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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 notesNDA 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±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B.
  • cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B (note the flipped sign).

Sum and difference

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B,cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\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°.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Trigonometric IdentitiesMODERATE
What is sin⁡105°+cos⁡105°\sin 105° + \cos 105° equal to?

[Q38 · Apr · 2018]

sin⁡(A+B)≠sin⁡A+sin⁡B\sin(A+B)\neq\sin A+\sin B — sine is NOT linear

The single most common error in the chapter: treating sin⁡(A+B)\sin(A+B) as sin⁡A+sin⁡B\sin A+\sin B (or cos⁡(A+B)\cos(A+B) as cos⁡A+cos⁡B\cos A+\cos B). A quick disproof: sin⁡(30°+60°)=sin⁡90°=1\sin(30°+60°)=\sin 90°=1, but sin⁡30°+sin⁡60°=12+32≈1.37\sin 30°+\sin 60°=\tfrac12+\tfrac{\sqrt3}{2}\approx1.37. You must expand: sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B)=\sin A\cos B+\cos A\sin B.
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±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}. A useful corollary: tan⁡(45°+θ)=1+tan⁡θ1−tan⁡θ\tan(45°+\theta)=\dfrac{1+\tan\theta}{1-\tan\theta}. For cotangent, cot⁡(A±B)=cot⁡Acot⁡B∓1cot⁡B±cot⁡A\cot(A\pm B)=\dfrac{\cot A\cot B\mp 1}{\cot B\pm\cot A}.

Tangent of a sum/difference

tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}

Worked example

Find tan⁡75°\tan 75° using the tangent formula.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Trigonometric IdentitiesMODERATE
tan⁡54°\tan 54° can be expressed as

[Q32 · Apr · 2019]

The tan⁡(A+B)\tan(A+B) denominator is 1−tan⁡Atan⁡B1-\tan A\tan B, and the signs OPPOSE

Two errors cluster here: forgetting the denominator entirely (writing tan⁡(A+B)=tan⁡A+tan⁡B\tan(A+B)=\tan A+\tan B), and getting the denominator sign wrong. The rule is tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} — the bottom sign is opposite the top, so a *sum* on top means a *minus* on the bottom. (And tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}.)
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:

  • cos⁡θ−sin⁡θcos⁡θ+sin⁡θ=tan⁡(45°−θ)\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=\tan(45°-\theta) (divide by cos⁡θ\cos\theta).
  • acos⁡θ+bsin⁡θ=Rcos⁡(θ−φ)a\cos\theta+b\sin\theta=R\cos(\theta-\varphi) with R=a2+b2R=\sqrt{a^2+b^2} (e.g. 3cos⁡θ−sin⁡θ=2cos⁡(θ+30°)\sqrt3\cos\theta-\sin\theta=2\cos(\theta+30°)).
  • sin⁡2A−sin⁡2B=sin⁡(A+B)sin⁡(A−B)\sin^2 A-\sin^2 B=\sin(A+B)\sin(A-B); cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B)\cos^2 A-\sin^2 B=\cos(A+B)\cos(A-B).
  • Complementary pairs: tan⁡θ⋅tan⁡(90°−θ)=1\tan\theta\cdot\tan(90°-\theta)=1.

Worked example

Simplify sin⁡2 ⁣(π4+θ)−sin⁡2 ⁣(π4−θ)\sin^2\!\left(\tfrac{\pi}{4}+\theta\right)-\sin^2\!\left(\tfrac{\pi}{4}-\theta\right).
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Trigonometric IdentitiesEASY
What is cos⁡17°−sin⁡17°cos⁡17°+sin⁡17°\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 asin⁡2x+bcos⁡2x=ca\sin^2 x+b\cos^2 x=c, divide through to extract tan⁡2x\tan^2 x; or simplify a fearsome-looking product of (1±sec⁡)(1±csc⁡)(1\pm\sec)(1\pm\csc) expressions.

Definition

Standard moves: **divide by cos⁡2x\cos^2 x** to turn asin⁡2x+bcos⁡2x=ca\sin^2 x+b\cos^2 x=c into atan⁡2x+b=csec⁡2x=c(1+tan⁡2x)a\tan^2 x+b=c\sec^2 x=c(1+\tan^2 x), then solve for tan⁡2x\tan^2 x. Convert all functions to sin⁡/cos⁡\sin/\cos over a common denominator when a product looks unmanageable. 1+tan⁡2θ1+cot⁡2θ=tan⁡2θ\dfrac{1+\tan^2\theta}{1+\cot^2\theta}=\tan^2\theta.

Worked example

If 3sin⁡2x+cos⁡2x=23\sin^2 x+\cos^2 x=2, find tan⁡2x\tan^2 x.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Trigonometric IdentitiesMODERATE
What is 1+tan⁡2θ1+cot⁡2θ−(1−tan⁡θ1−cot⁡θ)2\frac{1+\tan^2\theta}{1+\cot^2\theta} - \left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2 equal to?

[Q25 · Apr · 2021]

cos⁡(A+B)cos⁡(A−B)=cos⁡2A−sin⁡2B\cos(A+B)\cos(A-B)=\cos^2 A-\sin^2 B, not cos⁡2A−cos⁡2B\cos^2 A-\cos^2 B

When collapsing a product like cos⁡(A+B)cos⁡(A−B)\cos(A+B)\cos(A-B), students misremember the result as cos⁡2A−cos⁡2B\cos^2 A-\cos^2 B. Expanding gives cos⁡2A−sin⁡2B\cos^2 A-\sin^2 B (equivalently cos⁡2B−sin⁡2A\cos^2 B-\sin^2 A). The companion is sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B\sin(A+B)\sin(A-B)=\sin^2 A-\sin^2 B. **The leftover square is sin⁡2B\sin^2 B, a different function from the cos⁡2A\cos^2 A it sits beside.**
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⁡α,tan⁡β\tan\alpha,\tan\beta are the roots of a quadratic, Vieta's formulas hand you tan⁡α+tan⁡β\tan\alpha+\tan\beta and tan⁡αtan⁡β\tan\alpha\tan\beta — exactly the ingredients of tan⁡(α+β)\tan(\alpha+\beta). For ratio equations, componendo-dividendo collapses sin(x+y)/sin(x−y) cleanly.

Definition

  • Roots → compound: if tan⁡α,tan⁡β\tan\alpha,\tan\beta are roots of x2−bx+c=0x^2-bx+c=0, then tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β=b1−c\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\dfrac{b}{1-c}.
  • Componendo-dividendo: from sin⁡(x+y)sin⁡(x−y)=ab\dfrac{\sin(x+y)}{\sin(x-y)}=\dfrac{a}{b}, apply componendo-dividendo and expand to get tan⁡xtan⁡y=a+ba−b\dfrac{\tan x}{\tan y}=\dfrac{a+b}{a-b}.

Worked example

If tan⁡α,tan⁡β\tan\alpha,\tan\beta are roots of x2−6x+8=0x^2-6x+8=0, find tan⁡(α+β)\tan(\alpha+\beta).
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Trigonometric IdentitiesMODERATE
If tan⁡α\tan\alpha and tan⁡β\tan\beta are the roots of the equation x2−6x+8=0x^2-6x+8=0, then what is the value of cos⁡(2α+2β)\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±B)=sin⁡Acos⁡B±cos⁡Asin⁡B,cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\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±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}

Watch out for (3)

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