MHT-CET Maths · Trigonometry - II

Double, Triple and Half Angles

The formulas for 2A, 3A and A/2, the sign of a half angle read from its own quadrant, and the standard values at π/8, 18° and 36° that most of the HARD questions reduce to.

All Trigonometry - II notes MHT-CET Maths strategy

Why this matters

13 PYQs, nine HARD — the hardest page in the chapter. Eight use the double- and half-angle formulas: tan(π/8), a half angle with its sign, sin(x/4) from sec x + tan x, cos 2θ from three sines in H.P. Five are evaluations at π/8, 10° or 20°, done with the triple-angle formulas and angle pairing. Two cards.

Concept 1 of 2: Double and Half Angles, and the Sign of a Half Angle

Put B = A in the compound formulas and you get the double angles. cos 2A has three forms; pick the one that cancels what the question already has. Read backwards, the same forms give the half angle: cos²(x/2) = (1 + cos x)/2. The square root leaves a sign to choose, and it comes from where x/2 lies, not where x lies.

Definition

$\sin 2A = 2\sin A\cos A = \dfrac{2\tan A}{1 + \tan^2 A}$ .
$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A = \dfrac{1 - \tan^2 A}{1 + \tan^2 A}$ .
$\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$ .
Half angles: $\cos\frac{x}{2} = \pm\sqrt{\dfrac{1 + \cos x}{2}}$ , $\sin\frac{x}{2} = \pm\sqrt{\dfrac{1 - \cos x}{2}}$ , $\tan\frac{x}{2} = \dfrac{1 - \cos x}{\sin x}$ .
The sign: if $\pi < x < \frac{3\pi}{2}$ , then $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$ , the second quadrant, where cosine is negative.
$\tan\frac{\pi}{8} = \sqrt2 - 1$ , from $\dfrac{1 - \cos\frac{\pi}{4}}{\sin\frac{\pi}{4}}$ .

Double and half angles

\cos 2A = 2\cos^2 A - 1 = 1 - 2\sin^2 A \qquad \cos\frac{x}{2}=\pm\sqrt{\frac{1+\cos x}{2}} \qquad \tan\frac{x}{2}=\frac{1-\cos x}{\sin x}

Worked example

\cos x = -\frac{7}{25}

and

\pi < x < \frac{3\pi}{2}

, find

\sin\frac{x}{2}

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q136Moderate

Example 1 · Trigonometry - II · Multiple and Sub-multiple Angles

\tan x=\frac{3}{4}

and

\pi<x<\frac{3\pi}{2}

, then

\cos\frac{x}{2}=

Taking the sign from the quadrant of x

With x in the third quadrant, cos x is negative, but x/2 is in the second quadrant and so is cos(x/2) — negative too. With x in the fourth, x/2 is in the second again. Always halve the interval first.

Using the wrong form of cos 2A

2cos²A − 1 and 1 − 2sin²A are both right; the choice decides whether the question collapses. For cos 2θ from sin²θ, use 1 − 2sin²θ.

The negative root of tan(π/8)

tan(π/8) solves t² + 2t − 1 = 0, whose roots are √2 − 1 and −1 − √2. π/8 is acute, so only the positive root is right; the other is an option.

Drill 7 more on double and half angles, and the sign of a half angle

Concept 2 of 2: Triple Angles, Standard Values and Pairing

Most HARD evaluations here are not algebra. They pair angles that add to 90° or 180°, use a triple-angle formula at an angle whose triple is standard (10° to 30°, 20° to 60°), or use a product that collapses to a triple angle. Learn the values and look for the pairing before expanding anything.

Definition

$\sin 3A = 3\sin A - 4\sin^3 A$ , $\cos 3A = 4\cos^3 A - 3\cos A$ , $\tan 3A = \dfrac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$ .
$\tan\theta\tan(60^\circ - \theta)\tan(60^\circ + \theta) = \tan 3\theta$ : so $\tan 20^\circ\tan 40^\circ\tan 80^\circ = \sqrt3$ .
$\sin 18^\circ = \dfrac{\sqrt5 - 1}{4}$ , $\cos 36^\circ = \dfrac{\sqrt5 + 1}{4}$ .
Pairing: $\cos\frac{5\pi}{8} = -\cos\frac{3\pi}{8}$ , $\cos\frac{7\pi}{8} = -\cos\frac{\pi}{8}$ , and $\cos\frac{3\pi}{8} = \sin\frac{\pi}{8}$ .
Squaring a triple-angle identity at a known angle gives a polynomial identity: at $\theta = 10^\circ$ , $3\tan^6\theta - 27\tan^4\theta + 33\tan^2\theta = 1$ .

Triple angles

\sin 3A = 3\sin A - 4\sin^3 A \qquad \cos 3A = 4\cos^3 A - 3\cos A \qquad \tan 3A=\frac{3\tan A-\tan^3 A}{1-3\tan^2 A}

Worked example

Find

\sin^4\frac{\pi}{8} + \sin^4\frac{3\pi}{8} + \sin^4\frac{5\pi}{8} + \sin^4\frac{7\pi}{8}

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift II · Q122Moderate

Example 2 · Trigonometry - II · Multiple and Sub-multiple Angles

The value of

\tan20^{\circ}\tan80^{\circ}\cot50^{\circ}=

Multiplying out four brackets

(1 + cos π/8)(1 + cos 3π/8)(1 + cos 5π/8)(1 + cos 7π/8) pairs into (1 − cos²π/8)(1 − cos²3π/8) = sin²(π/8) cos²(π/8) = 1/8. Expanding all four brackets is slow and error-prone.

Mixing up sin 18° and cos 36°

sin 18° = (√5 − 1)/4 and cos 36° = (√5 + 1)/4. They differ only in one sign, and both are printed as options.

Drill 4 more on triple angles, standard values and pairing

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)

Double and Half Angles, and the Sign of a Half Angle
Double and half angles
$\cos 2A = 2\cos^2 A - 1 = 1 - 2\sin^2 A \qquad \cos\frac{x}{2}=\pm\sqrt{\frac{1+\cos x}{2}} \qquad \tan\frac{x}{2}=\frac{1-\cos x}{\sin x}$
Triple Angles, Standard Values and Pairing
Triple angles
$\sin 3A = 3\sin A - 4\sin^3 A \qquad \cos 3A = 4\cos^3 A - 3\cos A \qquad \tan 3A=\frac{3\tan A-\tan^3 A}{1-3\tan^2 A}$

Watch out for (5)

Taking the sign from the quadrant of x
→ Double and Half Angles, and the Sign of a Half Angle
Using the wrong form of cos 2A
→ Double and Half Angles, and the Sign of a Half Angle
The negative root of tan(π/8)
→ Double and Half Angles, and the Sign of a Half Angle
Multiplying out four brackets
→ Triple Angles, Standard Values and Pairing
Mixing up sin 18° and cos 36°
→ Triple Angles, Standard Values and Pairing

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.

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