The equation 2 sin θ + 3 cos θ = 4 has:

Exactly two solutions in [0, 2π]

While solving sin θ cos θ = sin θ, dividing both sides by sin θ:

Loses the solutions where sin θ = 0

Is valid and gives all the solutions

Loses the solutions where sin θ = 0

Introduces extraneous solutions

Changes the equation into an identity

Trigonometry — Identities, Equations and Inverse Functions

Trigonometry is the topic that most rewards knowing a small number of identities extremely well rather than a large number vaguely. Three of them — the Pythagorean identity, the compound-angle formulae and the factorisation formulae — generate almost everything else, and it is worth being able to derive the double-angle and half-angle results rather than storing them separately. The second half of the chapter, trigonometric equations and inverse functions, is where marks are actually lost, and for a single reason: a trigonometric equation has infinitely many solutions, and an inverse trigonometric function has exactly one. Keeping those two facts apart is most of the skill.

The Identities Worth Knowing Cold

The generating identities
Family	Identity
Pythagorean	sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ
Compound angle	sin(A ± B) = sinA cosB ± cosA sinB; cos(A ± B) = cosA cosB ∓ sinA sinB
Double angle	sin2θ = 2sinθcosθ; cos2θ = 1 − 2sin²θ = 2cos²θ − 1
Triple angle	sin3θ = 3sinθ − 4sin³θ; cos3θ = 4cos³θ − 3cosθ
Sum to product	sinA + sinB = 2 sin((A+B)/2) cos((A−B)/2)
Range of a sinθ + b cosθ	−√(a² + b²) to +√(a² + b²)

🧠 Any a sinθ + b cosθ is a single sine wave, which settles all its extrema

Write a sinθ + b cosθ as R sin(θ + φ) with R = √(a² + b²), and the whole expression becomes one sine curve of amplitude R. That immediately gives its maximum as +R, its minimum as −R and the values of θ where each occurs — no calculus needed. So the maximum of 3 sinθ + 4 cosθ is 5, and the maximum of sinθ + cosθ is √2. The same conversion tells you when an equation like a sinθ + b cosθ = c has no solution at all, namely when |c| > √(a² + b²), which is a favourite way of setting a question that looks solvable and is not.

For a triangle, the sine rule a/sinA = b/sinB = c/sinC = 2R and the cosine rule a² = b² + c² − 2bc cosA cover most problems, with area given by ½ab sinC or by Heron's formula. Note that the sine rule can be ambiguous when you are given two sides and a non-included angle, because two different triangles may fit — a genuine ambiguity, not a mistake, and one that questions occasionally exploit.

Trigonometric Equations

General solutions
Equation	General solution
sin θ = 0	θ = nπ
cos θ = 0	θ = (2n + 1)π/2
tan θ = 0	θ = nπ
sin θ = sin α	θ = nπ + (−1)ⁿα
cos θ = cos α	θ = 2nπ ± α
tan θ = tan α	θ = nπ + α

⚠️ Squaring or cancelling a trigonometric equation changes its solution set

Two manipulations regularly cost marks. Squaring both sides introduces extraneous roots, because it cannot distinguish +x from −x, so every solution obtained must be substituted back into the original equation and checked. Cancelling a common factor loses roots: dividing sin θ cos θ = sin θ by sin θ discards all the solutions with sin θ = 0, and those are usually among the options. The safe procedure is to factorise rather than cancel — write sin θ (cos θ − 1) = 0 and set each factor to zero — so that no branch is silently thrown away. Questions asking for "the number of solutions in [0, 2π]" are built precisely to catch a lost or an extra root.

Inverse Trigonometric Functions

Domains and principal ranges
Function	Domain	Principal range
sin⁻¹x	[−1, 1]	[−π/2, π/2]
cos⁻¹x	[−1, 1]	[0, π] — never negative
tan⁻¹x	All reals	(−π/2, π/2), open
cot⁻¹x	All reals	(0, π)
sec⁻¹x	\|x\| ≥ 1	[0, π] excluding π/2
cosec⁻¹x	\|x\| ≥ 1	[−π/2, π/2] excluding 0

🎯 Why sin⁻¹(sin x) is not always x

The inverse sine returns a value in [−π/2, π/2] and nothing else, so sin⁻¹(sin x) equals x only when x already lies in that range. For x = 3π/4, sin x = 1/√2 and sin⁻¹ of that is π/4, not 3π/4 — the function has folded the input back into its principal range. The general rule is to reduce the angle to the principal range first, using sin(π − x) = sin x or whichever identity applies, and only then apply the inverse. This is the single most examined subtlety in the topic, and the same caution applies to cos⁻¹(cos x), which returns a value in [0, π] and so maps 7π/4 to π/4 rather than to itself.

Three relations recur: sin⁻¹x + cos⁻¹x = π/2, tan⁻¹x + cot⁻¹x = π/2 and sec⁻¹x + cosec⁻¹x = π/2, all valid on the appropriate domains. The addition formula tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy)) holds only when xy < 1; if xy > 1 you must add or subtract π to land back in the principal range, and forgetting that correction is the second most common error here after the sin⁻¹(sin x) folding.

Key takeaways

Learn the Pythagorean, compound-angle and factorisation identities well; derive the rest from them.
a sinθ + b cosθ is a single wave of amplitude √(a² + b²), so its range is −R to +R.
a sinθ + b cosθ = c has no solution when |c| > √(a² + b²) — check before solving.
sin θ = sin α gives nπ + (−1)ⁿα, cos θ = cos α gives 2nπ ± α, tan θ = tan α gives nπ + α.
Never cancel a factor in a trigonometric equation — factorise, or you lose whole families of roots.
Squaring introduces extraneous roots, so every solution must be checked in the original equation.
cos⁻¹ returns a value in [0, π] and is never negative; sin⁻¹ returns one in [−π/2, π/2].
sin⁻¹(sin x) = x only if x is already in the principal range — reduce the angle first.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The maximum value of 3 sin θ + 4 cos θ is:
1. 3
2. 4
3. 5
4. 7
Show answer
Answer: C — 5
Any expression a sin θ + b cos θ can be written as R sin(θ + φ) with R = √(a² + b²) = √(9 + 16) = 5, so the value oscillates between −5 and 5. The answer 7 comes from adding the coefficients as though sine and cosine could both reach 1 at the same θ, which they cannot — the two peaks are a quarter period apart.
sin⁻¹(sin 3π/4) equals:
1. 3π/4
2. π/4
3. −π/4
4. π/2
Show answer
Answer: B — π/4
The inverse sine only ever returns values in [−π/2, π/2], and 3π/4 lies outside that interval, so the identity sin⁻¹(sin x) = x does not apply. Using sin(3π/4) = sin(π − 3π/4) = sin(π/4) reduces the angle to the principal range, giving π/4. Answering 3π/4 is the standard error and comes from treating the two functions as unconditional inverses.
The equation 2 sin θ + 3 cos θ = 4 has:
1. Exactly two solutions in [0, 2π]
2. Infinitely many solutions
3. No solution
4. Exactly one solution
Show answer
Answer: C — No solution
The left side can only take values between −√(4 + 9) = −√13 ≈ −3.61 and +√13 ≈ 3.61, so it can never equal 4 and the equation has no solution. Checking whether |c| exceeds √(a² + b²) takes one line and rules the question out before any algebra — a habit that also protects against answers obtained by manipulating an inconsistent equation.
While solving sin θ cos θ = sin θ, dividing both sides by sin θ:
1. Is valid and gives all the solutions
2. Loses the solutions where sin θ = 0
3. Introduces extraneous solutions
4. Changes the equation into an identity
Show answer
Answer: B — Loses the solutions where sin θ = 0
Dividing by sin θ silently assumes sin θ ≠ 0, so all the solutions θ = nπ are discarded. The correct route is to factorise as sin θ (cos θ − 1) = 0 and set each factor to zero, which keeps both families: θ = nπ from the first and θ = 2nπ from the second. Squaring is what introduces extraneous roots; cancelling does the opposite and removes genuine ones.
The principal range of cos⁻¹x is:
1. [−π/2, π/2]
2. [0, π]
3. (−π/2, π/2)
4. (0, π)
Show answer
Answer: B — [0, π]
The inverse cosine is defined to take values in [0, π], with both endpoints included since cos⁻¹1 = 0 and cos⁻¹(−1) = π. That range is never negative, which is why cos⁻¹ of a negative number is an obtuse angle rather than a negative one. The interval [−π/2, π/2] belongs to sin⁻¹ and the open interval (−π/2, π/2) to tan⁻¹.
sin⁻¹x + cos⁻¹x equals:
1. π
2. π/2
3. 0
4. It depends on x
Show answer
Answer: B — π/2
For every x in [−1, 1] the two inverse functions are complementary and sum to exactly π/2, which is why either can always be rewritten in terms of the other. The same complementary relation holds for tan⁻¹x + cot⁻¹x and for sec⁻¹x + cosec⁻¹x on their domains, and using it is often the quickest way to simplify an expression containing two different inverse functions.
The formula tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy)) is valid when:
1. Always
2. xy < 1
3. x and y are both positive
4. x = y
Show answer
Answer: B — xy < 1
When xy > 1 the sum of the two angles leaves the principal range (−π/2, π/2), so the plain formula lands on the wrong branch and π must be added or subtracted to correct it. The condition xy < 1 is therefore part of the statement, not a technicality, and questions that supply values with xy > 1 are testing whether the correction is applied.
In any triangle, a/sin A equals:
1. The circumradius R
2. 2R, twice the circumradius
3. The inradius r
4. The area of the triangle
Show answer
Answer: B — 2R, twice the circumradius
The sine rule states a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle, so the common ratio is twice the circumradius rather than the circumradius itself. This form is useful because it converts between sides and angles and simultaneously gives R, which several triangle problems ask for directly.