2-sqrt(3)" hint="Use

\tan(45^\circ-30^\circ)

." solution="Using

\qquad \tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}

, we get $\qquad \tan 15^\circ = \tan(45^\circ-30^\circ) = \dfrac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}}$ Multiply numerator and denominator by

\sqrt3

\qquad \tan 15^\circ = \dfrac{\sqrt3-1}{\sqrt3+1}

Now rationalize: $\qquad \dfrac{\sqrt3-1}{\sqrt3+1}\cdot \dfrac{\sqrt3-1}{\sqrt3-1} = \dfrac{(\sqrt3-1)^2}{3-1} = \dfrac{3-2\sqrt3+1}{2} = 2-\sqrt3$ Hence the answer is

\boxed{2-\sqrt3}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["

\sin(A+B)=\sin A\cos B+\cos A\sin B

","

\cos(A+B)=\cos A\cos B-\sin A\sin B

","

\cos(A-B)=\cos A\cos B+\sin A\sin B

","

\tan(A+B)=\dfrac{\tan A+\tan B}{1+\tan A\tan B}

"] answer="A,B,C" hint="Check the tangent denominator sign carefully." solution="1. True.

True.

False. The correct formula is

\qquad \tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}

Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="Using compound angle formulas, prove that

\sin 2x = 2\sin x \cos x

." answer="Set

A=B=x

in the sine addition formula." hint="Apply

\sin(A+B)

with

A=x

and

B=x

." solution="Using the sine addition formula,

\qquad \sin(A+B)=\sin A\cos B+\cos A\sin B

Set

\qquad A=x,\quad B=x

Then

\qquad \sin(2x)=\sin(x+x)=\sin x\cos x+\cos x\sin x

\qquad \sin 2x = 2\sin x\cos x

Hence the identity is proved." ::: ---

Summary

❗ Key Takeaways for CMI

Compound-angle formulas are essential for exact values, identities, and equations.

Sine sum has a plus; cosine sum has a minus.

Many special-angle values come from $45^\circ\pm 30^\circ$ or $60^\circ\pm 30^\circ$ .

Double-angle formulas are special cases of compound-angle formulas.

Sign accuracy is as important as the formula itself.

---

💡 Next Up

Proceeding to Double angle formulas.

---

Part 2: Double angle formulas

Double Angle Formulas

Overview

Double-angle formulas are among the most important trigonometric transformation tools. They convert expressions involving

2x

into expressions involving

x

, and vice versa. In exam problems, they are used for simplification, solving equations, proving identities, and converting between

\sin x

\cos x

, and

\tan x

efficiently. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Recall the standard double-angle formulas for sine, cosine, and tangent.

Use equivalent forms of $\cos 2x$ strategically.

Apply double-angle identities in simplification and equation solving.

Move between angle-doubling and half-angle style expressions.

Avoid common sign and denominator mistakes.

---

Core Formulas

📐 Main Double-Angle Identities

$\qquad \sin 2x = 2\sin x \cos x$

$\qquad \cos 2x = \cos^2 x - \sin^2 x$

$\qquad \tan 2x = \dfrac{2\tan x}{1-\tan^2 x}$ , provided $1-\tan^2 x\ne 0$

These are the standard double-angle identities and must be memorized exactly. ---

Equivalent Forms of $\cos 2x$

📐 Three Useful Forms of

\cos 2x

Starting from
$\qquad \cos 2x = \cos^2 x - \sin^2 x$

we get:

$\qquad \cos 2x = 2\cos^2 x - 1$

$\qquad \cos 2x = 1 - 2\sin^2 x$

These alternative forms are extremely useful depending on whether the expression contains only

\sin x

or only

\cos x

. ---

Why the Tangent Formula Needs Care

⚠️ Domain Restriction

The formula

$\qquad \tan 2x = \dfrac{2\tan x}{1-\tan^2 x}$

is valid only when:

$\tan x$ is defined

$1-\tan^2 x\ne 0$

So it cannot be used blindly for every

x

---

Standard Uses

💡 Where Double-Angle Formulas Are Used

simplifying expressions like $2\sin x\cos x$

rewriting $\sin^2 x$ or $\cos^2 x$ in terms of $\cos 2x$

solving equations such as $\sin 2x = a$ or $\cos 2x = b$

proving identities

comparing trigonometric expressions

---

Solving Equations with Double Angles

📐 Basic Equation Patterns

$\sin 2x = a$

$\cos 2x = a$

$\tan 2x = a$

The standard method is:

solve for $2x$

then divide the final angle solutions by $2$

This is a very common exam step that students often forget. ---

Minimal Worked Examples

Example 1 If

\sin x=\dfrac{3}{5}

and

x

is acute, find

\sin 2x

. Since

x

is acute,

\qquad \cos x=\dfrac{4}{5}

\qquad \sin 2x = 2\sin x\cos x = 2\cdot \dfrac{3}{5}\cdot \dfrac{4}{5} = \dfrac{24}{25}

--- Example 2 Express

\cos 2x

only in terms of

\sin x

. Use

\qquad \cos 2x = 1-2\sin^2 x

---

Standard Transformations

📐 Very Useful Rearrangements

From the double-angle identities:

$\qquad \sin^2 x = \dfrac{1-\cos 2x}{2}$

$\qquad \cos^2 x = \dfrac{1+\cos 2x}{2}$

$\qquad 1-\cos 2x = 2\sin^2 x$

$\qquad 1+\cos 2x = 2\cos^2 x$

These are often hidden inside simplification and proof problems. ---

Standard Patterns

📐 High-Value Patterns

$2\sin x\cos x \to \sin 2x$

$\cos^2 x-\sin^2 x \to \cos 2x$

$1-2\sin^2 x \to \cos 2x$

$2\cos^2 x-1 \to \cos 2x$

$\dfrac{2\tan x}{1-\tan^2 x} \to \tan 2x$

---

Common Mistakes

⚠️ Avoid These Errors

❌ Writing $\cos 2x=\cos^2 x+\sin^2 x$

✅ Correct:

\cos 2x=\cos^2 x-\sin^2 x

❌ Forgetting the factor 2 in $\sin 2x$

✅

\sin 2x=2\sin x\cos x

❌ Using the tangent formula when $1-\tan^2 x=0$

✅ Check denominator first

❌ Solving $\sin 2x=a$ and forgetting to divide the angle by 2 at the end

---

CMI Strategy

💡 How to Attack These Questions

Look first for one of the standard patterns.

Choose the form of $\cos 2x$ that simplifies the expression most.

In equation solving, solve for $2x$ first.

Track domain restrictions carefully for tangent.

In proofs, convert both sides toward the same pattern rather than expanding everything blindly.

---

Practice Questions

:::question type="MCQ" question="Which of the following is equal to

\sin 2x

?" options=["

\sin^2 x+\cos^2 x

","

2\sin x\cos x

","

\cos^2 x-\sin^2 x

","

1-2\sin^2 x

"] answer="B" hint="Recall the standard double-angle formula for sine." solution="The standard identity is

\qquad \sin 2x = 2\sin x\cos x

Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="If

\sin x=\dfrac{5}{13}

and

x

is acute, find

\sin 2x

." answer="120/169" hint="Find

\cos x

first." solution="Since

x

is acute,

\qquad \cos x=\dfrac{12}{13}

So $\qquad \sin 2x=2\sin x\cos x =2\cdot \dfrac{5}{13}\cdot \dfrac{12}{13} =\dfrac{120}{169}$ Hence the answer is

\boxed{\dfrac{120}{169}}

." ::: :::question type="MSQ" question="Which of the following are equal to

\cos 2x

?" options=["

\cos^2 x-\sin^2 x

","

2\cos^2 x-1

","

1-2\sin^2 x

","

2\sin x\cos x

"] answer="A,B,C" hint="One option is actually

\sin 2x

." solution="The equivalent forms of

\cos 2x

are:

\qquad \cos^2 x-\sin^2 x,\quad 2\cos^2 x-1,\quad 1-2\sin^2 x

The expression

2\sin x\cos x

\sin 2x

, not

\cos 2x

. Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="If

\tan x=\dfrac{1}{2}

, find

\tan 2x

." answer="4/3" hint="Use the tangent double-angle formula." solution="Using

\qquad \tan 2x=\dfrac{2\tan x}{1-\tan^2 x}

we substitute

\tan x=\dfrac{1}{2}

: $\qquad \tan 2x = \dfrac{2\cdot \frac{1}{2}}{1-\left(\frac{1}{2}\right)^2} = \dfrac{1}{1-\frac{1}{4}} = \dfrac{1}{\frac{3}{4}} = \dfrac{4}{3}$ Hence the answer is

\boxed{\dfrac{4}{3}}

." ::: ---

Summary

❗ Key Takeaways for CMI

The core identities are $\sin 2x$ , $\cos 2x$ , and $\tan 2x$ .

$\cos 2x$ has three very useful equivalent forms.

Double-angle formulas are often used to simplify, transform, and solve equations.

The tangent formula needs denominator care.

