Trigonometry
Master JEE trigonometry — ratios, identities, equations, inverse functions, properties of triangles, and heights & distances with competitive-level tricks and shortcuts.
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01
Trigonometric Ratios & Identities
Compound angles, multiple angles, half angles, sum-to-product, and product-to-sum formulas — the identity toolkit for JEE.
Fundamental & Compound Angle Identities
Fundamental Identities
$\sin^2\theta + \cos^2\theta = 1$
$1 + \tan^2\theta = \sec^2\theta$
$1 + \cot^2\theta = \csc^2\theta$
$1 + \tan^2\theta = \sec^2\theta$
$1 + \cot^2\theta = \csc^2\theta$
Compound Angle Formulas
$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$
$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$
$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$
$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$
$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$
Multiple Angle Formulas
$\sin 2A = 2\sin A\cos A = \frac{2\tan A}{1+\tan^2 A}$
$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A = \frac{1-\tan^2 A}{1+\tan^2 A}$
$\tan 2A = \frac{2\tan A}{1-\tan^2 A}$
$\sin 3A = 3\sin A - 4\sin^3 A$
$\cos 3A = 4\cos^3 A - 3\cos A$
$\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$
$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A = \frac{1-\tan^2 A}{1+\tan^2 A}$
$\tan 2A = \frac{2\tan A}{1-\tan^2 A}$
$\sin 3A = 3\sin A - 4\sin^3 A$
$\cos 3A = 4\cos^3 A - 3\cos A$
$\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$
Half Angle & Auxiliary Angle
$\sin^2\frac{A}{2} = \frac{1-\cos A}{2}$, $\cos^2\frac{A}{2} = \frac{1+\cos A}{2}$, $\tan\frac{A}{2} = \frac{\sin A}{1+\cos A} = \frac{1-\cos A}{\sin A}$.
JEE Trick — Auxiliary angle: $a\sin\theta+b\cos\theta = \sqrt{a^2+b^2}\sin(\theta+\phi)$ where $\tan\phi = \frac{b}{a}$. Max value $= \sqrt{a^2+b^2}$, min $= -\sqrt{a^2+b^2}$.
JEE Trick — Auxiliary angle: $a\sin\theta+b\cos\theta = \sqrt{a^2+b^2}\sin(\theta+\phi)$ where $\tan\phi = \frac{b}{a}$. Max value $= \sqrt{a^2+b^2}$, min $= -\sqrt{a^2+b^2}$.
Sum/Product Formulas
$\sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}$
$\cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$
$2\sin A\cos B = \sin(A+B)+\sin(A-B)$
$2\cos A\cos B = \cos(A-B)+\cos(A+B)$
$\cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$
$2\sin A\cos B = \sin(A+B)+\sin(A-B)$
$2\cos A\cos B = \cos(A-B)+\cos(A+B)$
📝 Example
Find the maximum value of $3\sin\theta + 4\cos\theta$.
Using auxiliary angle: max $= \sqrt{3^2+4^2} = \sqrt{25} = \boxed{5}$.
02
Trigonometric Equations
General solutions for trig equations — systematic methods and special cases critical for JEE.
General Solutions
Standard General Solutions
$\sin\theta = \sin\alpha \Rightarrow \theta = n\pi + (-1)^n\alpha$, $n \in \mathbb{Z}$.
$\cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha$, $n \in \mathbb{Z}$.
$\tan\theta = \tan\alpha \Rightarrow \theta = n\pi + \alpha$, $n \in \mathbb{Z}$.
$\cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha$, $n \in \mathbb{Z}$.
$\tan\theta = \tan\alpha \Rightarrow \theta = n\pi + \alpha$, $n \in \mathbb{Z}$.
