Free practice › Class 11 › Mathematics › Chapter 3
Trigonometric Functions — JEE MCQs with answers
Topics in this chapter
- 3.1 Introduction
- 3.2 Angles
- 3.3 Trigonometric Functions
- 3.4 Trigonometric Functions of Sum and Difference of Two Angles
Free practice questions
In Example 5 the second arc subtends $110°$ at the centre. In lowest terms its radian measure is
- A. $\dfrac{11\pi}{36}$
- B. $\dfrac{11\pi}{18}$
- C. $\dfrac{18\pi}{11}$
- D. $\dfrac{11\pi}{9}$
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Answer: B — $110\times\dfrac{\pi}{180}=\dfrac{110\pi}{180}=\dfrac{11\pi}{18}$; NCERT writes this as $\dfrac{22\pi}{36}$, which reduces to the same thing. The value $\dfrac{11\pi}{36}$ is $55°$; $\dfrac{18\pi}{11}$ inverts the fraction; and $\dfrac{11\pi}{9}$ is $220°$, twice the required angle.
NCERT Class 11 Maths Ch 3 Example 5 p.48
The value of $2\sin 75^\circ\sin 15^\circ$ is
- A. $\dfrac{1}{4}$
- B. $\dfrac{1}{2}$
- C. $-\dfrac{1}{2}$
- D. $\dfrac{\sqrt3}{2}$
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Answer: B — Result 21(ii) reads $-2\sin x\sin y=\cos(x+y)-\cos(x-y)$, so $2\sin x\sin y=\cos(x-y)-\cos(x+y)$. With $x=75^\circ$, $y=15^\circ$ this is $\cos 60^\circ-\cos 90^\circ=\dfrac12-0=\dfrac12$, the second term being exactly zero. Numerically $2\times0.9659\times0.2588=0.5000$. The option $\dfrac{\sqrt3}{2}$ is $\cos 30^\circ$, which matches neither the difference $60^\circ$ nor the sum $90^\circ$ of the two angles. The option $-\dfrac12$ comes from forgetting to multiply result 21(ii) by $-1$, and is impossible since both sines are positive. The option $\dfrac14$ is the value of $\sin 75^\circ\sin 15^\circ$ without the factor $2$ required by the question.
NCERT Class 11 Maths Ch 3 Trigonometric Functions, Sec 3.4, p.64
If $\cos x>0$ and $\operatorname{cosec} x<0$, then $x$ lies in the
- A. fourth quadrant
- B. second quadrant
- C. first quadrant
- D. third quadrant
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Answer: A — $\operatorname{cosec} x=\frac{1}{\sin x}$ is negative exactly when $\sin x<0$, i.e. in the third or fourth quadrant. Combined with $\cos x>0$, which holds in the first and fourth quadrants, the only possibility is the fourth quadrant. The first quadrant fails because there $\operatorname{cosec} x>0$; the second and third fail because there $\cos x<0$.
NCERT Class 11 Maths Ch 3 §3.3.1 Sign of trigonometric functions - sign table p.52
The word 'trigonometry' is derived from which pair of Greek words?
- A. 'trigon' and 'gonia'
- B. 'tri' and 'metron'
- C. 'trigon' and 'metron'
- D. 'triangle' and 'metron'
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Answer: C — The section opens by stating that the word 'trigonometry' is derived from the Greek words 'trigon' and 'metron'. Option A ('trigon' and 'gonia') replaces the second word with one the text never prints. Option B ('tri' and 'metron') truncates the first word, which the text gives in full as 'trigon'. Option D ('triangle' and 'metron') substitutes the English word 'triangle' for the Greek 'trigon'; 'triangle' appears in the section only inside the meaning of the word, not as one of its Greek roots.
NCERT Class 11 Maths Ch 3 §3.1 Introduction p.43
Using $\pi=\dfrac{22}{7}$, the radian measure of $1°$ is approximately
- A. $0.00556$ radian
- B. $0.03492$ radian
- C. $0.01746$ radian
- D. $57.2727$ radian
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Answer: C — $1°=\dfrac{\pi}{180}$ radian $=\dfrac{22}{7\times 180}=\dfrac{22}{1260}=0.01746$ radian, the value printed in Section 3.2.4. The value $57.2727$ comes from multiplying by $\dfrac{180}{\pi}$ instead of $\dfrac{\pi}{180}$ (that is $1$ radian in degrees); $0.00556=\dfrac{1}{180}$ divides by $180$ but drops the factor $\pi$; and $0.03492=\dfrac{2\pi}{180}$ uses a full revolution $2\pi$ where a straight angle $\pi$ is required.