Most mistakes come from sign errors or missing the factor 2.

---

💡 Next Up

Proceeding to Triple angle formulas.

---

Part 3: Triple angle formulas

Triple Angle Formulas

Overview

Triple-angle formulas connect

\sin 3x

\cos 3x

, and

\tan 3x

with functions of

x

. They are essential in simplification, equation solving, polynomial-trigonometric conversion, and factorisation. In exam problems, these formulas are especially powerful because they reduce higher-angle expressions to algebraic expressions in

\sin x

\cos x

, or

\tan x

. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Recall and use the formulas for $\sin 3x$ , $\cos 3x$ , and $\tan 3x$ .

Convert trigonometric equations into algebraic equations using triple-angle identities.

Factor and simplify expressions involving $\sin 3x$ and $\cos 3x$ .

Solve standard exam problems using triple-angle substitutions.

Track domain restrictions in tangent-based formulas.

---

Core Formulas

📐 Triple Angle Formulas

$\qquad \sin 3x = 3\sin x - 4\sin^3 x$

$\qquad \cos 3x = 4\cos^3 x - 3\cos x$

$\qquad \tan 3x = \dfrac{3\tan x-\tan^3 x}{1-3\tan^2 x}$

These are the standard forms to memorise. ---

How They Are Used

❗ Main Roles

Triple-angle formulas are used to:

rewrite higher-angle expressions in terms of $\sin x$ or $\cos x$

convert trigonometric equations into cubic equations

derive factorisations

simplify expressions involving $\sin 3x$ or $\cos 3x$

solve tangent equations with rational structure

---

Very Useful Rearrangements

📐 Derived Factor Forms

From
$\qquad \sin 3x = 3\sin x - 4\sin^3 x$

we get

$\qquad \sin 3x - \sin x = 2\sin x(1-2\sin^2 x)$

and

$\qquad \sin 3x = \sin x(3-4\sin^2 x)$

From
$\qquad \cos 3x = 4\cos^3 x - 3\cos x$

we get

$\qquad \cos 3x - \cos x = 2\cos x(2\cos^2 x-2)$

and more cleanly

$\qquad \cos 3x - \cos x = 2\cos x(2\cos^2 x-1)-2\cos x$

but usually it is better to keep
$\qquad \cos 3x=4\cos^3 x-3\cos x$
directly.

---

Minimal Worked Examples

Example 1 If

\qquad \cos x=\dfrac{3}{5}

, find

\cos 3x

. Using

\qquad \cos 3x = 4\cos^3 x - 3\cos x

\qquad \cos 3x = 4\left(\dfrac{3}{5}\right)^3 - 3\left(\dfrac{3}{5}\right)

$\qquad = 4\cdot \dfrac{27}{125} - \dfrac{9}{5} = \dfrac{108}{125} - \dfrac{225}{125} = -\dfrac{117}{125}$ So

\qquad \cos 3x=\boxed{-\dfrac{117}{125}}

--- Example 2 Solve

\qquad \sin 3x = \sin x

Using the triple-angle formula:

\qquad 3\sin x - 4\sin^3 x = \sin x

\qquad 2\sin x - 4\sin^3 x = 0

\qquad 2\sin x(1-2\sin^2 x)=0

Hence

$\qquad \sin x = 0$ , or

$\qquad \sin^2 x = \dfrac{1}{2}$

Thus the solution set can be found from elementary angles. ::: ---

Tangent Triple Angle

📐 Important Domain Reminder

The formula

$\qquad \tan 3x = \dfrac{3\tan x-\tan^3 x}{1-3\tan^2 x}$

is valid whenever both sides are defined.

So always check:

denominator

\qquad 1-3\tan^2 x \ne 0

original tangent expressions are defined

---

Common Equation Types

💡 Typical Exam Patterns

Solve

\qquad \sin 3x = c

Solve

\qquad \cos 3x = \cos x

Convert

\qquad 4\cos^3 x - 3\cos x

into a simpler trig expression

\qquad \tan 3x

to solve cubic-looking tangent equations

Factor

\qquad \sin 3x - \sin x

\qquad \cos 3x - \cos x

---

Common Mistakes

⚠️ Avoid These Errors

❌ Writing $\sin 3x = 3\sin x$

✅ The correct formula is

\qquad \sin 3x = 3\sin x - 4\sin^3 x

❌ Writing $\cos 3x = 3\cos x - 4\cos^3 x$

✅ The correct formula is

\qquad \cos 3x = 4\cos^3 x - 3\cos x

❌ Using the tangent formula without checking where it is defined

✅ Watch the denominator and tangent domain

❌ Solving the cubic-looking equation partially and stopping too early

✅ Factor completely and solve all cases

---

CMI Strategy

💡 How to Attack Triple-Angle Questions

Decide whether the expression should be rewritten in $\sin x$ , $\cos x$ , or $\tan x$ .

