← Mathematics — Std 11
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Trigonometry — II

Ch. 3Std 11

Easy Overview

So you know sin θ, cos θ, and the unit circle. You can find any trig ratio, you know the identities — great. Now it's time to level up. What happens when you add two angles inside a trig function? Spoiler: sin(A + B) is NOT sin A + sin B. That would be too easy. Instead, there's a beautiful pattern that connects the trig ratios of sums and differences to products of individual ratios. This is where trigonometry becomes a powerful toolkit. The compound angle formulas are the foundation. sin(A ± B) = sin A cos B ± cos A sin B. Cos(A ± B) = cos A cos B ∓ sin A sin B. Notice the pattern — sin keeps the same sign, cos flips it. Tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B). These formulas let you find exact trig values for angles like 75° (45° + 30°) without a calculator. Once you have compound angles, you can set B = A and get double angle formulas: sin 2A = 2 sin A cos A, and cos 2A = cos²A − sin²A = 2 cos²A − 1 = 1 − 2 sin²A. These three forms of cos 2A are incredibly useful because they let you switch between squares and linear terms — something you'll rely on in calculus when integrating or differentiating. Triple angle formulas come next: sin 3A = 3 sin A − 4 sin³A and cos 3A = 4 cos³A − 3 cos A. These have a rhythmic pattern that's easy to memorise with practice. Half-angle formulas let you express sin(A/2) and cos(A/2) in terms of cos A, using those cos 2A forms in reverse. Then there are the factorisation (sum-to-product) and defactorisation (product-to-sum) formulas. These convert sums of trig functions into products and vice versa. They're the key to solving many equations and simplifying complex expressions. For example, sin C + sin D = 2 sin((C+D)/2) cos((C−D)/2) — this turns an addition into a multiplication, often making the problem much simpler. You'll also learn to express a sin θ + b cos θ as R sin(θ + α) or R cos(θ + α), where R = √(a² + b²). This is a game-changer for solving equations, finding maxima and minima, and modelling real-world oscillations. Finally, we tackle inverse trigonometric functions — the 'undo' buttons for sin, cos, and tan. If sin θ = 0.5, what's θ? The inverse sine function, written sin⁻¹(0.5) or arcsin(0.5), tells you the angle. But it only gives one answer — the principal value. You need general solutions to capture all possible angles. By the end, you'll be solving complex trig equations, proving compound-angle identities in your sleep, and using these tools as second nature. This is the trigonometry that powers calculus, physics, and engineering.

Compound angle formulas for sin — the sum and difference

sin(A + B) = sin A cos B + cos A sin B. sin(A − B) = sin A cos B − cos A sin B. The pattern: sin-cos-cos-sin with the same sign in the middle. These formulas let you find sin of a non-standard angle by splitting it into two standard ones. For example, sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°.

Compound angle formulas for cos

cos(A + B) = cos A cos B − sin A sin B. cos(A − B) = cos A cos B + sin A sin B. Note the sign flip — the plus on the left becomes minus on the right, and vice versa. The pattern: cos-cos-sin-sin, with the 'different' sign. For cos 15° = cos(45° − 30°) = cos 45° cos 30° + sin 45° sin 30°.

Compound angle formulas for tan

tan(A + B) = (tan A + tan B)/(1 − tan A tan B). tan(A − B) = (tan A − tan B)/(1 + tan A tan B). The numerator has the same sign as the compound; the denominator has the opposite sign. These are derived by dividing sin(A+B) by cos(A+B). They're useful when you're given values of tan rather than sin or cos.

Double angle formulas — setting B = A

Put B = A in the compound formulas and you get sin 2A = 2 sin A cos A. For cos 2A, you get three equivalent forms: cos²A − sin²A, 2 cos²A − 1, and 1 − 2 sin²A. Tan 2A = 2 tan A/(1 − tan²A). The three forms of cos 2A are especially important because they let you express sin²A or cos²A in terms of cos 2A — invaluable in integration.

Triple angle formulas

sin 3A = 3 sin A − 4 sin³A. cos 3A = 4 cos³A − 3 cos A. These can be derived by applying compound formulas twice: sin(2A + A). They're useful in solving cubic trigonometric equations and in proofs. The pattern: sin bounces from 3 to 4, cos from 4 to 3.

Half angle formulas — working backwards from cos 2A

Using cos 2A = 1 − 2 sin²A, let A = θ/2, so cos θ = 1 − 2 sin²(θ/2). Rearranging: sin(θ/2) = ±√((1 − cos θ)/2). Similarly, cos(θ/2) = ±√((1 + cos θ)/2). And tan(θ/2) = ±√((1 − cos θ)/(1 + cos θ)) = sin θ/(1 + cos θ) = (1 − cos θ)/sin θ. The sign depends on which quadrant θ/2 falls in.

Sum-to-product (factorisation) formulas

These convert sums into products: sin C + sin D = 2 sin((C+D)/2) cos((C−D)/2). sin C − sin D = 2 cos((C+D)/2) sin((C−D)/2). cos C + cos D = 2 cos((C+D)/2) cos((C−D)/2). cos C − cos D = −2 sin((C+D)/2) sin((C−D)/2). They're derived by setting A+B = C and A−B = D in the compound formulas and adding/subtracting.

