NCERT Class 11 Trigonometric Functions Exercise 3.2 Guide

Understanding Signs of Trigonometric Functions: The ASTC Rule

To evaluate trigonometric functions for any angle $x$, we map them to a unit circle where coordinates of a point $P(x,y)$ are defined as $(\cos \theta, \sin \theta)$. Because coordinates change signs depending on the quadrant, trigonometric functions do too! We remember this with the famous mnemonic ASTC (All Silver Tea Cups or Add Sugar To Coffee):

• Quadrant I ($0$ to $\pi/2$): All trigonometric functions are positive ($x > 0, y > 0$).
• Quadrant II ($\pi/2$ to $\pi$): Only Sine and its reciprocal Cosecant ($\csc$) are positive ($x < 0, y > 0$).
• Quadrant III ($\pi$ to $3\pi/2$): Only Tangent and its reciprocal Cotangent ($\cot$) are positive ($x < 0, y < 0$).
• Quadrant IV ($3\pi/2$ to $2\pi$): Only Cosine and its reciprocal Secant ($\sec$) are positive ($x > 0, y < 0$).

Additionally, trigonometric functions are periodic. Sine and Cosine repeat their values after an interval of $2\pi$ radians (or $360^\circ$). Therefore, $\sin(2n\pi + x) = \sin x$ and $\cos(2n\pi + x) = \cos x$ for any integer $n$.

Step-by-Step Method for Ex 3.2 Problems

1. Identify Quadrant and Assign Sign Limits — Read the quadrant of the angle carefully (e.g., Quadrant III implies both $\sin x$ and $\cos x$ are negative, while $\tan x$ is positive).
2. Apply Pythagorean Identities — Use $\sin^2 x + \cos^2 x = 1$, $1 + \tan^2 x = \sec^2 x$, or $1 + \cot^2 x = \csc^2 x$ to find the missing primary function.
3. Determine Square Root Sign — When taking the square root (e.g., $\cos x = \pm \sqrt{1 - \sin^2 x}$), choose the positive or negative sign explicitly based on the given quadrant.
4. Calculate Reciprocal and Quotient Ratios — Compute the rest of the ratios using definitions: $\csc x = 1/\sin x$, $\sec x = 1/\cos x$, $\tan x = \sin x / \cos x$, and $\cot x = 1/\tan x$.
5. For Large Angles, Express as $2n\pi + \theta$ — Divide the angle by $2\pi$ (or $360^\circ$) to isolate the excess angle $\theta$ within $[0, 2\pi]$, then evaluate $f(\theta)$.

Avoiding the 'Square Root Sign' Trap

One of the most frequent marks-deduction areas in CBSE evaluations is the automatic assumption that square roots are positive.

When writing $\sin x = \sqrt{1 - \cos^2 x}$, remember that algebra dictates $\sin x = \pm \sqrt{1 - \cos^2 x}$.

Board Exam Tip: Always write a line of justification in your answer paper, such as: 'Since $x$ lies in the third quadrant, $\sin x$ must be negative. Thus, $\sin x = -\sqrt{...}

Practice Questions with Solutions

• Q: If $\cos x = -\frac{1}{2}$ and $x$ lies in the third quadrant, find the values of the other five trigonometric functions. A: Step 1: Identify the signs for Quadrant III. In Quadrant III, only tangent and cotangent are positive; sine, cosecant, and secant are negative. Step 2: Find secant directly as it is the reciprocal of cosine: $\sec x = \frac{1}{\cos x} = \frac{1}{-1/2} = -2$. Step 3: Use the identity $\sin^2 x + \cos^2 x = 1 \implies \sin^2 x = 1 - (-\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$. Step 4: Since $x$ lies in the third quadrant, $\sin x$ is negative: $\sin x = -\frac{\sqrt{3}}{2}$. Step 5: Find the reciprocal cosecant: $\csc x = \frac{1}{\sin x} = -\frac{2}{\sqrt{3}}$. Step 6: Compute tangent and cotangent: $\tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$ and $\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$. Final answer: $\sin x = -\frac{\sqrt{3}}{2}$, $\csc x = -\frac{2}{\sqrt{3}}$, $\sec x = -2$, $\tan x = \sqrt{3}$, and $\cot x = \frac{1}{\sqrt{3}}$.
• Q: Find the value of the trigonometric function $\sin \frac{31\pi}{3}$. A: Step 1: Divide $31$ by $3$ to express the fraction as a multiple of $2\pi$. We observe that $\frac{31\pi}{3} = 10\pi + \frac{\pi}{3}$. Step 2: Rewrite this in terms of $2n\pi$: $\frac{31\pi}{3} = 5(2\pi) + \frac{\pi}{3}$. Step 3: Use the periodicity property of sine: $\sin(2n\pi + \theta) = \sin \theta$. Thus, $\sin \left(10\pi + \frac{\pi}{3}\right) = \sin \frac{\pi}{3}$. Step 4: Recall the standard exact value: $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$. Final answer: $\frac{\sqrt{3}}{2}$
• Q: Find the value of $\cot \left(-\frac{15\pi}{4}\right)$. A: Step 1: Use the negative angle identity $\cot(-\theta) = -\cot \theta$ to rewrite the expression: $\cot \left(-\frac{15\pi}{4}\right) = -\cot \left(\frac{15\pi}{4}\right)$. Step 2: Express $\frac{15\pi}{4}$ as a multiple of $2\pi$. Note that $\frac{15\pi}{4} = 4\pi - \frac{\pi}{4} = 2(2\pi) - \frac{\pi}{4}$. Step 3: Since cotangent is periodic (it repeats every $\pi$ and thus every $2\pi$), $\cot \left(2(2\pi) - \frac{\pi}{4}\right) = \cot \left(-\frac{\pi}{4}\right) = -\cot \frac{\pi}{4}$. Step 4: Combine the negative signs: $-\cot \left(\frac{15\pi}{4}\right) = -(-\cot \frac{\pi}{4}) = \cot \frac{\pi}{4}$. Step 5: Substitute the standard value $\cot \frac{\pi}{4} = 1$. Final answer: $1$
• Q: If $\tan x = -\frac{5}{12}$ and $x$ lies in the second quadrant, find the values of the other five trigonometric functions. A: Step 1: In Quadrant II, only sine and cosecant are positive; cosine, secant, tangent, and cotangent are negative. Step 2: Find cotangent: $\cot x = \frac{1}{\tan x} = -\frac{12}{5}$. Step 3: Use the identity $\sec^2 x = 1 + \tan^2 x = 1 + \left(-\frac{5}{12}\right)^2 = 1 + \frac{25}{144} = \frac{169}{144}$. Step 4: Since $x$ is in the second quadrant, $\sec x$ is negative: $\sec x = -\sqrt{\frac{169}{144}} = -\frac{13}{12}$. Step 5: Find cosine: $\cos x = \frac{1}{\sec x} = -\frac{12}{13}$. Step 6: Compute sine using $\sin x = \tan x \cdot \cos x = \left(-\frac{5}{12}\right) \cdot \left(-\frac{12}{13}\right) = \frac{5}{13}$. Step 7: Find cosecant: $\csc x = \frac{1}{\sin x} = \frac{13}{5}$. Final answer: $\sin x = \frac{5}{13}$, $\csc x = \frac{13}{5}$, $\cos x = -\frac{12}{13}$, $\sec x = -\frac{13}{12}$, and $\cot x = -\frac{12}{5}$.

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