JEE Corner — Trigonometry for JEE Main and Advanced

Why This Section Exists

The rationalised textbook stops at identities — but the official JEE Main syllabus explicitly includes trigonometrical equations, and JEE constantly tests the range of $a\sin x + b\cos x$, special values like sin 18°, and conditional identities. This section restores the full JEE toolkit:

1. Trigonometric equations — principal solutions and general solutions;
2. The $a\sin x + b\cos x$ machine — range, maxima-minima, solvability;
3. Special values — sin 18°, cos 36° and the 15°-75° family collected;
4. Conditional identities — the $A + B + C = \pi$ family;
5. Assorted JEE patterns — telescoping products and quadratics in sin/cos.

The worked examples and practice questions here are modelled on the JEE Main and Advanced pattern and difficulty; they are practice questions in the exam style, not reproductions of specific past papers.

[Board Note] CBSE students: trigonometric equations were removed from the rationalised syllabus — treat this section as JEE preparation. The $a\sin x + b\cos x$ range technique, however, quietly helps even in Board-level maximum-minimum questions.

Trigonometric Equations — Principal and General Solutions

An equation like $\sin x = \frac{1}{2}$ has infinitely many solutions (the sine wave crosses the level $\frac{1}{2}$ forever). Two vocabularies organise them:

• Principal solutions: those in $[0, 2\pi)$. For $\sin x = \frac{1}{2}$: $x = \frac{\pi}{6}$ and $\frac{5\pi}{6}$.
• General solution: a formula with an integer parameter n capturing ALL solutions.

Key Point (the three master formulas, $n \in \mathbb{Z}$):

• $\sin x = \sin y \Rightarrow x = n\pi + (-1)^n y$
• $\cos x = \cos y \Rightarrow x = 2n\pi \pm y$
• $\tan x = \tan y \Rightarrow x = n\pi + y$

Special cases worth quoting directly: $\sin x = 0 \Rightarrow x = n\pi$; $\cos x = 0 \Rightarrow x = (2n+1)\frac{\pi}{2}$; $\tan x = 0 \Rightarrow x = n\pi$; $\sin x = 1 \Rightarrow x = 2n\pi + \frac{\pi}{2}$; $\cos x = 1 \Rightarrow x = 2n\pi$; $\cos x = -1 \Rightarrow x = (2n+1)\pi$.

The solving protocol

1. Reduce the equation to the form (trig function) = (known value), by factoring or identities — never divide away a factor that can vanish.
2. Name a convenient y with that value (e.g. $\sin y = \frac{1}{2}$, take $y = \frac{\pi}{6}$).
3. Quote the master formula.

[JEE Tip] The $(-1)^n$ in the sine formula alternates the sign of y as n steps — check your general solution by extracting n = 0, 1, 2 and comparing with the principal solutions. A general solution that cannot reproduce the principal ones is wrong.

The Range of a sin x + b cos x

Divide by $R = \sqrt{a^2 + b^2}$: the coefficients $\frac{a}{R}, \frac{b}{R}$ have squares summing to 1, so they are $\cos\phi$ and $\sin\phi$ for some $\phi$, and

$$a\sin x + b\cos x = R\sin(x + \phi), \qquad R = \sqrt{a^2 + b^2}, \quad \tan\phi = \frac{b}{a}$$

Key Point: The range of $a\sin x + b\cos x$ is $\left[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}\right]$.

Instant consequences:

• max of $3\sin x + 4\cos x$ is 5; min is $-5$.
• $\sin x - \cos x \in [-\sqrt{2}, \sqrt{2}]$.
• The equation $a\sin x + b\cos x = c$ has solutions iff $|c| \leq \sqrt{a^2 + b^2}$.
• Range of $a\sin x + b\cos x + c$: $\left[c - R, c + R\right]$.

Special values the exam assumes

$$\sin 18° = \frac{\sqrt{5} - 1}{4} \qquad \cos 36° = \frac{\sqrt{5} + 1}{4}$$

(Derivation sketch for sin 18°: let $\theta = 18°$; then $5\theta = 90°$, so $2\theta = 90° - 3\theta$, giving $\sin 2\theta = \cos 3\theta$; expanding both sides in $s = \sin\theta$ yields $4s^2 + 2s - 1 = 0$, whose positive root is $\frac{\sqrt{5} - 1}{4}$.)

Companions: $\cos 18° = \frac{\sqrt{10 + 2\sqrt{5}}}{4}$, $\sin 36° = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$, and the memorable product $\sin 18° \cos 36° = \frac{1}{4}$.