Use the exact memorised triple-angle formula.

Factor the resulting algebraic expression.

Solve the resulting elementary trig cases carefully.

In tangent problems, always track undefined points.

---

Practice Questions

:::question type="MCQ" question="The expression

4\cos^3 x-3\cos x

is equal to" options=["

\cos 2x

","

\cos 3x

","

\sin 3x

","

\tan 3x

"] answer="B" hint="Recall the standard triple-angle formula for cosine." solution="The standard triple-angle identity is

\qquad \cos 3x = 4\cos^3 x - 3\cos x

Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="If

\sin x=\dfrac{1}{2}

, find

\sin 3x

." answer="1" hint="Use the triple-angle formula for sine." solution="Using

\qquad \sin 3x = 3\sin x - 4\sin^3 x

, $\qquad \sin 3x = 3\cdot \dfrac{1}{2} - 4\left(\dfrac{1}{2}\right)^3 = \dfrac{3}{2}-\dfrac{1}{2}=1$ Hence the answer is

\boxed{1}

." ::: :::question type="MSQ" question="Which of the following are correct?" options=["

\sin 3x = 3\sin x - 4\sin^3 x

","

\cos 3x = 4\cos^3 x - 3\cos x

","

\tan 3x = \dfrac{3\tan x-\tan^3 x}{1-3\tan^2 x}

","

\cos 3x = 3\cos x - 4\cos^3 x

"] answer="A,B,C" hint="One option has the cosine signs reversed." solution="1. True.

True.

False. The correct formula is

\qquad \cos 3x = 4\cos^3 x - 3\cos x

. Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="Solve

\sin 3x=\sin x

for

x\in[0,2\pi)

." answer="

x=0,\dfrac{\pi}{4},\dfrac{3\pi}{4},\pi,\dfrac{5\pi}{4},\dfrac{7\pi}{4}

" hint="Use the triple-angle formula and factor." solution="Use

\qquad \sin 3x = 3\sin x - 4\sin^3 x

Then

\qquad 3\sin x - 4\sin^3 x = \sin x

\qquad 2\sin x - 4\sin^3 x = 0

\qquad 2\sin x(1-2\sin^2 x)=0

Hence either

\qquad \sin x=0

, giving

\qquad x=0,\pi

\qquad 1-2\sin^2 x=0

\qquad \sin^2 x=\dfrac{1}{2}

hence

\qquad \sin x=\pm \dfrac{1}{\sqrt2}

This gives

\qquad x=\dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{5\pi}{4},\dfrac{7\pi}{4}

Therefore the full solution set in

[0,2\pi)

\qquad \boxed{0,\dfrac{\pi}{4},\dfrac{3\pi}{4},\pi,\dfrac{5\pi}{4},\dfrac{7\pi}{4}}

" ::: ---

Summary

❗ Key Takeaways for CMI

Triple-angle formulas reduce higher-angle expressions to algebraic expressions in one trig function.

They are especially useful in equation solving and factorisation.

Sign accuracy matters a lot, especially in $\cos 3x$ .

The tangent formula requires domain care.

Most exam problems become straightforward once the correct formula is applied cleanly.

---

💡 Next Up

Proceeding to Product-sum transformations.

---

Part 4: Product-sum transformations

Product-Sum Transformations

Overview

Product-sum transformations convert products of trigonometric functions into sums or differences. These identities are extremely useful in simplification, equation solving, telescoping expressions, and integration-type manipulations. In exam problems, the real skill is knowing which pair to combine and which identity matches the sign pattern. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Convert products of sines and cosines into sums.

Recognise the correct product-to-sum identity from the sign pattern.

Use these identities to simplify expressions and solve equations.

Detect cancellations after transformation.

Handle mixed-angle products in exam-style manipulations.

---

Core Identities

📐 Product-to-Sum Formulas

For angles $A$ and $B$ ,

$\qquad 2\sin A\cos B = \sin(A+B)+\sin(A-B)$

$\qquad 2\cos A\sin B = \sin(A+B)-\sin(A-B)$

$\qquad 2\cos A\cos B = \cos(A+B)+\cos(A-B)$

$\qquad 2\sin A\sin B = \cos(A-B)-\cos(A+B)$

These are the main formulas for this topic. ---

How to Choose the Right Formula

💡 Pattern Recognition

Look at the product:

$\sin \cdot \cos$ usually becomes a sum of sines

$\cos \cdot \cos$ usually becomes a sum of cosines

$\sin \cdot \sin$ usually becomes a difference of cosines

Also check whether the coefficient

2

is present. If not, create it by multiplying and dividing appropriately.

