Class 11 FSc Math Chapter 9 Fundamentals of Trigonometry Proofs & Solved Exercise 9.3 Notes PDF - FBISE & Punjab Boards

Theoretical Foundations & Core Theorems

Trigonometry, derived from the Greek words trigonon (triangle) and metron (measure), forms the mathematical backbone of calculus, classical mechanics, wave motion, and electrical engineering. In Class 11 FSc Mathematics (Chapter 9), Exercise 9.3 specifically addresses the evaluation of trigonometric functions at standard special angles ($0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ or $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}$ radians), quadrantal angles, and coterminal angles.

1. Geometrical Derivations of Special Angles

To establish rigor in mathematical proofs, trigonometric ratios for standard values are derived geometrically rather than memorized empirically.

A. Derivation for Angle $45^\circ$ ($\frac{\pi}{4}$ radians)

Consider an isosceles right-angled triangle $\Delta ABC$ where $\angle C = 90^\circ$ and the two legs are of equal length: $AC = BC = 1$.

By the Pythagorean Theorem:

$$AB^2 = AC^2 + BC^2 = 1^2 + 1^2 = 2 \implies AB = \sqrt{2}$$

Since $\Delta ABC$ is isosceles and right-angled, $\angle A = \angle B = 45^\circ$. From the definitions of trigonometric ratios:

$$\sin 45^\circ = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}}$$ $$\cos 45^\circ = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}}$$ $$\tan 45^\circ = \frac{\text{Perpendicular}}{\text{Base}} = \frac{1}{1} = 1$$

B. Derivation for Angles $30^\circ$ ($\frac{\pi}{6}$) and $60^\circ$ ($\frac{\pi}{3}$)

Consider an equilateral triangle $\Delta ABC$ with side length $2$. Drop a perpendicular AD from vertex $A$ to side $BC$. This altitude bisects the base $BC$ and the vertex angle $\angle A$.

Therefore, $BD = CD = 1$, $\angle B = 60^\circ$, and $\angle BAD = 30^\circ$.

Using the Pythagorean theorem on right-angled triangle $\Delta ABD$:

$$AD = \sqrt{AB^2 - BD^2} = \sqrt{2^2 - 1^2} = \sqrt{3}$$

Now, evaluating ratios for $30^\circ$ from $\Delta ABD$:

$$\sin 30^\circ = \frac{BD}{AB} = \frac{1}{2}, \quad \cos 30^\circ = \frac{AD}{AB} = \frac{\sqrt{3}}{2}, \quad \tan 30^\circ = \frac{BD}{AD} = \frac{1}{\sqrt{3}}$$

Evaluating ratios for $60^\circ$ from $\Delta ABD$:

$$\sin 60^\circ = \frac{AD}{AB} = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{BD}{AB} = \frac{1}{2}, \quad \tan 60^\circ = \frac{AD}{BD} = \sqrt{3}$$

2. Trigonometric Functions of Quadrantal Angles

A quadrantal angle is an angle in standard position whose terminal side lies on either the x-axis or y-axis. These include $0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ$ ($0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$).

Consider a point $P(x,y)$ on the unit circle ($r = 1$). The trigonometric functions are defined as:

$$\cos \theta = x, \quad \sin \theta = y, \quad \tan \theta = \frac{y}{x} \quad (x \neq 0)$$

• At $\theta = 0^\circ \ (1, 0)$: $\cos 0^\circ = 1$, $\sin 0^\circ = 0$, $\tan 0^\circ = 0$.
• At $\theta = 90^\circ \ (0, 1)$: $\cos 90^\circ = 0$, $\sin 90^\circ = 1$, $\tan 90^\circ = \text{undefined}$.
• At $\theta = 180^\circ \ (-1, 0)$: $\cos 180^\circ = -1$, $\sin 180^\circ = 0$, $\tan 180^\circ = 0$.
• At $\theta = 270^\circ \ (0, -1)$: $\cos 270^\circ = 0$, $\sin 270^\circ = -1$, $\tan 270^\circ = \text{undefined}$.

3. Coterminal Angles & General Periodicity Principle

Two or more angles in standard position having the same terminal side are called coterminal angles. For any angle $\theta$ and integer $k \in \mathbb{Z}$:

$$f(\theta + 2k\pi) = f(\theta)$$

This fundamental periodicity allows us to evaluate trigonometric functions of arbitrarily large negative or positive angles by reducing them to coterminal angles in the domain $[0, 2\pi)$ or $[0^\circ, 360^\circ)$.

