Class 11 FSc Math Chapter 9 Fundamentals of Trigonometry Proofs and Solved Exercise 9.3 Study Guide PDF | FBISE & Punjab Boards
Trigonometry, derived from the Greek words trigonon (triangle) and metron (measure), forms the backbone of higher mathematical analysis, calculus, and theoretical physics. In FSc Part 1 (Class 11), Chapter 9 shifts the concept of trigonometric ratios from right-angled triangles to general angles in the Cartesian plane using the unit circle model.
Consider a unit circle centered at the origin $(0,0)$ in the $xy$-plane with an equation given by $x^2 + y^2 = 1$. Let a ray originating from the origin rotate counterclockwise by an angle $\theta$ in standard position, intersecting the unit circle at a point $P(x,y)$.
By basic geometric definitions in a right triangle formed by dropping a perpendicular from $P(x,y)$ to the x-axis:
$$\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{x}{1} = x$$ $$\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{y}{1} = y$$ $$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{y}{x} \quad (x \neq 0)$$Thus, every point on the unit circle has coordinates $P(\cos \theta, \sin \theta)$. Substituting these into the unit circle equation yields the fundamental Pythagorean Identity:
$$\cos^2 \theta + \sin^2 \theta = 1$$In Federal Board (FBISE) and Punjab Board examinations (BISE Lahore, Rawalpindi, Multan, etc.), numerical answers derived directly from geometric figures receive full credit, whereas mere decimal values from calculators are awarded zero marks.
An angle is called a quadrantal angle if its terminal side lies on either the x-axis or the y-axis. The standard quadrantal angles are $0^\circ, 90^\circ, 180^\circ, 270^\circ,$ and $360^\circ$ (or $0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ radians).
Using coordinates of intersection on the unit circle:
The signs of the trigonometric functions depend entirely on the algebraic signs of $x$ and $y$ in each quadrant:
Mnemonic Rule: "Add Sugar To Coffee" (ASTC: Quadrants I, II, III, IV respectively).
| Key Concept / Angle | $\sin \theta$ | $\cos \theta$ | $\tan \theta$ | Reciprocal & Quotient Rules |
|---|---|---|---|---|
| $0^\circ$ ($0$ rad) | $0$ | $1$ | $0$ | $\csc 0^\circ = \text{Undefined}, \, \sec 0^\circ = 1, \, \cot 0^\circ = \text{Undefined}$ |
| $30^\circ$ ($\frac{\pi}{6}$ rad) | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{3}}$ | $\csc 30^\circ = 2, \, \sec 30^\circ = \frac{2}{\sqrt{3}}, \, \cot 30^\circ = \sqrt{3}$ |
| $45^\circ$ ($\frac{\pi}{4}$ rad) | $\frac{1}{\sqrt{2}}$ | $\frac{1}{\sqrt{2}}$ | $1$ | $\csc 45^\circ = \sqrt{2}, \, \sec 45^\circ = \sqrt{2}, \, \cot 45^\circ = 1$ |
| $60^\circ$ ($\frac{\pi}{3}$ rad) | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ | $\csc 60^\circ = \frac{2}{\sqrt{3}}, \, \sec 60^\circ = 2, \, \cot 60^\circ = \frac{1}{\sqrt{3}}$ |
| $90^\circ$ ($\frac{\pi}{2}$ rad) | $1$ | $0$ | Undefined | $\csc 90^\circ = 1, \, \sec 90^\circ = \text{Undefined}, \, \cot 90^\circ = 0$ |
| $180^\circ$ ($\pi$ rad) | $0$ | $-1$ | $0$ | $\csc 180^\circ = \text{Undefined}, \, \sec 180^\circ = -1, \, \cot 180^\circ = \text{Undefined}$ |