---

Angle Structure

❗ The Two New Angles

Every product-to-sum formula creates the angles

$\qquad A+B \quad \text{and} \quad A-B$

So after transformation, always simplify these new angles carefully.

Examples:

$2\sin 5x\cos x = \sin 6x+\sin 4x$

$2\cos 7x\cos 3x = \cos 10x+\cos 4x$

$2\sin 4x\sin x = \cos 3x-\cos 5x$

---

Minimal Worked Examples

Example 1 Simplify

\qquad 2\sin 7x\cos x

Using

\qquad 2\sin A\cos B = \sin(A+B)+\sin(A-B)

, we get

\qquad 2\sin 7x\cos x = \sin 8x+\sin 6x

--- Example 2 Simplify

\qquad \sin 5x\sin x

Use

\qquad 2\sin A\sin B = \cos(A-B)-\cos(A+B)

\qquad 2\sin 5x\sin x = \cos 4x-\cos 6x

Hence

\qquad \sin 5x\sin x = \dfrac{\cos 4x-\cos 6x}{2}

---

Why These Formulas Matter

❗ Main Uses

Product-to-sum identities are especially useful when:

products are hard to compare directly

several transformed terms cancel

you need to solve a trigonometric equation

you want to express everything in one trig family

the angles are in arithmetic progression

---

Common Mistakes

⚠️ Avoid These Errors

❌ Mixing the $\cos\cos$ and $\sin\sin$ identities

✅

\cos\cos

gives a sum of cosines, while

\sin\sin

gives a difference of cosines

❌ Forgetting the factor $2$

✅ If your expression has no leading

2

, divide the transformed result by

2

❌ Making sign errors in $A-B$

✅ Compute the angle difference carefully

❌ Replacing a product by a wrong sum family

✅

\sin\cos

leads to sines, not cosines

---

CMI Strategy

💡 How to Attack These Questions

Identify the product type first.

Write the matching identity from memory.

Expand using $A+B$ and $A-B$ .

Look for cancellation or factorisation after transformation.

In equations, reduce everything to standard angles before solving.

---

Practice Questions

:::question type="MCQ" question="The expression

2\cos 5x\cos x

is equal to" options=["

\sin 6x+\sin 4x

","

\cos 6x+\cos 4x

","

\cos 6x-\cos 4x

","

\sin 6x-\sin 4x

"] answer="B" hint="Use the

\cos\cos

product-to-sum identity." solution="Using

\qquad 2\cos A\cos B = \cos(A+B)+\cos(A-B)

, we get

\qquad 2\cos 5x\cos x = \cos 6x+\cos 4x

. Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="Write

\sin 3x\sin x

in the form

\dfrac{\cos(\alpha x)-\cos(\beta x)}{2}

. What is

\alpha+\beta

?" answer="4" hint="Use the

\sin\sin

identity." solution="Using

\qquad 2\sin A\sin B=\cos(A-B)-\cos(A+B)

, we get

\qquad 2\sin 3x\sin x=\cos 2x-\cos 4x

. So

\qquad \sin 3x\sin x=\dfrac{\cos 2x-\cos 4x}{2}

. Hence

\alpha=2

and

\beta=4

, so

\alpha+\beta=\boxed{6}

." ::: :::question type="MSQ" question="Which of the following are correct?" options=["

2\sin A\cos B=\sin(A+B)+\sin(A-B)

","

2\sin A\sin B=\cos(A-B)-\cos(A+B)

","

2\cos A\cos B=\cos(A+B)+\cos(A-B)

","

2\cos A\sin B=\cos(A+B)-\cos(A-B)

"] answer="A,B,C" hint="Check which family appears after transformation." solution="1. True.

True.

