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CDS Mathematics · Trigonometric Ratios and Identities

Compound & Multiple Angles

The addition formulas for sin(A ± B), cos(A ± B) and tan(A ± B), and the double-angle forms that follow from them.

All Trigonometric Ratios and Identities notesCDS Mathematics strategy

Why this matters

Only eight PYQs, because CDS keeps compound angles light: recognise an expansion and collapse it to a standard angle, or use 2 sin θ cos θ = sin 2θ. They are worth knowing mainly because the maximum-and-minimum page leans on sin 2θ.

Concept 1 of 2: The addition formulas

The paper uses these backwards far more than forwards: it prints sin⁡46∘cos⁡44∘+cos⁡46∘sin⁡44∘\sin 46^\circ\cos 44^\circ + \cos 46^\circ\sin 44^\circ and expects you to see sin⁡90∘\sin 90^\circ. Recognising the pattern is the whole skill.

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 sign flips)
  • 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}

When a stem supplies tan⁡(α+β)\tan(\alpha + \beta) and tan⁡(α−β)\tan(\alpha - \beta) as standard values, find α+β\alpha + \beta and α−β\alpha - \beta first and solve for the angles.

Addition formulas

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B,cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B, \quad \cos(A+B) = \cos A\cos B - \sin A\sin B

Worked example

Evaluate cos⁡70∘cos⁡10∘+sin⁡70∘sin⁡10∘\cos 70^\circ\cos 10^\circ + \sin 70^\circ\sin 10^\circ.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (II) 2019 — Elementary Mathematics · Q52Easy

Example 1 · Trigonometric Ratios and Identities · Compound and Multiple Angles

What is the value of sin⁡46∘cos⁡44∘+cos⁡46∘sin⁡44∘\sin 46^\circ \cos 44^\circ + \cos 46^\circ \sin 44^\circ ?

cos(A + B) has a minus sign

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A\cos B - \sin A\sin B. The sign in the cosine formula is the opposite of the sign inside the bracket, and writing it with a plus reproduces cos⁡(A−B)\cos(A - B) instead.
Drill 4 more on the addition formulas

Concept 2 of 2: Double-angle forms

Put B=AB = A into the addition formulas and you get the double-angle forms. Their main use in CDS is recognition: 2sin⁡θcos⁡θ2\sin\theta\cos\theta is sin⁡2θ\sin 2\theta, 1−2sin⁡2θ1 - 2\sin^2\theta is cos⁡2θ\cos 2\theta, and a fraction 2tan⁡θ1−tan⁡2θ\dfrac{2\tan\theta}{1 - \tan^2\theta} is tan⁡2θ\tan 2\theta.

Definition

  • sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta
  • cos⁡2θ=cos⁡2θ−sin⁡2θ=1−2sin⁡2θ=2cos⁡2θ−1\cos 2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1
  • tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}

A fraction like sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ\dfrac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta} factors to sin⁡θcos⁡2θcos⁡θcos⁡2θ=tan⁡θ\dfrac{\sin\theta\cos 2\theta}{\cos\theta\cos 2\theta} = \tan\theta.

Double-angle forms

sin⁡2θ=2sin⁡θcos⁡θ,cos⁡2θ=1−2sin⁡2θ,tan⁡2θ=2tan⁡θ1−tan⁡2θ\sin 2\theta = 2\sin\theta\cos\theta, \quad \cos 2\theta = 1 - 2\sin^2\theta, \quad \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}

Worked example

If tan⁡θ=3\tan\theta = 3 with θ\theta acute, find tan⁡2θ\tan 2\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (II) 2018 — Elementary Mathematics · Q61Moderate

Example 2 · Trigonometric Ratios and Identities · Compound and Multiple Angles

If cos⁡θ=15\cos \theta = \frac{1}{\sqrt{5}}, where 0<θ<π20 < \theta < \frac{\pi}{2}, then 2tan⁡θ1−tan⁡2θ\frac{2 \tan \theta}{1 - \tan^2 \theta} is equal to

tan 2θ can be negative for an acute θ

If θ\theta is above 45∘45^\circ, then 2θ2\theta is above 90∘90^\circ and tan⁡2θ<0\tan 2\theta < 0. A positive option with the right size is the planted answer.
Drill 2 more on double-angle forms

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)

  • The addition formulas

    Addition formulas

    sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B,cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B, \quad \cos(A+B) = \cos A\cos B - \sin A\sin B
  • Double-angle forms

    Double-angle forms

    sin⁡2θ=2sin⁡θcos⁡θ,cos⁡2θ=1−2sin⁡2θ,tan⁡2θ=2tan⁡θ1−tan⁡2θ\sin 2\theta = 2\sin\theta\cos\theta, \quad \cos 2\theta = 1 - 2\sin^2\theta, \quad \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}

Watch out for (2)

Test yourself on Trigonometric Ratios and Identities

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Trigonometric Ratios and Identities chapter testDrill all 8 questions