Formula Summary & Quick Reference

Key Concept	Formula / Value Condition	Application & Board Note
Fundamental Identity I	$\sin^2 \theta + \cos^2 \theta = 1$	Valid for all real $\theta \in \mathbb{R}$. Used to convert sine to cosine.
Fundamental Identity II	$1 + \tan^2 \theta = \sec^2 \theta$	Valid for $\theta \neq (2k+1)\frac{\pi}{2}$. Crucial for integral proofs.
Fundamental Identity III	$1 + \cot^2 \theta = \csc^2 \theta$	Valid for $\theta \neq k\pi, k \in \mathbb{Z}$. Key identity in Ex 9.3 proofs.
Coterminal Angle Rule	$\theta_{\text{coterminal}} = \theta + 2k\pi \quad (k \in \mathbb{Z})$	Reduces large angles like $750^\circ \to 30^\circ$ before applying ratios.
Special Values ($30^\circ$)	$\sin 30^\circ = \frac{1}{2}, \cos 30^\circ = \frac{\sqrt{3}}{2}, \tan 30^\circ = \frac{1}{\sqrt{3}}$	Exact algebraic radical form required in FBISE/Punjab answers.
Special Values ($45^\circ$)	$\sin 45^\circ = \frac{1}{\sqrt{2}}, \cos 45^\circ = \frac{1}{\sqrt{2}}, \tan 45^\circ = 1$	Note that $\sin 45^\circ = \cos 45^\circ$.
Special Values ($60^\circ$)	$\sin 60^\circ = \frac{\sqrt{3}}{2}, \cos 60^\circ = \frac{1}{2}, \tan 60^\circ = \sqrt{3}$	Complementary relation: $\sin 60^\circ = \cos 30^\circ$.
Signs in Quadrants	Q1: All +, Q2: Sine +, Q3: Tan +, Q4: Cos +	Mnemonic: "Add Sugar To Coffee" or "ASTC".

Step-by-Step Solved Board Exam Questions

Problem 1 (Verification of Special Angle Identity)

Question: Prove that $\sin^2 \frac{\pi}{6} : \sin^2 \frac{\pi}{4} : \sin^2 \frac{\pi}{3} : \sin^2 \frac{\pi}{2} = 1 : 2 : 3 : 4$. (BISE Lahore 2019, FBISE 2021)

Solution:

Step 1: Write down the standard values in radian measure:

$$\sin \frac{\pi}{6} = \sin 30^\circ = \frac{1}{2}$$ $$\sin \frac{\pi}{4} = \sin 45^\circ = \frac{1}{\sqrt{2}}$$ $$\sin \frac{\pi}{3} = \sin 60^\circ = \frac{\sqrt{3}}{2}$$ $$\sin \frac{\pi}{2} = \sin 90^\circ = 1$$

Step 2: Compute the squares of each expression:

$$\sin^2 \frac{\pi}{6} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ $$\sin^2 \frac{\pi}{4} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$$ $$\sin^2 \frac{\pi}{3} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$$ $$\sin^2 \frac{\pi}{2} = (1)^2 = 1$$

Step 3: Construct the continued ratio:

$$\sin^2 \frac{\pi}{6} : \sin^2 \frac{\pi}{4} : \sin^2 \frac{\pi}{3} : \sin^2 \frac{\pi}{2} = \frac{1}{4} : \frac{1}{2} : \frac{3}{4} : 1$$

Step 4: Multiply the entire ratio by the common denominator $4$:

$$4 \times \left(\frac{1}{4} : \frac{1}{2} : \frac{3}{4} : 1\right) = 1 : 2 : 3 : 4$$

Conclusion:

$$\text{L.H.S.} = \text{R.H.S.} \quad \blacksquare$$

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Problem 2 (Evaluation of Fractional Trigonometric Expression)

Question: Evaluate $\frac{\tan \frac{\pi}{3} - \tan \frac{\pi}{6}}{1 + \tan \frac{\pi}{3} \tan \frac{\pi}{6}}$ and show that it equals $\tan \frac{\pi}{6}$. (BISE Multan 2018, Rawalpindi 2022)

Solution:

Step 1: State the values of involved functions:

$$\tan \frac{\pi}{3} = \sqrt{3}, \quad \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$$

Step 2: Substitute values into Left Hand Side (L.H.S.):

$$\text{L.H.S.} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + (\sqrt{3})\left(\frac{1}{\sqrt{3}}\right)}$$

Step 3: Simplify the numerator and denominator independently:

Numerator:

$$\sqrt{3} - \frac{1}{\sqrt{3}} = \frac{(\sqrt{3})(\sqrt{3}) - 1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$$

Denominator:

$$1 + 1 = 2$$

Step 4: Substitute numerator and denominator back into main fraction:

$$\text{L.H.S.} = \frac{\frac{2}{\sqrt{3}}}{2} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$$

Step 5: Compare with Right Hand Side (R.H.S.):

$$\text{R.H.S.} = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$$ $$\text{Since L.H.S.} = \text{R.H.S.}, \text{ the identity is verified.} \quad \blacksquare$$

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Problem 3 (Coterminal Angle Reduction for High-Order Angles)

Question: Find the exact values of all six trigonometric functions for $\theta = \frac{19\pi}{3}$. (BISE Gujranwala 2020, FBISE 2017)

Solution:

Step 1: Express the angle in terms of an integer multiple of $2\pi$:

$$\frac{19\pi}{3} = \frac{18\pi + \pi}{3} = \frac{18\pi}{3} + \frac{\pi}{3} = 6\pi + \frac{\pi}{3} = 3(2\pi) + \frac{\pi}{3}$$

Step 2: Apply periodicity principle ($k = 3 \in \mathbb{Z}$):

Since $3(2\pi)$ represents 3 complete revolutions, $\frac{19\pi}{3}$ is coterminal with $\frac{\pi}{3}$ ($60^\circ$).