False. The correct identity is

\qquad 2\cos A\sin B=\sin(A+B)-\sin(A-B)

. Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="Simplify

\sin 5x\sin x-\sin 3x\sin 2x

as far as possible." answer="

\dfrac{\cos 4x-\cos 5x-\cos 6x+\cos x}{2}

" hint="Apply product-to-sum separately to both products." solution="Use

\qquad 2\sin A\sin B=\cos(A-B)-\cos(A+B)

First,

\qquad 2\sin 5x\sin x=\cos 4x-\cos 6x

\qquad \sin 5x\sin x=\dfrac{\cos 4x-\cos 6x}{2}

Next,

\qquad 2\sin 3x\sin 2x=\cos x-\cos 5x

\qquad \sin 3x\sin 2x=\dfrac{\cos x-\cos 5x}{2}

Therefore, $\qquad \sin 5x\sin x-\sin 3x\sin 2x =\dfrac{\cos 4x-\cos 6x}{2}-\dfrac{\cos x-\cos 5x}{2}$

\qquad = \dfrac{\cos 4x-\cos 6x-\cos x+\cos 5x}{2}

Hence the simplified form is

\qquad \boxed{\dfrac{\cos 4x+\cos 5x-\cos 6x-\cos x}{2}}

." ::: ---

Summary

❗ Key Takeaways for CMI

Product-to-sum identities convert products into sums of standard angles.

The new angles are always $A+B$ and $A-B$ .

Correct sign handling is crucial.

These identities are especially useful in simplification and equation solving.

Most mistakes come from choosing the wrong identity family.

---

💡 Next Up

Proceeding to Sum-product transformations.

---

Part 5: Sum-product transformations

Sum-Product Transformations

Overview

Sum-product transformations are the reverse of product-sum identities. They convert sums or differences of sines and cosines into products. These identities are extremely important in simplification, solving equations, studying zeros, and extracting common factors from angle-progressions. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Convert sums and differences of trigonometric functions into products.

Use these identities to factor trigonometric expressions.

Solve equations by turning sums into products.

Simplify trigonometric sums in arithmetic progression.

Track half-angle expressions carefully.

---

Core Identities

📐 Sum-to-Product Formulas

For angles $A$ and $B$ ,

$\qquad \sin A+\sin B = 2\sin\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)$

$\qquad \sin A-\sin B = 2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)$

$\qquad \cos A+\cos B = 2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)$

$\qquad \cos A-\cos B = -2\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)$

Equivalent form of the last identity:

\qquad \cos A-\cos B = 2\sin\left(\dfrac{B+A}{2}\right)\sin\left(\dfrac{B-A}{2}\right)

Both are the same identity written differently. ::: ---

Main Structure

❗ Half-Sum and Half-Difference

Every sum-to-product identity uses:

$\qquad \dfrac{A+B}{2}$

$\qquad \dfrac{A-B}{2}$

So the first step after spotting the formula is to compute these two angles accurately.

Examples:

$\sin 7x+\sin x = 2\sin 4x\cos 3x$

$\cos 5x+\cos x = 2\cos 3x\cos 2x$

$\cos 7x-\cos x = -2\sin 4x\sin 3x$

---

Why These Identities Matter

❗ Main Uses

Sum-to-product transformations are useful when:

you want to factor a trigonometric sum

you need to solve a trigonometric equation

a sum of terms in arithmetic progression appears

zeros are easier to read in product form

you want to compare two expressions term-by-term

---

Minimal Worked Examples

Example 1 Simplify

\qquad \sin 5x+\sin x

Using

\qquad \sin A+\sin B = 2\sin\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)

, we get

\qquad \sin 5x+\sin x = 2\sin 3x\cos 2x

--- Example 2 Simplify

\qquad \cos 7x-\cos x

Using

\qquad \cos A-\cos B = -2\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)

, we get

\qquad \cos 7x-\cos x = -2\sin 4x\sin 3x

---

Solving Equations by Factorisation

💡 Very Important Use

If an equation has a form like

$\qquad \sin A+\sin B=0$

convert it into product form and solve by zero-product logic.

Example:
$\qquad \sin 5x+\sin x=0$

becomes

$\qquad 2\sin 3x\cos 2x=0$

So solve:

$\qquad \sin 3x=0$ , or

$\qquad \cos 2x=0$

This is often much faster than direct manipulation. ::: ---

Common Mistakes

⚠️ Avoid These Errors

❌ Forgetting the factor $2$

✅ All standard sum-to-product formulas have a leading

2

❌ Using $A+B$ instead of $(A+B)/2$

✅ The transformed angles are half-sum and half-difference

❌ Mixing the signs in $\cos A-\cos B$

✅ Check the minus sign carefully

❌ Turning a sum into the wrong trig family

✅

\sin+\sin

gives

\sin\cos

, while

\cos+\cos

gives

\cos\cos

---

CMI Strategy

💡 How to Attack These Questions

Spot whether the expression is a sum or difference.

Match the identity by trig family and sign.

Compute the half-sum and half-difference carefully.

Factor the expression completely.

If solving an equation, split into factor cases.

