Class 11 FSc Math Chapter 9 Fundamentals of Trigonometry Proofs & Solved Exercise 9.3 Notes PDF - FBISE & Punjab Boards
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.
To establish rigor in mathematical proofs, trigonometric ratios for standard values are derived geometrically rather than memorized empirically.
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:
$$\sin 45^\circ = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}}$$ $$\cos 45^\circ = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}}$$ $$\tan 45^\circ = \frac{\text{Perpendicular}}{\text{Base}} = \frac{1}{1} = 1$$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 $\Delta ABD$:
$$\sin 60^\circ = \frac{AD}{AB} = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{BD}{AB} = \frac{1}{2}, \quad \tan 60^\circ = \frac{AD}{BD} = \sqrt{3}$$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:
$$\cos \theta = x, \quad \sin \theta = y, \quad \tan \theta = \frac{y}{x} \quad (x \neq 0)$$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 $[0^\circ, 360^\circ)$.
| 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". |
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:
$$\text{L.H.S.} = \text{R.H.S.} \quad \blacksquare$$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.):
$$\text{R.H.S.} = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$$ $$\text{Since L.H.S.} = \text{R.H.S.}, \text{ the identity is verified.} \quad \blacksquare$$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 $\frac{\pi}{3}$:
$$\sin\left(\frac{19\pi}{3}\right) = \sin\left(6\pi + \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$ $$\cos\left(\frac{19\pi}{3}\right) = \cos\left(6\pi + \frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$ $$\tan\left(\frac{19\pi}{3}\right) = \tan\left(6\pi + \frac{\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$$ $$\csc\left(\frac{19\pi}{3}\right) = \frac{1}{\sin(\pi/3)} = \frac{2}{\sqrt{3}}$$ $$\sec\left(\frac{19\pi}{3}\right) = \frac{1}{\cos(\pi/3)} = 2$$ $$\cot\left(\frac{19\pi}{3}\right) = \frac{1}{\tan(\pi/3)} = \frac{1}{\sqrt{3}} \quad \blacksquare$$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 $-$):
Step 4: Write the complete solution set:
$$\text{Solution Set} = \left\{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\} \quad \blacksquare$$Q1: What is the exact value of $\sin(-780^\circ)$?
Q2: If $\sin \theta < 0$ and $\cos \theta > 0$, in which quadrant does the terminal arm of angle $\theta$ lie?
Q3: The domain of the function $y = \sec \theta$ is:
Q4: Value of $\cos \left(-\frac{7\pi}{4}\right)$ is:
Q5: The terminal side of angle $\theta = -450^\circ$ lies on which axis?
Q6: Evaluation of $\cos^2 30^\circ - \sin^2 30^\circ$ gives:
Q7: What is the primary period of $\tan \theta$?
Q8: The expression $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ}$ simplifies to:
Negative angles indicate clockwise rotation from the positive x-axis. Using the fundamental parity properties of trigonometric functions:
Always extract or eliminate the negative sign first before performing coterminal reduction.
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.
Follow this 3-step board algorithm: