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CLASS 11 PHYSICS • BASIC MATHEMATICS — PART 3 OF 7
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1: Algebra
2: Logarithms
3: Trigonometry
4: Coord. Geometry
5: Differentiation
6: Integration
7: Vectors
TRIGONOMETRY
WHY THIS MATTERS
Trigonometry is the language of angles and directions . Resolving forces into components, calculating work done by a force, expressing Simple Harmonic Motion, finding the path of projectiles — all require trigonometry. It is literally impossible to solve Class 11 mechanics without it.
§1. Angle Measurement: Degrees & Radians
Angles can be measured in two systems. Physics equations involving calculus and oscillations use Radians .
CONVERSION FORMULA
$$\pi \text{ radians} = 180°$$
$1 \text{ rad} = \frac{180°}{\pi} \approx 57.3°$ $1° = \frac{\pi}{180} \text{ rad}$
$30° = \frac{\pi}{6}$ rad
$45° = \frac{\pi}{4}$ rad
$60° = \frac{\pi}{3}$ rad
$90° = \frac{\pi}{2}$ rad
$180° = \pi$ rad
$360° = 2\pi$ rad
§2. Trigonometric Ratios (Right-Angle Triangle)
Base (Adjacent) = b
Perp = p
Hypotenuse = h
θ
$\sin\theta = \frac{p}{h}$ (Perp/Hyp)
$\cos\theta = \frac{b}{h}$ (Base/Hyp)
$\tan\theta = \frac{p}{b} = \frac{\sin\theta}{\cos\theta}$
$\csc\theta = \frac{h}{p} = \frac{1}{\sin\theta}$
$\sec\theta = \frac{h}{b} = \frac{1}{\cos\theta}$
$\cot\theta = \frac{b}{p} = \frac{1}{\tan\theta}$
Mnemonic
"SOH-CAH-TOA" : S ine = O pposite/H ypotenuse | C osine = A djacent/H ypotenuse | T angent = O pposite/A djacent
§3. Standard Trigonometric Values
Angle (θ) 0° 30° 45° 60° 90° 120° 180° 270° 360°
sin θ 0 1/2 1/√2 √3/2 1 √3/2 0 -1 0
cos θ 1 √3/2 1/√2 1/2 0 -1/2 -1 0 1
tan θ 0 1/√3 1 √3 ∞ -√3 0 ∞ 0
Memory Trick
For sin: 0°, 30°, 45°, 60°, 90° → $\sqrt{0/4}, \sqrt{1/4}, \sqrt{2/4}, \sqrt{3/4}, \sqrt{4/4}$ = 0, ½, 1/√2, √3/2, 1. For cos, read the sin table backwards!
The 37°-53° Special Physics Triangle (3-4-5)
⭐ USED EVERYWHERE IN PHYSICS
$\sin 37° = \cos 53° = \frac{3}{5} = 0.6$
$\cos 37° = \sin 53° = \frac{4}{5} = 0.8$
$\tan 37° = \frac{3}{4} = 0.75$
$\tan 53° = \frac{4}{3} \approx 1.33$
The sides of the right triangle are in ratio 3 : 4 : 5 .
§4. ASTC Rule — Signs in Quadrants
II
I
III
IV
sin +
(S)
All +
(A)
tan +
(T)
cos +
(C)
0
90°/270°
0°
180°
Mnemonic: A–S–T–C going from Quadrant I → II → III → IV = "All Students Take Calculus"
Allied Angle Reductions
$\sin(180° - \theta) = \sin\theta$
$\cos(180° - \theta) = -\cos\theta$
$\sin(90° + \theta) = \cos\theta$
$\cos(90° + \theta) = -\sin\theta$
$\sin(-\theta) = -\sin\theta$
$\cos(-\theta) = \cos\theta$
$\sin(270° - \theta) = -\cos\theta$
$\cos(270° - \theta) = -\sin\theta$
§5. Fundamental Identities
THREE PYTHAGOREAN IDENTITIES
$\sin^2\theta + \cos^2\theta = 1$
$1 + \tan^2\theta = \sec^2\theta$
$1 + \cot^2\theta = \csc^2\theta$
§6. Addition & Double Angle Formulas
COMPOUND ANGLES
$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$
$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$
$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$
DOUBLE ANGLE FORMULAS
$\sin(2\theta) = 2\sin\theta\cos\theta$
$\cos(2\theta) = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$
$\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}$
⚛️ Physics Use — Projectile Motion: Range $R = \frac{u^2\sin 2\theta}{g}$. Maximum range at $\theta = 45°$ because $\sin 90° = 1$. This formula comes directly from the double angle identity!