| $270^\circ$ ($\frac{3\pi}{2}$ rad) | $-1$ | $0$ | Undefined | $\csc 270^\circ = -1, \, \sec 270^\circ = \text{Undefined}, \, \cot 270^\circ = 0$ |
Problem: Evaluate the following expression without using a mathematical table or calculator:
$$\frac{\sin^2 30^\circ + \sin^2 60^\circ + \tan^2 45^\circ}{\sec^2 45^\circ - \csc^2 60^\circ + \cot^2 30^\circ}$$Solution:
Step 1: List the exact trigonometric values for each required term:
$$\sin 30^\circ = \frac{1}{2} \implies \sin^2 30^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ $$\sin 60^\circ = \frac{\sqrt{3}}{2} \implies \sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$$ $$\tan 45^\circ = 1 \implies \tan^2 45^\circ = (1)^2 = 1$$ $$\sec 45^\circ = \sqrt{2} \implies \sec^2 45^\circ = (\sqrt{2})^2 = 2$$ $$\csc 60^\circ = \frac{2}{\sqrt{3}} \implies \csc^2 60^\circ = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}$$ $$\cot 30^\circ = \sqrt{3} \implies \cot^2 30^\circ = (\sqrt{3})^2 = 3$$Step 2: Substitute these values into the numerator:
$$\text{Numerator} = \frac{1}{4} + \frac{3}{4} + 1 = \frac{1 + 3}{4} + 1 = \frac{4}{4} + 1 = 1 + 1 = 2$$Step 3: Substitute these values into the denominator:
$$\text{Denominator} = 2 - \frac{4}{3} + 3 = (2 + 3) - \frac{4}{3} = 5 - \frac{4}{3} = \frac{15 - 4}{3} = \frac{11}{3}$$Step 4: Combine numerator and denominator to get the final fraction:
$$\text{Value} = \frac{2}{\left(\frac{11}{3}\right)} = 2 \times \frac{3}{11} = \frac{6}{11}$$Final Answer: $\frac{6}{11}$
Problem: Verify that $\cos 2\theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta$ when $\theta = 30^\circ$.
Solution:
Part A: Evaluate Left-Hand Side ($\cos 2\theta$)
$$\text{LHS} = \cos(2 \times 30^\circ) = \cos 60^\circ = \frac{1}{2}$$Part B: Evaluate Middle Term ($2\cos^2 \theta - 1$)
$$\text{Middle Term} = 2\cos^2 30^\circ - 1$$ $$\text{Since } \cos 30^\circ = \frac{\sqrt{3}}{2}:$$ $$\text{Middle Term} = 2\left(\frac{\sqrt{3}}{2}\right)^2 - 1 = 2\left(\frac{3}{4}\right) - 1 = \frac{3}{2} - 1 = \frac{1}{2}$$Part C: Evaluate Right-Hand Side ($1 - 2\sin^2 \theta$)
$$\text{RHS} = 1 - 2\sin^2 30^\circ$$ $$\text{Since } \sin 30^\circ = \frac{1}{2}:$$ $$\text{RHS} = 1 - 2\left(\frac{1}{2}\right)^2 = 1 - 2\left(\frac{1}{4}\right) = 1 - \frac{1}{2} = \frac{1}{2}$$Conclusion:
$$\text{LHS} = \text{Middle Term} = \text{RHS} = \frac{1}{2}$$Hence, verified for $\theta = 30^\circ$.
Problem: Prove that $\frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ} = \tan 30^\circ$.
Solution:
Step 1: Consider the Left-Hand Side (LHS):
$$\text{LHS} = \frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}$$Step 2: Substitute exact values $\tan 60^\circ = \sqrt{3}$ and $\tan 30^\circ = \frac{1}{\sqrt{3}}$:
$$\text{Numerator} = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{(\sqrt{3})(\sqrt{3}) - 1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$$ $$\text{Denominator} = 1 + (\sqrt{3})\left(\frac{1}{\sqrt{3}}\right) = 1 + 1 = 2$$Step 3: Simplify the fraction:
$$\text{LHS} = \frac{\left(\frac{2}{\sqrt{3}}\right)}{2} = \frac{2}{\sqrt{3}} \times \frac{1}{2} = \frac{1}{\sqrt{3}}$$Step 4: Evaluate the Right-Hand Side (RHS):
$$\text{RHS} = \tan 30^\circ = \frac{1}{\sqrt{3}}$$Conclusion:
$$\text{LHS} = \text{RHS} = \frac{1}{\sqrt{3}}$$Hence proved.