NCERT Class 11 Maths Ch 3 Sec 3.2.4 'Relation between degree and radian' p.46
The point of the unit circle corresponding to $x = \pi$ has coordinates
- A. $(0,\ -1)$
- B. $(1,\ 0)$
- C. $(-1,\ 0)$
- D. $(0,\ 1)$
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Answer: C — $\angle AOC = \pi$, and NCERT lists the coordinates of $C$ as $(-1,\,0)$; consistently $\cos\pi=-1$ and $\sin\pi=0$. $(0,\,-1)$ is the point for $x=\frac{3\pi}{2}$, $(1,\,0)$ is the starting point $A$ ($x=0$ or $2\pi$) and $(0,\,1)$ is the point for $x=\frac{\pi}{2}$.
NCERT Class 11 Maths Ch 3 §3.3 Trigonometric Functions - quadrantal angles p.50
The value of $\cos 40^\circ+\cos 80^\circ+\cos 160^\circ$ is
- A. $-\dfrac{1}{2}$
- B. $\dfrac{3}{2}$
- C. $0$
- D. $\cos 20^\circ$
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Answer: C — Pair the first two terms: half-sum $60^\circ$, half-difference $-20^\circ$, so $\cos 40^\circ+\cos 80^\circ=2\cos 60^\circ\cos 20^\circ=\cos 20^\circ$ (the cosine is even, so the negative half-difference is harmless). The expression becomes $\cos 20^\circ+\cos 160^\circ=2\cos 90^\circ\cos(-70^\circ)$, and $\cos 90^\circ$ is **exactly** $0$, so the total is exactly $0$. Numerically $0.7660+0.1736-0.9397=0$. The option $\cos 20^\circ$ is the running total before the last term is included. The option $-\dfrac12$ would follow from misreading $2\cos 60^\circ$ as $-1$. The option $\dfrac{3}{2}$ ignores that $\cos 160^\circ$ is negative.
NCERT Class 11 Maths Ch 3 Trigonometric Functions, Sec 3.4, p.63
Which of the following best describes the shift in the treatment of trigonometry that the Introduction announces for this chapter?
- A. From functions of a real number back to ratios of the sides of a triangle
- B. From trigonometric identities to problems on heights and distances
- C. From ratios of the sides of a right angled triangle to trigonometric functions and their properties
- D. From measuring angles of a triangle to measuring areas of a triangle
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Answer: C — The Introduction contrasts the earlier study of trigonometric ratios of acute angles, defined as ratios of the sides of a right angled triangle, with the plan to 'generalise the concept of trigonometric ratios to trigonometric functions and study their properties'. Option C captures exactly this movement. Option A reverses it, going from functions back to ratios. Option B names two topics the section places in EARLIER classes, not in this chapter. Option D speaks of measuring areas, which the section never mentions.
NCERT Class 11 Maths Ch 3 §3.1 Introduction p.43
How many integers fail to belong to the range of $y = \operatorname{cosec} x$?
- A. exactly two (namely $1$ and $-1$)
- B. none -- every integer occurs
- C. exactly three (namely $-1,\ 0,\ 1$)
- D. exactly one (namely $0$)
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Answer: D — The range is $\{y : y \in R,\ y \ge 1$ or $y \le -1\}$, so the only values missing are those strictly between $-1$ and $1$. The one integer in that open gap is $0$, so exactly one integer is excluded. 'Exactly two (namely $1$ and $-1$)' is never right: $1$ and $-1$ are genuinely attained ($\operatorname{cosec}\frac{\pi}{2} = 1$ and $\operatorname{cosec}\frac{3\pi}{2} = -1$), because the printed endpoints are **closed**. 'Exactly three ($-1,\ 0,\ 1$)' is what the strict reading $y > 1$ or $y < -1$ would give, but that is not what the text prints. 'None' ignores the gap around $0$ altogether.
NCERT Class 11 Mathematics Ch 3 §3.3.2 p.53 ("range is the set {y : y in R, y >= 1 or y <= - 1}")
The radian measure of $22°30'$ is
- A. $\dfrac{\pi}{16}$
- B. $\dfrac{\pi}{4}$
- C. $\dfrac{\pi}{8}$
- D. $\dfrac{\pi}{12}$
Show answer
Answer: C — $22°30'=22.5°=\dfrac{45}{2}$ degree, so the radian measure is $\dfrac{45}{2}\times\dfrac{\pi}{180}=\dfrac{45\pi}{360}=\dfrac{\pi}{8}$. The value $\dfrac{\pi}{16}$ is $11°15'$, obtained by halving once too often; $\dfrac{\pi}{4}$ is $45°$, twice the angle; and $\dfrac{\pi}{12}$ is $15°$.
NCERT Class 11 Maths Ch 3 Sec 3.2.4 'Relation between degree and radian' p.46
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