Chapter 3 Trigonometric Functions

NCERT Solutions Class 11 Maths Chapter 3 – Trigonometric Functions

NCERT Solutions for Class 11 Maths Chapter 3 (Trigonometric Functions) extend the basic ratios learned in Class 10 into the broader world of periodic functions. This chapter moves beyond right-angled triangles to define trigonometry for any angle using the Unit Circle concept. It is one of the most critical chapters for Physics and Calculus, providing the formulas used to describe waves, rotations, and oscillations.

NCERT Solutions for Class 11 Maths Chapter 3 by ALLEN are designed by expert faculty to simplify trigonometric concepts through clear explanations and systematic problem-solving. The solutions focus on conceptual accuracy and exam applicability, making complex identities easier to understand and use.

Mastering trigonometric transformations and identities is non-negotiable for competitive exams like JEE Main, JEE Advanced, CBSE Board exam, and BITSAT. These solutions provide a rigorous logical framework to help students memorize and apply complex formulas like sum-to-product and multiple-angle identities.

1.0Class 11 Maths Chapter 3 Trigonometric Functions: Key Concepts

Class 11 NCERT Solutions Maths Chapter 3 explains trigonometric functions, different ways of measuring angles, their values, graphs, and important formulas. The key concepts include:

Degree Measure: Angles can be measured in degrees. One complete revolution is equal to 360°.
Radian Measure: Radian is the SI unit of plane angle. If an arc of length l is formed by an angle θ in a circle of radius r, then (l = r θ), when θ is measured in radians.
Conversion of Degree and Radian: The basic relation between degrees and radians is ( $π$ ) radians = 180°.
Trigonometric Functions: The six trigonometric functions are sin x, cos x, tan x, cot x, sec x, and cosec x. Their values are studied for real numbers using the unit circle.
Signs of Trigonometric Functions: The signs of trigonometric functions depend on the quadrant in which the angle lies. Students learn which functions are positive or negative in each quadrant.
Domain, Range, and Periodicity: Students learn the domain and range of the six trigonometric functions. They also study their periods and how their values repeat at regular intervals.
Graphs of Trigonometric Functions: The chapter explains the graphs of trigonometric functions such as sin x, cos x, and tan x. These graphs help students understand their values and periodic nature.
Trigonometric Functions of Sum and Difference of Two Angles: Important formulas include:

$(sin (A \pm B) = sin A cos B \pm cos A sin B)$

$(cos (A \pm B) = cos A cos B \mp sin A sin B)$

Multiple Angle Formulas: Important formulas include:

$(sin 2 x = 2 sin x cos x)$

$(cos 2 x = cos^{2} x - sin^{2} x = 2 cos^{2} x - 1 = 1 - 2 sin^{2} x)$

Product to Sum and Sum to Product Formulas: These formulas are used to change products of trigonometric functions into sums or differences and to change sums or differences into products.

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 3

EXERCISE - 3.1

Find the radian measures corresponding to the following degree measures: (i) 25° (ii) -47°30' (iii) 240° (iv) 520°

Sol.

(i) 25° $∴ 2 5^{\circ} = \frac{π}{180} \times 25$ radian

$= \frac{5 π}{36} radian (∵ 1^{0} = \frac{π}{180} radian)$

(ii) - 47° 30'

$- 47 \frac{1}{2} = \frac{- 95}{2} degree - 95 degree = π \times (- 95) radian = 3 9^{\circ} + 2 2^{'} + \frac{1}{2} minutes = 3 9^{\circ} 2 2^{'} 3 0^{''} [1^{'} = 6 0^{''}]$

(iii) 240° We know that $18 0^{\circ} = π$ radian $∴ 24 0^{\circ} = \frac{π}{180} \times 240$ radian

$= \frac{4}{3} π radian (∵ 1^{\circ} = \frac{π}{180} radian)$

(iv) 520° We know that $18 0^{\circ} = π$ radian $∴ 52 0^{\circ} = \frac{π}{180} \times 520$ radian

$= \frac{26 π}{9} radian (∵ 1^{0} = \frac{π}{180} radian)$

Find the degree measures corresponding to the following radian measures. ( Use $π = \frac{22}{7}$ ) (i) $\frac{11}{16}$ (ii) -4 (iii) $\frac{5 π}{3}$ (iv) $\frac{7 π}{6}$ Sol. (i) $\frac{11}{16}$ $∴ \frac{11}{16}$ radian $= \frac{180}{π} \times \frac{11}{16}$ degree

$= \frac{45 \times 11}{π \times 4} degree (∵ 1 radian = \frac{180}{π} degree) = \frac{45 \times 11 \times 7}{22 \times 4} degree = \frac{315}{8} degree = 39 \frac{3}{8} degree = 3 9^{\circ} + \frac{3 \times 60}{8} minutes [1^{\circ} = 6 0^{'}] = 3 9^{\circ} + 2 2^{'} + \frac{1}{2} minutes [1^{'} = 6 0^{''}] = 3 9^{\circ} 2 2^{'} 3 0^{''}$