---

Practice Questions

:::question type="MCQ" question="The expression

\cos 5x+\cos x

is equal to" options=["

2\cos 3x\cos 2x

","

2\sin 3x\sin 2x

","

2\sin 3x\cos 2x

","

2\cos 2x\sin 3x

"] answer="A" hint="Use the

\cos+\cos

identity." solution="Using

\qquad \cos A+\cos B = 2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)

, we get

\qquad \cos 5x+\cos x = 2\cos 3x\cos 2x

. Hence the correct option is

\boxed{A}

." ::: :::question type="NAT" question="Write

\sin 7x+\sin x

in the form

2\sin(\alpha x)\cos(\beta x)

. What is

\alpha+\beta

?" answer="7" hint="Use half-sum and half-difference." solution="Using

\qquad \sin A+\sin B = 2\sin\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)

, we get

\qquad \sin 7x+\sin x = 2\sin 4x\cos 3x

. So

\alpha=4

and

\beta=3

, hence

\alpha+\beta=\boxed{7}

." ::: :::question type="MSQ" question="Which of the following are correct?" options=["

\sin A+\sin B = 2\sin\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)

","

\cos A+\cos B = 2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)

","

\sin A-\sin B = 2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)

","

\cos A-\cos B = 2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)

"] answer="A,B,C" hint="Check the sign and function family carefully." solution="1. True.

True.

False. The correct form is

\qquad \cos A-\cos B = -2\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)

. Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="Factorise

\cos x+\cos 3x+\cos 5x+\cos 7x

as a product." answer="

4\cos 4x\cos 2x\cos x

" hint="Group symmetrically and use sum-to-product twice." solution="Group the terms as

\qquad (\cos x+\cos 7x)+(\cos 3x+\cos 5x)

Now use

\qquad \cos A+\cos B = 2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)

\qquad \cos x+\cos 7x = 2\cos 4x\cos 3x

and

\qquad \cos 3x+\cos 5x = 2\cos 4x\cos x

Thus $\qquad \cos x+\cos 3x+\cos 5x+\cos 7x =2\cos 4x(\cos 3x+\cos x)$ Again,

\qquad \cos 3x+\cos x = 2\cos 2x\cos x

Hence $\qquad \cos x+\cos 3x+\cos 5x+\cos 7x = 2\cos 4x \cdot 2\cos 2x\cos x$

\qquad = \boxed{4\cos 4x\cos 2x\cos x}

" ::: ---

Summary

❗ Key Takeaways for CMI

Sum-to-product identities turn sums into factors.

The transformed angles are half-sum and half-difference.

These formulas are especially useful for solving equations.

Grouping terms cleverly is often the key.

Sign discipline is essential, especially in $\cos A-\cos B$ .

---

Chapter Summary

❗ Trigonometric transformations — Key Points

Mastery of Fundamental Identities: Thorough understanding and recall of compound angle and double angle formulas are essential as they form the basis for most transformations.

Strategic Application: Develop proficiency in identifying when and how to apply specific transformations (e.g., product-to-sum, sum-to-product, or power reduction) to simplify expressions or solve equations efficiently.

Exact Value Derivations: Utilize these transformations to derive exact trigonometric values for non-standard angles (e.g., $15^\circ$ , $75^\circ$ , $105^\circ$ ) by expressing them as sums or differences of standard angles.

Equation Solving: Transformations are critical tools for converting complex trigonometric equations into simpler forms that can be solved using basic trigonometric principles.

Proof Techniques: Leverage identities to prove other trigonometric relations and establish equivalences, a common requirement in CMI problems.

Interconversion: Practice converting between different forms (e.g., sum-to-product and product-to-sum) to adapt to various problem requirements.

---

Chapter Review Questions

:::question type="MCQ" question="The exact value of $\sin 15^\circ$ is:" options=[" $\frac{\sqrt{6}-\sqrt{2}}{4}$ "," $\frac{\sqrt{6}+\sqrt{2}}{4}$ "," $\frac{\sqrt{3}-1}{2\sqrt{2}}$ "," $\frac{\sqrt{3}+1}{2\sqrt{2}}$ "] answer=" $\frac{\sqrt{6}-\sqrt{2}}{4}$ " hint="Consider $\sin(45^\circ - 30^\circ)$ and apply the compound angle formula." solution="Using the compound angle formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$ :
$\sin 15^\circ = \sin(45^\circ - 30^\circ)$
$= \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$
$= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)$
$= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$
$= \frac{\sqrt{6}-\sqrt{2}}{4}$ "
:::