Step 3: Compute all six trigonometric functions using $\frac{\pi}{3}$:

$$\sin\left(\frac{19\pi}{3}\right) = \sin\left(6\pi + \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$ $$\cos\left(\frac{19\pi}{3}\right) = \cos\left(6\pi + \frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$ $$\tan\left(\frac{19\pi}{3}\right) = \tan\left(6\pi + \frac{\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$$ $$\csc\left(\frac{19\pi}{3}\right) = \frac{1}{\sin(\pi/3)} = \frac{2}{\sqrt{3}}$$ $$\sec\left(\frac{19\pi}{3}\right) = \frac{1}{\cos(\pi/3)} = 2$$ $$\cot\left(\frac{19\pi}{3}\right) = \frac{1}{\tan(\pi/3)} = \frac{1}{\sqrt{3}} \quad \blacksquare$$

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Problem 4 (Evaluation of Complex Quadrantal Angle Expression)

Question: Find the value of $\theta \in [0, 2\pi]$ satisfying the equation $2\sin^2 \theta - 1 = 0$. (BISE Sargodha 2021)

Solution:

Step 1: Isolate $\sin \theta$:

$$2\sin^2 \theta = 1 \implies \sin^2 \theta = \frac{1}{2}$$ $$\sin \theta = \pm \frac{1}{\sqrt{2}}$$

Step 2: Determine the reference angle $\theta_r$:

$$\sin \theta_r = \left|\pm \frac{1}{\sqrt{2}}\right| = \frac{1}{\sqrt{2}} \implies \theta_r = \frac{\pi}{4}$$

Step 3: Find solutions in all four quadrants (since $\sin \theta$ is both $+$ and $-$):

• Quadrant I ($\sin \theta > 0$): $\theta = \theta_r = \frac{\pi}{4}$
• Quadrant II ($\sin \theta > 0$): $\theta = \pi - \theta_r = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$
• Quadrant III ($\sin \theta < 0$): $\theta = \pi + \theta_r = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$
• Quadrant IV ($\sin \theta < 0$): $\theta = 2\pi - \theta_r = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$

Step 4: Write the complete solution set:

$$\text{Solution Set} = \left\{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\} \quad \blacksquare$$

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

1. How do I systematically handle angles with negative signs like $\sin(-\theta)$ or $\cos(-\theta)$?

Negative angles indicate clockwise rotation from the positive x-axis. Using the fundamental parity properties of trigonometric functions:

• $\cos(-\theta) = \cos \theta$ (Even function)
• $\sec(-\theta) = \sec \theta$ (Even function)
• $\sin(-\theta) = -\sin \theta$ (Odd function)
• $\csc(-\theta) = -\csc \theta$ (Odd function)
• $\tan(-\theta) = -\tan \theta$ (Odd function)
• $\cot(-\theta) = -\cot \theta$ (Odd function)

Always extract or eliminate the negative sign first before performing coterminal reduction.

2. Why are $\tan 90^\circ$ and $\sec 90^\circ$ undefined, while $\sin 90^\circ$ is defined?

On the unit circle, the coordinates of an angle are $(x, y) = (\cos \theta, \sin \theta)$. At $\theta = 90^\circ$, the point is $(0, 1)$, meaning $x = 0$ and $y = 1$. By definition:

$$\tan 90^\circ = \frac{y}{x} = \frac{1}{0} \quad \text{(Undefined / Division by Zero)}$$ $$\sec 90^\circ = \frac{1}{x} = \frac{1}{0} \quad \text{(Undefined / Division by Zero)}$$

Conversely, $\sin 90^\circ = y = 1$, which is perfectly well-defined.

3. What is the step-by-step algorithm to evaluate trigonometric ratios for large angles like $\theta = 4050^\circ$?

Follow this 3-step board algorithm:

1. Divide by $360^\circ$: $\frac{4050}{360} = 11.25$. The integer part is $11$.
2. Subtract multiple of $360^\circ$: $4050^\circ - 11(360^\circ) = 4050^\circ - 3960^\circ = 90^\circ$.
3. Evaluate at coterminal remainder: $f(4050^\circ) = f(90^\circ)$. For example, $\cos 4050^\circ = \cos 90^\circ = 0$.