§7. Small Angle Approximations
⭐ CRITICAL PHYSICS TOOL
When $\theta$ is very small (measured in radians):
$$\sin\theta \approx \theta \qquad \tan\theta \approx \theta \qquad \cos\theta \approx 1$$
Used in: Pendulum SHM, Optics (small angles), Elasticity.
Simple Pendulum Derivation:
Restoring force $F = -mg\sin\theta$. For small angles $\sin\theta \approx \theta = \frac{x}{L}$.
$\therefore F \approx -mg\frac{x}{L} = -\left(\frac{mg}{L}\right)x$. This is SHM!
§8. Graphs of Trig Functions
Graphs of sin x and cos x over [0, 2π]
Practice Questions
DRILL 1 — VALUES & CONVERSION
1. Convert 240° to radians.
2. Find the value of $\sin 150° + \cos 120°$.
3. Find $\tan 315°$.
4. If $\sin\theta = 3/5$ and $\theta$ is in Quadrant II, find $\cos\theta$ and $\tan\theta$.
5. Find the value of $\cos(-60°)$.
DRILL 2 — IDENTITIES & FORMULAS
6. Prove: $\sin^4\theta - \cos^4\theta = \sin^2\theta - \cos^2\theta$
7. Simplify: $\frac{1 - \cos^2\theta}{\sin\theta}$
8. Find $\sin 75°$ using addition formula.
9. Find $\cos 2\theta$ if $\sin\theta = 0.6$.
10. Prove: $\frac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta$
DRILL 3 — PHYSICS APPLICATIONS
11. A force of 50 N acts at 37° to the horizontal. Find its horizontal and vertical components.
12. A projectile is launched at 60° with speed 20 m/s. Find the horizontal range ($g = 10 \text{ m/s}^2$).
13. Using small angle approx, find $\sin(0.1 \text{ rad})$ approximately.
14. The angle of a pendulum is $\theta = 0.1\sin(2t)$ radians. Find the approximate restoring force if $mg = 5$ N and $L = 0.5$ m.
15. A ramp makes 53° with horizontal. A box of mass 10 kg slides on it. Find the component of gravity along and perpendicular to the ramp.
DRILL 4 — ADVANCED PROBLEMS
16. If $\tan A = 1/2$ and $\tan B = 1/3$, show that $A + B = 45°$.
17. Find all angles in $[0°, 360°]$ satisfying $2\sin^2\theta - \sin\theta - 1 = 0$.
18. Prove: $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$
19. In a triangle, $A = 60°$, $B = 75°$. Find $\sin C$ and then $\cos C$.
20. The intensity of light through a polarizer is $I = I_0\cos^2\theta$. At what angle is $I = I_0/4$?
Answer Key
Q
Answer & Method
1 $240° = \frac{4\pi}{3}$ rad
2 $\sin 150° + \cos 120° = 1/2 + (-1/2) = \mathbf{0}$
3 $\tan 315° = \tan(360°-45°) = -\tan 45° = \mathbf{-1}$
4 Q II: $\cos\theta = -4/5$, $\tan\theta = -3/4$
5 $\cos(-60°) = \cos 60° = \mathbf{1/2}$
7 $\frac{\sin^2\theta}{\sin\theta} = \mathbf{\sin\theta}$
8 $\sin 75° = \sin(45°+30°) = \frac{\sqrt{6}+\sqrt{2}}{4} \approx \mathbf{0.966}$
9 $\cos 2\theta = 1 - 2\sin^2\theta = 1 - 2(0.36) = \mathbf{0.28}$
11 $F_x = 50\cos 37° = 50(0.8) = \mathbf{40\text{ N}}$, $F_y = 50\sin 37° = 50(0.6) = \mathbf{30\text{ N}}$
12 $R = \frac{400 \sin 120°}{10} = 40 \times \frac{\sqrt{3}}{2} \approx \mathbf{34.6\text{ m}}$
13 $\sin(0.1) \approx \mathbf{0.1}$ rad
15 Along ramp: $mg\sin 53° = 10(10)(0.8) = \mathbf{80\text{ N}}$, Perpendicular: $mg\cos 53° = 10(10)(0.6) = \mathbf{60\text{ N}}$
16 $\tan(A+B) = \frac{1/2+1/3}{1-1/6} = \frac{5/6}{5/6} = 1 \implies A+B = 45°$
17 $(2\sin\theta+1)(\sin\theta-1)=0 \implies \theta = \mathbf{90°, 210°, 330°}$
19 $C = 180°-60°-75°=45°$. $\sin C = 1/\sqrt{2}$, $\cos C = 1/\sqrt{2}$
20 $\cos^2\theta = 1/4 \implies \cos\theta = 1/2 \implies \theta = \mathbf{60°}$