Special Cases
$\sin\theta = 0 \Rightarrow \theta = n\pi$
$\cos\theta = 0 \Rightarrow \theta = (2n+1)\frac{\pi}{2}$
$\tan\theta = 0 \Rightarrow \theta = n\pi$
$\sin\theta = 1 \Rightarrow \theta = 2n\pi + \frac{\pi}{2}$
$\cos\theta = 1 \Rightarrow \theta = 2n\pi$
$\cos\theta = 0 \Rightarrow \theta = (2n+1)\frac{\pi}{2}$
$\tan\theta = 0 \Rightarrow \theta = n\pi$
$\sin\theta = 1 \Rightarrow \theta = 2n\pi + \frac{\pi}{2}$
$\cos\theta = 1 \Rightarrow \theta = 2n\pi$
JEE Tricks for Solving Trig Equations
1. Always check for extraneous solutions when squaring both sides.
2. For $a\sin\theta + b\cos\theta = c$: solution exists iff $|c| \leq \sqrt{a^2+b^2}$.
3. For equations like $\sin^2\theta + \sin^2 2\theta = k$, use double angle and convert to a single trig function.
4. In the interval $[0, 2\pi)$: $\sin\theta = k$ has 2 solutions if $|k| < 1$, 1 solution if $|k|=1$, 0 if $|k|>1$.
2. For $a\sin\theta + b\cos\theta = c$: solution exists iff $|c| \leq \sqrt{a^2+b^2}$.
3. For equations like $\sin^2\theta + \sin^2 2\theta = k$, use double angle and convert to a single trig function.
4. In the interval $[0, 2\pi)$: $\sin\theta = k$ has 2 solutions if $|k| < 1$, 1 solution if $|k|=1$, 0 if $|k|>1$.
📝 Example
Solve $2\cos^2\theta + 3\sin\theta = 0$ for general solution.
$2(1-\sin^2\theta)+3\sin\theta=0 \Rightarrow 2\sin^2\theta-3\sin\theta-2=0 \Rightarrow (2\sin\theta+1)(\sin\theta-2)=0$.
$\sin\theta = -\frac{1}{2}$ (since $\sin\theta = 2$ is impossible).
$\theta = n\pi+(-1)^n\left(-\frac{\pi}{6}\right) = n\pi-(-1)^n\frac{\pi}{6}$.
$\sin\theta = -\frac{1}{2}$ (since $\sin\theta = 2$ is impossible).
$\theta = n\pi+(-1)^n\left(-\frac{\pi}{6}\right) = n\pi-(-1)^n\frac{\pi}{6}$.
03
Inverse Trigonometric Functions
Domain, range, principal values, and key properties of inverse trig functions — high-weightage in both Mains and Advanced.
Domain, Range & Properties
Principal Value Branches
$\sin^{-1}x$: domain $[-1,1]$, range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
$\cos^{-1}x$: domain $[-1,1]$, range $[0, \pi]$.
$\tan^{-1}x$: domain $\mathbb{R}$, range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
$\cot^{-1}x$: domain $\mathbb{R}$, range $(0, \pi)$.
$\sec^{-1}x$: domain $|x|\geq1$, range $[0,\pi]\setminus\{\frac{\pi}{2}\}$.
$\csc^{-1}x$: domain $|x|\geq1$, range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\setminus\{0\}$.
$\cos^{-1}x$: domain $[-1,1]$, range $[0, \pi]$.
$\tan^{-1}x$: domain $\mathbb{R}$, range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
$\cot^{-1}x$: domain $\mathbb{R}$, range $(0, \pi)$.
$\sec^{-1}x$: domain $|x|\geq1$, range $[0,\pi]\setminus\{\frac{\pi}{2}\}$.
$\csc^{-1}x$: domain $|x|\geq1$, range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\setminus\{0\}$.
Key Identities
$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1,1]$.
$\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ for all $x$.
$\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}$ for $|x|\geq1$.
$\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ for all $x$.
$\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}$ for $|x|\geq1$.
Addition Formulas
$\tan^{-1}x + \tan^{-1}y = \begin{cases}\tan^{-1}\frac{x+y}{1-xy} & xy < 1 \\ \pi+\tan^{-1}\frac{x+y}{1-xy} & x>0, xy>1 \\ -\pi+\tan^{-1}\frac{x+y}{1-xy} & x<0, xy>1\end{cases}$
JEE Trick: $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x-y}{1+xy}$ when $xy > -1$.
JEE Trick: $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x-y}{1+xy}$ when $xy > -1$.
JEE Shortcut — Double Angle Forms
$2\tan^{-1}x = \sin^{-1}\frac{2x}{1+x^2} = \cos^{-1}\frac{1-x^2}{1+x^2} = \tan^{-1}\frac{2x}{1-x^2}$ (for $|x|\leq1$).
Caution: These hold only in specific domains — always verify the range condition.
Caution: These hold only in specific domains — always verify the range condition.
📝 Example
Simplify $\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}$.
$xy = \frac{1}{6} < 1$. So $\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3} = \tan^{-1}\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}} = \tan^{-1}\frac{\frac{5}{6}}{\frac{5}{6}} = \tan^{-1}1 = \boxed{\frac{\pi}{4}}$.
04
Properties of Triangles
Sine rule, cosine rule, area formulas, circumradius, inradius — the complete triangle toolkit for JEE.
Sine Rule, Cosine Rule & Area
Sine Rule
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$
where $R$ is the circumradius. JEE Trick: Use $a = 2R\sin A$ to convert between sides and angles.
Cosine Rule
$\cos A = \frac{b^2+c^2-a^2}{2bc}$, $\cos B = \frac{a^2+c^2-b^2}{2ac}$, $\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Equivalently: $a^2 = b^2+c^2-2bc\cos A$.
Equivalently: $a^2 = b^2+c^2-2bc\cos A$.
Area of Triangle
$\Delta = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B$.
Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$.
Also: $\Delta = \frac{abc}{4R}$ and $\Delta = rs$, where $r$ is the inradius.
Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$.
Also: $\Delta = \frac{abc}{4R}$ and $\Delta = rs$, where $r$ is the inradius.
Circumradius & Inradius
$R = \frac{a}{2\sin A} = \frac{abc}{4\Delta}$.
$r = \frac{\Delta}{s} = (s-a)\tan\frac{A}{2} = 4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$.
$r = \frac{\Delta}{s} = (s-a)\tan\frac{A}{2} = 4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$.
Half-Angle Formulas
$\sin\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$, $\cos\frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}$, $\tan\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$.
JEE Shortcut: $\tan\frac{A}{2} = \frac{r}{s-a} = \frac{\Delta}{s(s-a)}$.
JEE Shortcut: $\tan\frac{A}{2} = \frac{r}{s-a} = \frac{\Delta}{s(s-a)}$.
📝 Example
In a triangle $ABC$, $a=7$, $b=8$, $c=9$. Find the circumradius $R$.
$s=12$. $\Delta = \sqrt{12 \cdot 5 \cdot 4 \cdot 3} = \sqrt{720} = 12\sqrt{5}$.
$R = \frac{abc}{4\Delta} = \frac{7 \cdot 8 \cdot 9}{4 \cdot 12\sqrt{5}} = \frac{504}{48\sqrt{5}} = \frac{21}{2\sqrt{5}} = \frac{21\sqrt{5}}{10}$.
$R = \frac{abc}{4\Delta} = \frac{7 \cdot 8 \cdot 9}{4 \cdot 12\sqrt{5}} = \frac{504}{48\sqrt{5}} = \frac{21}{2\sqrt{5}} = \frac{21\sqrt{5}}{10}$.
05
Heights & Distances
Angle of elevation, angle of depression, and multi-step height/distance problems — a JEE Mains favourite.
Heights & Distances
Key Concepts
Angle of elevation: The angle above the horizontal from the observer to the object.
Angle of depression: The angle below the horizontal from the observer to the object.
Note: Angle of depression from A to B = Angle of elevation from B to A (alternate interior angles).
Angle of depression: The angle below the horizontal from the observer to the object.
Note: Angle of depression from A to B = Angle of elevation from B to A (alternate interior angles).
JEE Strategy for Heights & Distances
1. Draw a clear diagram marking all known angles and distances.
2. Identify right triangles and apply $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$.
3. For problems with two observation points, set up two equations and eliminate the unknown distance.
4. When the base is not directly below, use the sine or cosine rule in the non-right triangle.
2. Identify right triangles and apply $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$.
3. For problems with two observation points, set up two equations and eliminate the unknown distance.
4. When the base is not directly below, use the sine or cosine rule in the non-right triangle.
Common Configuration
Tower of height $h$ on a horizontal plane. From point $A$ at distance $d$, angle of elevation is $\alpha$. Then $h = d\tan\alpha$.
