NDA Maths · Trigonometric Identities

Compound Angles — sin/cos/tan(A ± B)

The sin(A±B), cos(A±B), tan(A±B) formulas — the base identity that double-angle, product-to-sum, and most manipulation are built on. The skill is spotting when an expression is a disguised compound angle.

All Trigonometric Identities notes NDA Maths strategy

Why this matters

This is the highest-leverage subtopic in the chapter: the NDA hides compound angle behind expressions like (cos 17° − sin 17°)/(cos 17° + sin 17°), √3 cos 10° − sin 10°, or tan α, tan β as roots of a quadratic. Each collapses to one application of a compound-angle formula.

Concept 1 of 5

sin(A ± B) and cos(A ± B)

Intuition

The two workhorses. Sine of a sum keeps the function order and adds; cosine of a sum swaps to the opposite sign. Get the sign rule right and these unlock the rest.

Definition

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ .
$\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ (note the flipped sign).

Sum and difference

\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,\quad \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B

Worked example

Find the exact value of

\cos 75°

$\cos 75°=\cos(45°+30°)=\cos 45°\cos 30°-\sin 45°\sin 30°$ .
$=\tfrac{1}{\sqrt2}\cdot\tfrac{\sqrt3}{2}-\tfrac{1}{\sqrt2}\cdot\tfrac12=\dfrac{\sqrt3-1}{2\sqrt2}$ .

Answer:

\cos 75°=\dfrac{\sqrt3-1}{2\sqrt2}

Practice this conceptself-check · 4 quick reps

Try it yourself

Find

\sin 105° + \cos 105°

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\sin(A+B)=?$
2.
$\cos(A+B)=?$
3.
$\sin 75°$ ?
4.
$\cos(A-B)-\cos(A+B)=?$

From the bank · past-year question

Example 1Trigonometric IdentitiesMODERATE

What is

\sin 105° + \cos 105°

equal to?

[Q38 · Apr · 2018]

Drill 3 more on sin(a ± b) and cos(a ± b)

Concept 2 of 5

tan(A ± B) and cot(A ± B)

Intuition

The tangent sum formula is the engine behind every "tan of a non-standard angle" question — write the angle as a sum or difference of known angles.

Definition

$\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}$ . A useful corollary: $\tan(45°+\theta)=\dfrac{1+\tan\theta}{1-\tan\theta}$ . For cotangent, $\cot(A\pm B)=\dfrac{\cot A\cot B\mp 1}{\cot B\pm\cot A}$ .

Tangent of a sum/difference

\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}

Worked example

Find

\tan 75°

using the tangent formula.

$\tan 75°=\tan(45°+30°)=\dfrac{\tan 45°+\tan 30°}{1-\tan 45°\tan 30°}=\dfrac{1+\tfrac{1}{\sqrt3}}{1-\tfrac{1}{\sqrt3}}$ .
Multiply by $\tfrac{\sqrt3}{\sqrt3}$ : $\dfrac{\sqrt3+1}{\sqrt3-1}=2+\sqrt3$ .

Answer:

\tan 75°=2+\sqrt3

Practice this conceptself-check · 4 quick reps

Try it yourself

Express

\tan 54°

in terms of

\tan 9°

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\tan(A+B)=?$
2.
$\tan(45°+\theta)=?$
3.
$\tan 75°$ ?
4.
If $\tan A\tan B=1$ , $\tan(A+B)$ is?

From the bank · past-year question

Example 2Trigonometric IdentitiesMODERATE

\tan 54°

can be expressed as

[Q32 · Apr · 2019]

Drill 2 more on tan(a ± b) and cot(a ± b)

Concept 3 of 5

Spotting a disguised compound angle

Intuition

The marks are in recognition, not the formula. A ratio like (cos−sin)/(cos+sin), a combination √3 cos θ − sin θ, or a difference of squares of sines — each is one compound-angle step in disguise.

Definition

Common disguises:

$\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=\tan(45°-\theta)$ (divide by $\cos\theta$ ).
$a\cos\theta+b\sin\theta=R\cos(\theta-\varphi)$ with $R=\sqrt{a^2+b^2}$ (e.g. $\sqrt3\cos\theta-\sin\theta=2\cos(\theta+30°)$ ).
$\sin^2 A-\sin^2 B=\sin(A+B)\sin(A-B)$ ; $\cos^2 A-\sin^2 B=\cos(A+B)\cos(A-B)$ .
Complementary pairs: $\tan\theta\cdot\tan(90°-\theta)=1$ .

Worked example

Simplify

\sin^2\!\left(\tfrac{\pi}{4}+\theta\right)-\sin^2\!\left(\tfrac{\pi}{4}-\theta\right)

Use $\sin^2 A-\sin^2 B=\sin(A+B)\sin(A-B)$ with $A=\tfrac\pi4+\theta,\ B=\tfrac\pi4-\theta$ .
$A+B=\tfrac\pi2,\ A-B=2\theta\Rightarrow\sin\tfrac\pi2\,\sin 2\theta=\sin 2\theta$ .

Answer:

\sin 2\theta

Practice this conceptself-check · 4 quick reps

Try it yourself

Express

\sqrt3\sin\theta+\cos\theta

as a single sine.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=?$
2.
$\sqrt3\cos\theta-\sin\theta=?$
3.
$\sin^2 A-\sin^2 B=?$
4.
$\tan 1°\tan 2°\cdots\tan 89°=?$

From the bank · past-year question

Example 3Trigonometric IdentitiesEASY

What is

\dfrac{\cos17°-\sin17°}{\cos17°+\sin17°}

equal to?

[Q30 · Sep · 2024]

Drill 13 more on spotting a disguised compound angle

Concept 4 of 5

Conditional ratio manipulation (a sin²+b cos²=c, etc.)