:::question type="NAT" question="If $\sin x + \sin y = \frac{1}{2}$ and $\cos x + \cos y = \frac{\sqrt{3}}{2}$ , what is the value of $\cos(x-y)$ ?" answer="-0.5" hint="Square both given equations and add them. Recall $\cos(x-y) = \cos x \cos y + \sin x \sin y$ ." solution="Given:

\sin x + \sin y = \frac{1}{2}

\cos x + \cos y = \frac{\sqrt{3}}{2}

Squaring equation (1):
$(\sin x + \sin y)^2 = \sin^2 x + \sin^2 y + 2 \sin x \sin y = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$ (Equation 3)

Squaring equation (2):
$(\cos x + \cos y)^2 = \cos^2 x + \cos^2 y + 2 \cos x \cos y = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$ (Equation 4)

Adding Equation 3 and Equation 4:
$(\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) + 2(\sin x \sin y + \cos x \cos y) = \frac{1}{4} + \frac{3}{4}$
$1 + 1 + 2(\cos x \cos y + \sin x \sin y) = 1$
$2 + 2 \cos(x-y) = 1$
$2 \cos(x-y) = 1 - 2$
$2 \cos(x-y) = -1$
$\cos(x-y) = -\frac{1}{2} = -0.5$ "
:::

:::question type="MCQ" question="Which of the following expressions is equivalent to $\frac{\sin 3A}{\sin A} - \frac{\cos 3A}{\cos A}$ ?" options=[" $1$ "," $2$ "," $2\cos 2A$ "," $2\sin 2A$ "] answer=" $2$ " hint="Combine the fractions using a common denominator and apply a compound angle formula in the numerator." solution="Combine the fractions:

\frac{\sin 3A}{\sin A} - \frac{\cos 3A}{\cos A} = \frac{\sin 3A \cos A - \cos 3A \sin A}{\sin A \cos A}

The numerator is in the form

\sin X \cos Y - \cos X \sin Y = \sin(X-Y)

.
Here,

X=3A

and

Y=A

\frac{\sin(3A - A)}{\sin A \cos A} = \frac{\sin 2A}{\sin A \cos A}

Now, apply the double angle formula for

\sin 2A = 2 \sin A \cos A

\frac{2 \sin A \cos A}{\sin A \cos A} = 2

"
:::

---

What's Next?

💡 Continue Your CMI Journey

Having mastered trigonometric transformations, you are now well-equipped to tackle more advanced topics. The principles learned here are fundamental for Solving Trigonometric Equations, where identities are often employed to simplify expressions before finding general solutions. Furthermore, these transformations form the bedrock for understanding Complex Numbers, particularly in deriving powers and roots using De Moivre's Theorem, and in exploring the relationships between trigonometric and exponential forms. A solid grasp of this chapter will significantly aid your progress in these interconnected areas.

Trigonometric transformations

Trigonometric transformations

Chapter Contents

| Topic |

Part 1: Compound angle formulas

Compound Angle Formulas

Overview

Learning Objectives

Main Formulas

Exact-Value Use

Minimal Worked Examples

Sign Discipline

Standard Derived Results

Common Patterns

Common Mistakes

CMI Strategy

Practice Questions

Summary

Part 2: Double angle formulas

Double Angle Formulas

Overview

Learning Objectives

Core Formulas

Equivalent Forms of cos⁡2x\cos 2xcos2x

Why the Tangent Formula Needs Care

Standard Uses

Solving Equations with Double Angles

Minimal Worked Examples

Standard Transformations

Standard Patterns

Common Mistakes

CMI Strategy

Practice Questions

Summary

Part 3: Triple angle formulas

Triple Angle Formulas

Overview

Learning Objectives

Core Formulas

How They Are Used

Very Useful Rearrangements

Minimal Worked Examples

Tangent Triple Angle

Common Equation Types

Common Mistakes

CMI Strategy

Practice Questions

Summary

Part 4: Product-sum transformations

Product-Sum Transformations

Overview

Learning Objectives

Core Identities

How to Choose the Right Formula

Angle Structure

Minimal Worked Examples

Why These Formulas Matter

Common Mistakes

CMI Strategy

Practice Questions

Summary

Part 5: Sum-product transformations

Sum-Product Transformations

Overview

Learning Objectives

Core Identities

Main Structure

Why These Identities Matter

Minimal Worked Examples

Solving Equations by Factorisation

Common Mistakes

CMI Strategy

Practice Questions

Summary

Chapter Summary

Chapter Review Questions

What's Next?

🎯 Key Points to Remember

Related Topics in Trigonometry and Complex Numbers

Geometry of complex numbers

Equivalent Forms of $\cos 2x$