From two points $A, B$ at distances $d_1, d_2$: $h = \frac{d_1 d_2(\tan\alpha-\tan\beta)}{d_1-d_2}$ (when both on same side).
JEE Trick: If angles of elevation from two points on opposite sides are $\alpha$ and $\beta$, and the distance between them is $d$: $h = \frac{d}{\cot\alpha+\cot\beta}$.
From two points $A, B$ at distances $d_1, d_2$: $h = \frac{d_1 d_2(\tan\alpha-\tan\beta)}{d_1-d_2}$ (when both on same side).
JEE Trick: If angles of elevation from two points on opposite sides are $\alpha$ and $\beta$, and the distance between them is $d$: $h = \frac{d}{\cot\alpha+\cot\beta}$.
📝 Example
The angles of elevation of the top of a tower from two points at distances $a$ and $b$ from the base (on the same side) are $\alpha$ and $\beta$ respectively. Prove that the height of the tower is $\sqrt{ab\tan\alpha\tan\beta}$ when $\alpha+\beta=90°$.
$h = a\tan\alpha$ and $h = b\tan\beta$. So $h^2 = ab\tan\alpha\tan\beta$.
When $\alpha+\beta=90°$: $\tan\beta = \cot\alpha$, so $h^2 = ab\tan\alpha\cot\alpha = ab$, giving $h = \sqrt{ab}$.
In general, $h = \sqrt{ab\tan\alpha\tan\beta}$.
When $\alpha+\beta=90°$: $\tan\beta = \cot\alpha$, so $h^2 = ab\tan\alpha\cot\alpha = ab$, giving $h = \sqrt{ab}$.
In general, $h = \sqrt{ab\tan\alpha\tan\beta}$.
★ Key Takeaways
- The auxiliary angle method $a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\sin(\theta+\phi)$ instantly gives max/min values — use it in JEE to save time.
- For trig equations, always check for extraneous solutions after squaring and verify solutions lie in the given domain.
- $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$ is the single most used inverse trig identity — it simplifies many JEE problems in one step.
- The relations $\Delta = rs = \frac{abc}{4R}$ connect area, inradius, circumradius, and sides — choose whichever form has the most known quantities.
- In heights and distances, always draw the diagram first and identify all right triangles before writing equations.
✎ Practice Problems
Problem 1
Prove that $\frac{\sin 5\theta - 2\sin 3\theta + \sin\theta}{\cos 5\theta - \cos\theta} = \tan\theta$.
Show Solution ▼
Numerator: $\sin5\theta+\sin\theta-2\sin3\theta = 2\sin3\theta\cos2\theta-2\sin3\theta = 2\sin3\theta(\cos2\theta-1) = -4\sin3\theta\sin^2\theta$.
Denominator: $\cos5\theta-\cos\theta = -2\sin3\theta\sin2\theta = -4\sin3\theta\sin\theta\cos\theta$.
Ratio $= \frac{-4\sin3\theta\sin^2\theta}{-4\sin3\theta\sin\theta\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta$.
Denominator: $\cos5\theta-\cos\theta = -2\sin3\theta\sin2\theta = -4\sin3\theta\sin\theta\cos\theta$.
Ratio $= \frac{-4\sin3\theta\sin^2\theta}{-4\sin3\theta\sin\theta\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta$.
Problem 2
Find the general solution of $\sin\theta + \sin3\theta + \sin5\theta = 0$.
Show Solution ▼
$(\sin\theta+\sin5\theta)+\sin3\theta = 2\sin3\theta\cos2\theta+\sin3\theta = \sin3\theta(2\cos2\theta+1)=0$.
Case 1: $\sin3\theta=0 \Rightarrow 3\theta=n\pi \Rightarrow \theta=\frac{n\pi}{3}$.
