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MHT-CET Maths · Trigonometry - II

Compound Angles and Conditional Identities

The sine, cosine and tangent of A + B and A − B, the maximum of a sin x + b cos x, and what a fixed sum such as A + B = 45° or A + B + C = π does to the tangent formula.

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Why this matters

13 PYQs, two HARD. Seven use the compound-angle formulas directly — a ratio from a given tangent, cot(A − B) from two differences, a maximum, an angle rewritten as α − π/4. Six fix a sum of angles and rearrange tan(A + B): tan 3A − tan 2A − tan A, A + B = 225°, the half angles of a triangle. Two cards.

Concept 1 of 2: The Compound-Angle Formulas and a sin x + b cos x

Every other identity in this chapter comes from four lines: the sine and cosine of A ± B. Sine keeps the sign of the angle sum; cosine flips it. Tangent is their quotient. The same formulas run backwards: a sin x + b cos x is one sine of a shifted angle, with amplitude a2+b2\sqrt{a^2 + b^2}, so that is its largest value.

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 — 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}.
  • asin⁡x+bcos⁡x=a2+b2 sin⁡(x+θ)a\sin x + b\cos x = \sqrt{a^2 + b^2}\,\sin(x + \theta), so it lies in [−a2+b2, a2+b2]\left[-\sqrt{a^2 + b^2},\ \sqrt{a^2 + b^2}\right].
  • sin⁡α−cos⁡αsin⁡α+cos⁡α=tan⁡α−1tan⁡α+1=tan⁡(α−π4)\dfrac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha} = \dfrac{\tan\alpha - 1}{\tan\alpha + 1} = \tan\left(\alpha - \frac{\pi}{4}\right).
  • A ratio of sines and cosines with a known tan⁡θ\tan\theta: divide top and bottom by cos⁡θ\cos\theta.

Compound angles

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

Worked example

Find the maximum of 3sin⁡x+4cos⁡x+23\sin x + 4\cos x + 2.
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 2 · Q150Moderate

Example 1 · Trigonometry - II · Compound Angles and Conditional Identities

If A>B and tan⁡A−tan⁡B=x\tan A - \tan B = x and cot⁡B−cot⁡A=y\cot B - \cot A = y, then cot⁡(A−B)=\cot(A-B)=

Keeping the plus sign in cos(A + B)

cos(A + B) = cos A cos B − sin A sin B. The sine formula keeps the sign of the angle sum; the cosine formula reverses it.

Taking a + b as the maximum of a sin x + b cos x

The two terms never peak together. The maximum is a2+b2\sqrt{a^2 + b^2}, not a+ba + b and not a2+b2a^2 + b^2.

Forgetting that cot is the reciprocal

cot B − cot A = (tan A − tan B)/(tan A tan B). Given tan A − tan B = x and cot B − cot A = y, the product tan A tan B is x/y, and cot(A − B) = 1/x + 1/y.
Drill 6 more on the compound-angle formulas and a sin x + b cos x

Concept 2 of 2: Fixed Angle Sums and the Tangent Formula

When a question fixes a sum of angles, write the tangent formula for that sum and clear the fraction. tan(A + B) = 1 becomes tan A + tan B + tan A tan B = 1. A triangle gives A/2 + B/2 = π/2 − C/2, so the tangent of the half-sum is the cotangent of the third half-angle. And 3A = 2A + A turns tan 3A into a relation between tan A, tan 2A and tan 3A.

Definition

  • A+B=π4A + B = \frac{\pi}{4} or 5π4\frac{5\pi}{4} (225°): (1+tan⁡A)(1+tan⁡B)=2(1 + \tan A)(1 + \tan B) = 2.
  • tan⁡3A−tan⁡2A−tan⁡A=tan⁡Atan⁡2Atan⁡3A\tan 3A - \tan 2A - \tan A = \tan A\tan 2A\tan 3A, from tan⁡3A=tan⁡(2A+A)\tan 3A = \tan(2A + A).
  • In a triangle: tan⁡A+B2=cot⁡C2\tan\frac{A + B}{2} = \cot\frac{C}{2}.
  • cot⁡(A+B)=0\cot(A + B) = 0 means A+B=π2A + B = \frac{\pi}{2} (up to nπn\pi), so sin⁡(A+2B)=sin⁡(π2+B)=cos⁡B=sin⁡A\sin(A + 2B) = \sin(\frac{\pi}{2} + B) = \cos B = \sin A.
  • To compare two angles given by tangents, compute tan⁡(A+B)\tan(A + B) and compare it with the third tangent.

Clearing the fraction

tan⁡(A+B)=k  ⟺  tan⁡A+tan⁡B=k (1−tan⁡Atan⁡B)\tan(A+B)=k \iff \tan A+\tan B = k\,(1-\tan A\tan B)

Worked example

If A+B=π4A + B = \frac{\pi}{4}, find (1+tan⁡A)(1+tan⁡B)(1 + \tan A)(1 + \tan B).
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 2 · Q150Moderate

Example 2 · Trigonometry - II · Compound Angles and Conditional Identities

If A+B=225∘A+B=225^\circ, then cot⁡A1+cot⁡A⋅cot⁡B1+cot⁡B\frac{\cot A}{1+\cot A}\cdot\frac{\cot B}{1+\cot B}, if it exists, is equal to

Reading 225° as a new case

tan 225° = tan 45° = 1, so A + B = 225° gives exactly the same identity as A + B = 45°: the product of (1 + tan A)(1 + tan B) is 2.

Dropping a sign in the triple rearrangement

tan 3A(1 − tan 2A tan A) = tan 2A + tan A, so tan 3A − tan 2A − tan A = +tan A tan 2A tan 3A. The negative of it is printed as an option.
Drill 5 more on fixed angle sums and the tangent formula

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 Compound-Angle Formulas and a sin x + b cos x

    Compound angles

    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\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B \qquad \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B \qquad \tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}
  • Fixed Angle Sums and the Tangent Formula

    Clearing the fraction

    tan⁡(A+B)=k  ⟺  tan⁡A+tan⁡B=k (1−tan⁡Atan⁡B)\tan(A+B)=k \iff \tan A+\tan B = k\,(1-\tan A\tan B)

Watch out for (5)

Test yourself on Trigonometry - II

15 past MHT-CET questions from this chapter, timed at 27 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 Trigonometry - II chapter testDrill all 13 questions