Product-to-sum (defactorisation) formulas

The reverse of the factorisation formulas: 2 sin A cos B = sin(A+B) + sin(A−B). 2 cos A sin B = sin(A+B) − sin(A−B). 2 cos A cos B = cos(A+B) + cos(A−B). 2 sin A sin B = cos(A−B) − cos(A+B). These are useful in integration (converting products to sums that are easier to integrate).

Expressing a sin θ + b cos θ in a single trig function

Any expression of the form a sin θ + b cos θ can be rewritten as R sin(θ + α) or R cos(θ + α), where R = √(a² + b²) and α = tan⁻¹(b/a) (with appropriate quadrant adjustments). For example, 3 sin θ + 4 cos θ = 5 sin(θ + 53.13°). This is the go-to technique for solving equations like 3 sin θ + 4 cos θ = 2.

General solutions for sin θ = k

If sin θ = k (where |k| ≤ 1), the general solution is θ = nπ + (−1)ⁿ α, where α = sin⁻¹(k) is the principal value (in [−π/2, π/2]) and n ∈ ℤ. The (−1)ⁿ factor alternates between 1 (for even n) and −1 (for odd n), which generates all solutions across multiple rotations.

General solutions for cos θ = k

If cos θ = k (where |k| ≤ 1), the general solution is θ = 2nπ ± α, where α = cos⁻¹(k) is the principal value (in [0, π]) and n ∈ ℤ. The ± reflects the symmetry of cos about the vertical axis — two solutions per cycle. For example, cos θ = ½ gives θ = 2nπ ± π/3.

General solutions for tan θ = k

If tan θ = k (any real k), the general solution is θ = nπ + α, where α = tan⁻¹(k) (principal value in (−π/2, π/2)) and n ∈ ℤ. Since tan has period π (180°), the solutions are equally spaced — no ± needed. For example, tan θ = 1 gives θ = nπ + π/4.

Principal values of inverse trigonometric functions

Inverse trig functions return a single 'principal' value. For sin⁻¹, range is [−π/2, π/2] (or [−90°, 90°]). For cos⁻¹, range is [0, π] (or [0°, 180°]). For tan⁻¹, range is (−π/2, π/2) (or (−90°, 90°)). These intervals are chosen to make the functions one-to-one and therefore properly invertible.

Properties of inverse trigonometric functions

Key properties: sin⁻¹(sin θ) = θ only when θ ∈ [−π/2, π/2]; otherwise you get a different angle in that range. Similarly for cos⁻¹ and tan⁻¹. The composition sin(sin⁻¹ x) = x for x ∈ [−1, 1]. Also: sin⁻¹(−x) = −sin⁻¹ x (odd), cos⁻¹(−x) = π − cos⁻¹ x, tan⁻¹(−x) = −tan⁻¹ x (odd). And sin⁻¹ x + cos⁻¹ x = π/2 — a very useful relation.

Trigonometric equations — practical solving strategies

To solve a trig equation: first simplify using identities. If multiple functions exist, try to express everything in one function (use sin² + cos² = 1 to substitute). Check domain restrictions. Apply general or particular solutions as needed. Factorise if possible. And always test your solutions — extraneous solutions can creep in from squaring or algebraic manipulation.

Inequalities involving trigonometric functions

Trigonometric inequalities like sin θ > ½ or cos θ ≤ 0 require understanding the graphs and unit circle. For sin θ > ½ in [0, 2π], the solution is (π/6, 5π/6). These problems ask you to find ranges of θ where the inequality holds, often combining algebraic solving with graphical reasoning.

Key Points

  • •sin(A ± B) = sin A cos B ± cos A sin B
  • •cos(A ± B) = cos A cos B ∓ sin A sin B
  • •tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)
  • •sin 2A = 2 sin A cos A
  • •cos 2A = cos²A − sin²A = 2 cos²A − 1 = 1 − 2 sin²A — three equivalent forms
  • •tan 2A = 2 tan A/(1 − tan²A)
  • •sin 3A = 3 sin A − 4 sin³A; cos 3A = 4 cos³A − 3 cos A
  • •Half-angle: sin(θ/2) = ±√((1 − cos θ)/2), cos(θ/2) = ±√((1 + cos θ)/2)
  • •sin C + sin D = 2 sin((C+D)/2) cos((C−D)/2)
  • •cos C + cos D = 2 cos((C+D)/2) cos((C−D)/2)
  • •a sin θ + b cos θ = R sin(θ + α) where R = √(a² + b²)
  • •General solution sin θ = k: θ = nπ + (−1)ⁿ α
  • •General solution cos θ = k: θ = 2nπ ± α
  • •General solution tan θ = k: θ = nπ + α
  • •sin⁻¹ x + cos⁻¹ x = π/2

Practice Questions

  • Prove that sin(A + B) sin(A − B) = sin²A − sin²B.
  • If tan A = 1/2 and tan B = 1/3, find tan(A + B). Hence find the value of A + B.
  • Solve the equation sin θ + cos θ = 1 for 0° ≤ θ ≤ 360°.
  • Prove that (sin 3θ)/(sin θ) − (cos 3θ)/(cos θ) = 2.
  • Find the principal value of sin⁻¹(−1/2), cos⁻¹(√3/2), and tan⁻¹(−1).
  • Solve 2 cos²θ + 3 sin θ = 0 for 0 ≤ θ ≤ 2π.
  • Express 4 sin θ − 3 cos θ in the form R sin(θ − α) and hence solve 4 sin θ − 3 cos θ = 2.
  • Prove that tan⁻¹(1/2) + tan⁻¹(1/3) = π/4.