Case 2: $\cos2\theta=-\frac{1}{2} \Rightarrow 2\theta=2n\pi\pm\frac{2\pi}{3} \Rightarrow \theta=n\pi\pm\frac{\pi}{3}$.
Combined: $\theta = \frac{n\pi}{3}$ (which includes both cases).
Case 1: $\sin3\theta=0 \Rightarrow 3\theta=n\pi \Rightarrow \theta=\frac{n\pi}{3}$.
Case 2: $\cos2\theta=-\frac{1}{2} \Rightarrow 2\theta=2n\pi\pm\frac{2\pi}{3} \Rightarrow \theta=n\pi\pm\frac{\pi}{3}$.
Combined: $\theta = \frac{n\pi}{3}$ (which includes both cases).
Problem 3
Find the value of $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$.
Show Solution ▼
$\tan^{-1}2+\tan^{-1}3$: here $xy=6>1$ and $x>0$, so $= \pi+\tan^{-1}\frac{2+3}{1-6}=\pi+\tan^{-1}(-1)=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$.
Total $= \frac{\pi}{4}+\frac{3\pi}{4} = \pi$.
Total $= \frac{\pi}{4}+\frac{3\pi}{4} = \pi$.
Problem 4
In triangle $ABC$, if $a=5$, $b=7$, $C=60°$, find $c$ and the area.
Show Solution ▼
$c^2 = a^2+b^2-2ab\cos C = 25+49-70\cdot\frac{1}{2} = 74-35 = 39$. So $c=\sqrt{39}$.
$\Delta = \frac{1}{2}ab\sin C = \frac{1}{2}\cdot5\cdot7\cdot\frac{\sqrt{3}}{2} = \frac{35\sqrt{3}}{4}$.
$\Delta = \frac{1}{2}ab\sin C = \frac{1}{2}\cdot5\cdot7\cdot\frac{\sqrt{3}}{2} = \frac{35\sqrt{3}}{4}$.
Problem 5
From the top of a cliff 200 m high, the angles of depression of two ships are $30°$ and $60°$. Find the distance between the ships if they are on the same side of the cliff.
Show Solution ▼
Let distances from base be $d_1$ and $d_2$ ($d_1 > d_2$).
$\tan30°=\frac{200}{d_1} \Rightarrow d_1=200\sqrt{3}$. $\tan60°=\frac{200}{d_2} \Rightarrow d_2=\frac{200}{\sqrt{3}}=\frac{200\sqrt{3}}{3}$.
Distance $= d_1-d_2 = 200\sqrt{3}-\frac{200\sqrt{3}}{3} = \frac{400\sqrt{3}}{3} \approx 230.9$ m.
$\tan30°=\frac{200}{d_1} \Rightarrow d_1=200\sqrt{3}$. $\tan60°=\frac{200}{d_2} \Rightarrow d_2=\frac{200}{\sqrt{3}}=\frac{200\sqrt{3}}{3}$.
Distance $= d_1-d_2 = 200\sqrt{3}-\frac{200\sqrt{3}}{3} = \frac{400\sqrt{3}}{3} \approx 230.9$ m.
🎯 Interactive MCQs
1. If $\sin\theta + \cos\theta = \sqrt{2}$, then $\tan\theta + \cot\theta$ equals:
A $\sqrt{2}$
B $2$
C $1$
D $2\sqrt{2}$
2. The number of solutions of $2\sin^2\theta - 5\sin\theta + 2 = 0$ in $[0, 2\pi)$ is:
A $1$
B $2$
C $3$
D $4$
3. The value of $\sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} + \sin^{-1}\frac{16}{65}$ is:
A $\frac{\pi}{4}$
B $\frac{\pi}{2}$
C $\frac{3\pi}{4}$
D $\pi$
4. In triangle $ABC$, if $\frac{a}{\cos A} = \frac{b}{\cos B} = \frac{c}{\cos C}$, then the triangle is:
A Isosceles
B Equilateral
C Right-angled
D Scalene
5. A tower subtends an angle of $60°$ at a point on the ground. On walking 100 m away, it subtends $30°$. The height of the tower is:
A $50\sqrt{3}$ m
B $100\sqrt{3}$ m
C $100$ m
D $75$ m