[JEE Tip] For $f(x) = a\sin x + b\cos x + c$ style questions asking "number of integer values in the range" — compute $[c - R, c + R]$ and count. R need not be an integer; the count is $\lfloor c + R \rfloor - \lceil c - R \rceil + 1$.

Conditional Identities — the A + B + C = π Family

When A, B, C are angles of a triangle ($A + B + C = \pi$), the constraint generates a whole catalogue. The two conversion facts powering every proof:

$$\sin(A + B) = \sin C, \qquad \cos(A + B) = -\cos C$$

and for half-angles, $\frac{A + B}{2} = \frac{\pi}{2} - \frac{C}{2}$, so $\sin\frac{A+B}{2} = \cos\frac{C}{2}$ and $\cos\frac{A+B}{2} = \sin\frac{C}{2}$.

The standard catalogue (all provable by pair → factor → convert):

• $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$
• $\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A\cos B\cos C$
• $\sin A + \sin B + \sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$
• $\cos A + \cos B + \cos C = 1 + 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$
• $\tan A + \tan B + \tan C = \tan A\tan B\tan C$
• $\tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1$

Proof of the tan identity (the shortest of the family): $A + B = \pi - C$ gives $\tan(A + B) = -\tan C$, i.e. $\frac{\tan A + \tan B}{1 - \tan A\tan B} = -\tan C$; cross-multiplying and rearranging: $\tan A + \tan B + \tan C = \tan A\tan B\tan C$. ∎

[JEE Tip] The tan identity means: for triangle angles, the SUM of tangents equals their PRODUCT. It reappears in coordinate geometry (slopes of triangle sides) and complex numbers — recognise it in disguise.

Solved Examples

Example 1: Principal solutions

Find the principal solutions of (i) $\sin x = \frac{\sqrt{3}}{2}$ (ii) $\tan x = -\frac{1}{\sqrt{3}}$.

Solution:

Step 1 — (i) Reference angle first. $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$, so the reference angle is $\frac{\pi}{3}$.

Step 2 — Place by sign. Sine is positive in Q I and Q II: $x = \frac{\pi}{3}$ and $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ — both in $[0, 2\pi)$.

Step 3 — (ii) Reference angle. $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$; the given value is negative.

Step 4 — Place by sign. tan is negative in Q II and Q IV: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ and $x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$.

Takeaway: Principal solutions = reference angle + quadrant placement; there are exactly two per basic equation (unless the value is 0 or $\pm 1$).

Example 2: General solution of a sine equation

Solve $\sin x = \frac{1}{2}$ in general.

Solution:

Step 1 — Name a convenient y. $\sin\frac{\pi}{6} = \frac{1}{2}$, so take $y = \frac{\pi}{6}$.

Step 2 — Quote the sine master formula. $x = n\pi + (-1)^n\frac{\pi}{6}$, $n \in \mathbb{Z}$.

Step 3 — Check against the principal solutions. n = 0: $\frac{\pi}{6}$ ✓; n = 1: $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$ ✓ — both reproduced.

Takeaway: Always run the n = 0, 1 check; it catches sign and formula slips instantly.

Example 3: General solution with cos

Solve $\cos x = -\frac{1}{2}$ in general.

Solution:

Step 1 — Name y. $\cos\frac{2\pi}{3} = -\frac{1}{2}$, so take $y = \frac{2\pi}{3}$.

Step 2 — Quote the cosine master formula. $x = 2n\pi \pm \frac{2\pi}{3}$, $n \in \mathbb{Z}$.

Step 3 — Check. n = 0 gives $\pm\frac{2\pi}{3}$; the minus branch equals $2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$ modulo $2\pi$ — both principal solutions covered. ✓

Takeaway: For cosine, the $\pm$ handles both principal solutions in one formula.

Example 4: A factorable equation

Solve $\sin 2x - \sin x = 0$ in general.

Solution:

Step 1 — Expand and factor. $2\sin x\cos x - \sin x = \sin x(2\cos x - 1) = 0$ — never divide by sin x.

Step 2 — Family 1. $\sin x = 0$: $x = n\pi$.

Step 3 — Family 2. $\cos x = \frac{1}{2} = \cos\frac{\pi}{3}$: $x = 2n\pi \pm \frac{\pi}{3}$.

Step 4 — State the union. $x = n\pi$ or $x = 2n\pi \pm \frac{\pi}{3}$, $n \in \mathbb{Z}$.

Takeaway: Factor, never divide — dividing by sin x silently deletes the entire $n\pi$ family.

Example 5: A quadratic in cos x

Solve $2\cos^2 x + 3\sin x = 0$ in general.

Solution:

Step 1 — Convert to one function. $\cos^2 x = 1 - \sin^2 x$: the equation becomes $2 - 2\sin^2 x + 3\sin x = 0$, i.e. $2\sin^2 x - 3\sin x - 2 = 0$.

Step 2 — Factor the quadratic. $(2\sin x + 1)(\sin x - 2) = 0$.

Step 3 — Reject the impossible root with a reason. $\sin x = 2$ is outside $[-1, 1]$ — rejected.

Step 4 — Solve the survivor. $\sin x = -\frac{1}{2} = \sin\left(-\frac{\pi}{6}\right)$: $x = n\pi + (-1)^n\left(-\frac{\pi}{6}\right) = n\pi - (-1)^n\frac{\pi}{6}$, $n \in \mathbb{Z}$.

Takeaway: Quadratics in sin/cos: substitute, factor, REJECT roots outside $[-1, 1]$ with a stated reason — that rejection carries a mark.

Example 6: Equation with equal functions of different angles

Solve $\sin 3x = \sin x$ in general.

Solution:

Step 1 — Apply the master formula to the pair (3x, x). $3x = n\pi + (-1)^n x$.

Step 2 — Split into even n. n = 2m: $3x = 2m\pi + x$, so $2x = 2m\pi$, giving $x = m\pi$.

Step 3 — Split into odd n. n = 2m + 1: $3x = (2m+1)\pi - x$, so $4x = (2m+1)\pi$, giving $x = (2m + 1)\frac{\pi}{4}$.

Step 4 — Union. $x = m\pi$ or $x = (2m + 1)\frac{\pi}{4}$, $m \in \mathbb{Z}$.

Takeaway: When both sides carry x, split the $(-1)^n$ into even/odd cases and solve each linear equation for x.

Example 7: The a sin + b cos equation

Solve $\sqrt{3}\cos x + \sin x = \sqrt{2}$.

Solution:

Step 1 — Compute the amplitude. $R = \sqrt{(\sqrt{3})^2 + 1^2} = 2$; since $|\sqrt{2}| \leq 2$, solutions exist.

Step 2 — Divide by R and compress. $\frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x = \frac{\sqrt{2}}{2}$, i.e. $\cos\left(x - \frac{\pi}{6}\right) = \frac{1}{\sqrt{2}}$ (recognising $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$, $\sin\frac{\pi}{6} = \frac{1}{2}$).

Step 3 — Apply the cosine master formula. $x - \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{4}$.

Step 4 — Solve for x. $x = 2n\pi + \frac{\pi}{6} + \frac{\pi}{4} = 2n\pi + \frac{5\pi}{12}$, or $x = 2n\pi + \frac{\pi}{6} - \frac{\pi}{4} = 2n\pi - \frac{\pi}{12}$.

Takeaway: Compress $a\sin x + b\cos x$ into a single cosine (or sine), then apply one master formula — the standard two-step for mixed equations.

Example 8: Max-min of a sin x + b cos x + c

Find the maximum and minimum values of $f(x) = 3\sin x + 4\cos x + 5$, and the general x at which the maximum occurs.

Solution:

Step 1 — Amplitude of the wave part. $R = \sqrt{9 + 16} = 5$, so $3\sin x + 4\cos x \in [-5, 5]$.

Step 2 — Shift by the constant. $f(x) \in [5 - 5, 5 + 5] = [0, 10]$: maximum 10, minimum 0.

Step 3 — Locate the maximum. Write $f = 5\sin(x + \phi) + 5$ with $\tan\phi = \frac{4}{3}$; the maximum needs $\sin(x + \phi) = 1$, i.e. $x + \phi = 2n\pi + \frac{\pi}{2}$, so $x = 2n\pi + \frac{\pi}{2} - \phi$.

Takeaway: The min of $3\sin x + 4\cos x + 5$ is exactly 0 — a designed coincidence JEE reuses to build always-non-negative expressions.

Example 9: Solvability test

For which values of k does $\sin x + \cos x = k$ have a solution?

Solution:

Step 1 — Compress. $\sin x + \cos x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)$.

Step 2 — Read off the range. The right side ranges over $[-\sqrt{2}, \sqrt{2}]$, so solutions exist iff $|k| \leq \sqrt{2}$.

Takeaway: "Has a solution" questions are range questions — no solving needed.

Example 10: sin 18° derivation

Prove that $\sin 18° = \frac{\sqrt{5} - 1}{4}$.

Solution:

Step 1 — Set up the angle equation. Let $\theta = 18°$; then $5\theta = 90°$, so $2\theta = 90° - 3\theta$.

Step 2 — Take sines of both sides. $\sin 2\theta = \sin(90° - 3\theta) = \cos 3\theta$, i.e. $2\sin\theta\cos\theta = 4\cos^3\theta - 3\cos\theta$.