Problem: If $\sin \theta = -\frac{\sqrt{3}}{2}$ and the terminal arm of angle $\theta$ lies in Quadrant III, find the values of all the remaining trigonometric functions of $\theta$.
Solution:
Step 1: Find $\cos \theta$ using fundamental Pythagorean Identity:
$$\sin^2 \theta + \cos^2 \theta = 1 \implies \cos^2 \theta = 1 - \sin^2 \theta$$ $$\cos^2 \theta = 1 - \left(-\frac{\sqrt{3}}{2}\right)^2 = 1 - \frac{3}{4} = \frac{1}{4}$$ $$\cos \theta = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$Since $\theta$ lies in Quadrant III, cosine is negative ($\cos \theta < 0$):
$$\cos \theta = -\frac{1}{2}$$Step 2: Calculate Quotient Functions ($\tan \theta$ and $\cot \theta$):
$$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \frac{\sqrt{3}}{1} = \sqrt{3}$$ $$\cot \theta = \frac{1}{\tan \theta} = \frac{1}{\sqrt{3}}$$Step 3: Calculate Reciprocal Functions ($\csc \theta$ and $\sec \theta$):
$$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}$$ $$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{1}{2}} = -2$$Final Summary of Functions:
$$\sin \theta = -\frac{\sqrt{3}}{2}, \quad \cos \theta = -\frac{1}{2}, \quad \tan \theta = \sqrt{3}, \quad \csc \theta = -\frac{2}{\sqrt{3}}, \quad \sec \theta = -2, \quad \cot \theta = \frac{1}{\sqrt{3}}$$Q1: What is the exact value of $\sin(-780^\circ)$?
Q2: If $\tan \theta > 0$ and $\csc \theta < 0$, in which quadrant does the terminal arm of angle $\theta$ lie?
Q3: The value of $\cos \pi + \sin \frac{3\pi}{2} - \tan 2\pi$ is equal to:
Q4: For which of the following quadrantal angles is $\sec \theta$ undefined?
Q5: What is the simplified numerical value of $\sin^2 \frac{\pi}{6} + \sin^2 \frac{\pi}{3} + \tan^2 \frac{\pi}{4}$?
Q6: Which of the following expressions is equivalent to $\frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ}$?
Q7: If the terminal arm of an angle $\theta$ lies on the negative y-axis, then $\theta$ is equal to:
Q8: The domain of $f(\theta) = \tan \theta$ excludes values of $\theta$ given by:
Answer: Intermediate boards (FBISE, Punjab Boards, Sindh Board, etc.) evaluate a student's conceptual understanding of exact geometric derivations and radical arithmetic. Decimal outputs generated via scientific calculators bypass the trigonometric proofs and fundamental algebraic identities required by the curriculum scheme. Unless explicitly asked for decimal approximations, students must present fractions and radical expressions (e.g., $\frac{\sqrt{3}}{2}$ instead of $0.8660$).
Answer: When an expression involves terms like $\tan 90^\circ = \frac{1}{0}$, it indicates that the function is undefined at that specific domain point. In standard board examination questions, expressions are constructed such that undefined conditions do not occur inside continuous equations. If a question leads to division by zero, double-check your algebraic steps or reciprocal substitutions (e.g., converting $\tan \theta$ to $\frac{\sin \theta}{\cos \theta}$ before evaluation).
Answer: Use the standard quadrant mnemonic **"Add Sugar To Coffee"** corresponding to Quadrants I, II, III, and IV:
Alternatively, remember coordinates on the unit circle: $x = \cos \theta$ and $y = \sin \theta$. The signs of $x$ and $y$ in Cartesian plane coordinates directly yield the signs of $\cos \theta$ and $\sin \theta$.