Intuition

A second family of compound-subtopic questions is pure algebraic manipulation of ratios: given a constraint like

a\sin^2 x+b\cos^2 x=c

, divide through to extract

\tan^2 x

; or simplify a fearsome-looking product of

(1\pm\sec)(1\pm\csc)

expressions.

Definition

Standard moves: **divide by $\cos^2 x$ ** to turn $a\sin^2 x+b\cos^2 x=c$ into $a\tan^2 x+b=c\sec^2 x=c(1+\tan^2 x)$ , then solve for $\tan^2 x$ . Convert all functions to $\sin/\cos$ over a common denominator when a product looks unmanageable. $\dfrac{1+\tan^2\theta}{1+\cot^2\theta}=\tan^2\theta$ .

Worked example

3\sin^2 x+\cos^2 x=2

, find

\tan^2 x

Divide by $\cos^2 x$ : $3\tan^2 x+1=2\sec^2 x=2(1+\tan^2 x)$ .
$3\tan^2 x+1=2+2\tan^2 x\Rightarrow\tan^2 x=1$ .

Answer:

\tan^2 x=1

Practice this conceptself-check · 4 quick reps

Try it yourself

Simplify

\dfrac{1+\tan^2\theta}{1+\cot^2\theta}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Turn $a\sin^2x+b\cos^2x=c$ into a $\tan$ equation by?
2.
$\dfrac{1+\tan^2\theta}{1+\cot^2\theta}=?$
3.
$\sqrt{\sec^2\alpha-1}$ for acute $\alpha$ ?
4.
$2\sin^2x+\cos^2x=\tfrac{3}{2}\Rightarrow\sin^2x=?$

From the bank · past-year question

Example 4Trigonometric IdentitiesMODERATE

What is

\frac{1+\tan^2\theta}{1+\cot^2\theta} - \left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2

equal to?

[Q25 · Apr · 2021]

Drill 8 more on conditional ratio manipulation (a sin²+b cos²=c, etc.)

Concept 5 of 5

Roots, componendo, and conditional compound angles

Intuition

When

\tan\alpha,\tan\beta

are the roots of a quadratic, Vieta's formulas hand you

\tan\alpha+\tan\beta

and

\tan\alpha\tan\beta

— exactly the ingredients of

\tan(\alpha+\beta)

. For ratio equations, componendo-dividendo collapses sin(x+y)/sin(x−y) cleanly.

Definition

Roots → compound: if $\tan\alpha,\tan\beta$ are roots of $x^2-bx+c=0$ , then $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\dfrac{b}{1-c}$ .
Componendo-dividendo: from $\dfrac{\sin(x+y)}{\sin(x-y)}=\dfrac{a}{b}$ , apply componendo-dividendo and expand to get $\dfrac{\tan x}{\tan y}=\dfrac{a+b}{a-b}$ .

Worked example

\tan\alpha,\tan\beta

are roots of

x^2-6x+8=0

, find

\tan(\alpha+\beta)

Vieta: $\tan\alpha+\tan\beta=6$ , $\tan\alpha\tan\beta=8$ .
$\tan(\alpha+\beta)=\dfrac{6}{1-8}=-\tfrac{6}{7}$ .

Answer:

\tan(\alpha+\beta)=-\tfrac67

Practice this conceptself-check · 4 quick reps

Try it yourself

\dfrac{\sin(x+y)}{\sin(x-y)}=\dfrac{a+b}{a-b}

, find

\dfrac{\tan x}{\tan y}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\tan\alpha,\tan\beta$ roots of $x^2-bx+c$ : $\tan(\alpha+\beta)$ ?
2.
Roots give which two quantities directly?
3.
Tool for $\sin(x+y)/\sin(x-y)$ equations?
4.
$\tan\alpha+\tan\beta=6,\tan\alpha\tan\beta=8$ : $\tan(\alpha+\beta)$ ?

From the bank · past-year question

Example 5Trigonometric IdentitiesMODERATE

\tan\alpha

and

\tan\beta

are the roots of the equation

x^2-6x+8=0

, then what is the value of

\cos(2\alpha+2\beta)

[Q48 · Apr · 2024]

Drill 7 more on roots, componendo, and conditional compound angles

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

sin(A ± B) and cos(A ± B)
Sum and difference
$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,\quad \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$
tan(A ± B) and cot(A ± B)
Tangent of a sum/difference
$\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Trigonometric IdentitiesHARD

\tan\alpha=\frac{1}{7},\ \sin\beta=\frac{1}{\sqrt{10}};\ 0<\alpha,\beta<\frac{\pi}{2}

, then what is the value of

\cos(\alpha+2\beta)

[Q30 · Sep · 2023]

Example 2Trigonometric IdentitiesMODERATE

\tan A - \tan B = x

and

\cot B - \cot A = y

, then what is the value of

\cot(A - B)

[Q63 · Sep · 2019]

Example 3Trigonometric IdentitiesHARD

What is the value of

\dfrac{\sin 34° \cos 236° - \sin 56° \sin 124°}{\cos 28° \cos 88° + \cos 178° \sin 208°}

[Q31 · Apr · 2019]

Example 4Trigonometric IdentitiesMODERATE

f(\theta)=\dfrac{1}{1+\tan\theta}

and

\alpha+\beta=\dfrac{5\pi}{4}

, then what is the value of

f(\alpha)\cdot f(\beta)

[Q47 · Apr · 2024]

Example 5Trigonometric IdentitiesHARD

\sec(\theta-\alpha)

\sec\theta

and

\sec(\theta+\alpha)

are in AP, where

\cos\alpha \neq 1

, then what is the value of

\sin^2\theta + \cos\alpha

[Q37 · Sep · 2018]

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