Step 3 — Divide by $\cos\theta$ (non-zero for 18°). $2\sin\theta = 4\cos^2\theta - 3 = 4(1 - \sin^2\theta) - 3 = 1 - 4\sin^2\theta$.

Step 4 — Solve the quadratic. With $s = \sin 18°$: $4s^2 + 2s - 1 = 0$, so $s = \frac{-2 + \sqrt{4 + 16}}{8} = \frac{-1 + \sqrt{5}}{4}$ — the positive root, since 18° is acute.

Takeaway: The 5θ = 90° trick converts a value question into a quadratic — same method yields cos 36° = $\frac{\sqrt{5}+1}{4}$.

Example 11: A value chain with 18°-36°

Evaluate $\sin 18° \cos 36°$ exactly.

Solution:

Step 1 — Substitute both surds. $\sin 18°\cos 36° = \frac{\sqrt{5} - 1}{4} \cdot \frac{\sqrt{5} + 1}{4}$.

Step 2 — Multiply the conjugates. $= \frac{(\sqrt{5})^2 - 1^2}{16} = \frac{4}{16} = \frac{1}{4}$.

Takeaway: The conjugate pair multiplies to $\frac{1}{4}$ — quote $\sin 18°\cos 36° = \frac{1}{4}$ as a standard result.

Example 12: Conditional identity — the cos family

If A + B + C = $\pi$, prove that $\cos A + \cos B + \cos C = 1 + 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$.

Solution:

Step 1 — Pair the first two. $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} = 2\sin\frac{C}{2}\cos\frac{A-B}{2}$, using $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.

Step 2 — Open the third with a half-angle. $\cos C = 1 - 2\sin^2\frac{C}{2}$.

Step 3 — Factor $2\sin\frac{C}{2}$. Total $= 1 + 2\sin\frac{C}{2}\left[\cos\frac{A-B}{2} - \sin\frac{C}{2}\right]$, and substitute $\sin\frac{C}{2} = \cos\frac{A+B}{2}$.

Step 4 — Sum-to-product on the bracket. $\cos\frac{A-B}{2} - \cos\frac{A+B}{2} = 2\sin\frac{A}{2}\sin\frac{B}{2}$. Total: $1 + 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$. ∎

Takeaway: The half-angle conversions $\sin\frac{C}{2} = \cos\frac{A+B}{2}$ are the hinge of every half-angle conditional identity.

Example 13: tan identity application

In a triangle, tan A = 1 and tan B = 2. Find tan C.

Solution:

Step 1 — Quote the triangle-tangent identity. $\tan A + \tan B + \tan C = \tan A\tan B\tan C$.

Step 2 — Substitute the data. $1 + 2 + \tan C = 1 \cdot 2 \cdot \tan C$, i.e. $3 + \tan C = 2\tan C$.

Step 3 — Solve. $\tan C = 3$.

Takeaway: With two tangents known, the identity is a one-line linear equation for the third — no angles needed.

Example 14: Counting solutions in an interval

How many solutions does $2\sin^2 x - \sin x - 1 = 0$ have in $[0, 2\pi]$?

Solution:

Step 1 — Factor. $(2\sin x + 1)(\sin x - 1) = 0$: $\sin x = -\frac{1}{2}$ or $\sin x = 1$.

Step 2 — Count the first family. $\sin x = -\frac{1}{2}$ in $[0, 2\pi]$: two crossings, at $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$.

Step 3 — Count the second. $\sin x = 1$: one touch, at $\frac{\pi}{2}$ (the peak counts once).

Step 4 — Total. $2 + 1 = 3$ solutions.

Takeaway: JEE asks for the COUNT more often than the list — factor, then count level-crossings per family on the sine graph.

Example 15: A range question with a square

Find the range of $f(x) = \sin^2 x + \sin x + 1$.

Solution:

Step 1 — Substitute. Let $t = \sin x \in [-1, 1]$: $f = t^2 + t + 1$.

Step 2 — Complete the square. $f = \left(t + \frac{1}{2}\right)^2 + \frac{3}{4}$ — a parabola with vertex at $t = -\frac{1}{2}$, which LIES INSIDE $[-1, 1]$.

Step 3 — Evaluate vertex and endpoints. Vertex: $\frac{3}{4}$; endpoints: $f(1) = 3$, $f(-1) = 1$.

Step 4 — Assemble the range. Minimum $\frac{3}{4}$ (vertex), maximum 3 (endpoint t = 1): range $\left[\frac{3}{4}, 3\right]$.

Takeaway: Quadratic-in-sine ranges = parabola on $[-1, 1]$: check the vertex if it lies inside, plus both endpoints. The maximum is at an ENDPOINT here, not the vertex.