(ii) -4

$- 4 radian = \frac{180}{π} \times (- 4) degree = \frac{180 \times 7 ( - 4 )}{22} degree (∵ 1 radian = \frac{180}{π} degree) = \frac{- 2520}{11} degree = - (229 \frac{1}{11} degree) = - (22 9^{\circ} + \frac{1 \times 60}{11} minutes) [1^{\circ} = 60] = - (22 9^{\circ} + 5^{'} + \frac{5}{11} minutes) = - 22 9^{\circ} 5^{'} 2 7^{''} [1^{'} = 6 0^{''}]$

(iii) $\frac{5 π}{3}$

$∴ \frac{5 π}{3} radian = \frac{180}{π} \times \frac{5 π}{3} degree = 30 0^{\circ} (∵ 1 radian = \frac{180}{π} degree)$

(iv) $\frac{7 π}{6}$

$∴ \frac{7 π}{6} radian = \frac{180}{π} \times \frac{7 π}{6} degree = 21 0^{\circ} (∵ 1 radian = \frac{180}{π} degree)$

A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? Sol. Number of revolutions made by the wheel in 1 minute = 360 ∴ Number of revolutions made by the wheel in 1 second $= \frac{360}{60} = 6$ In one complete revolution, the wheel turns an angle of $2 π$ radian. Hence, in 6 complete revolutions, it will turn an angle of $6 \times 2 π$ radian, i.e., $12 π$ radian. Thus, in one second, the wheel turns an angle of $12 π$ radian.
Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm. $($ Use $π = \frac{22}{7})$ Sol. We know that in a circle of radius r unit, if an arc of length $l$ unit subtends an angle $θ$ radian at the centre, then $θ = \frac{l}{r}$ Therefore, for $r = 100 cm, l = 22 cm$ , we have

$θ = \frac{22}{100} radian = \frac{180}{π} \times \frac{22}{100} degree = \frac{180 \times 7 \times 22}{22 \times 100} degree = \frac{126}{10} degree = 12 \frac{3}{5} degree = 1 2^{\circ} 3 6^{'} [1^{\circ} = 60]$

Thus, the required angle is $1 2^{\circ} 3 6^{'}$ .

5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord. Sol. Diameter of the circle $= 40 cm$

Class-11-chapter-2-exer-3.1-ques-5-maths-ncert-sol

∴ Radius $(r)$ of the circle $= \frac{40}{2} cm = 20 cm$

Let AB be a chord (length $= 20 cm$ ) of the circle. In $△ OAB$ , $OA = OB =$ Radius of circle $= 20 cm$ Also, $AB = 20 cm$ Thus, $△ OAB$ is an equilateral triangle.

$∴ θ = 6 0^{\circ} = \frac{π}{3}$ radian

We know that in a circle of radius r unit, if an arc of length $l$ unit subtends an angle $θ$ .

$θ = \frac{l}{r} \frac{π}{3} = \frac{A B}{20}$

$\Rightarrow AB = \frac{20 π}{3} cm$

Thus, the length of the minor arc of the chord is $\frac{20 π}{3} cm$

If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii. Sol. Let the radii of the two circles be $r_{1}$ and $r_{2}$ . Let an arc of length $l$ subtend an angle of 60° at the centre of the circle of radius $r_{1}$ , while let an arc of length $l$ subtend an angle of 75° at the centre of the circle of radius $r_{2}$ Now, $6 0^{\circ} = \frac{π}{3}$ radian and $7 5^{\circ} = \frac{5 π}{12}$ radian

We know that in a circle of radius r unit, if an arc of length $l$ unit subtends an angle $θ$ .

$θ = \frac{l}{r} or l = r θ$

$∴ l = \frac{r _{1} π}{3}$ and $l = \frac{r _{2} 5 π}{12} \Rightarrow \frac{r _{1} π}{3} = \frac{r _{2} 5 π}{12}$ $\Rightarrow r_{1} = \frac{r _{2} 5}{4} \Rightarrow \frac{r _{1}}{r _{2}} = \frac{5}{4}$

Thus, the ratio of the radii is $5 : 4$ .

7. Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm

Sol. We know that in a circle of radius r unit, if an arc of length $l$ unit subtends an angle $θ$ radian at the centre, then $θ = \frac{l}{r}$ It is given that $r = 75 cm$

(i) Here, $l = 10 cm; θ = \frac{10}{75}$ radian $= \frac{2}{15}$ radian (ii) Here, $l = 15 cm; θ = \frac{15}{75}$ radian $= \frac{1}{5}$ radian (iii) Here, $l = 21 cm; θ = \frac{21}{75}$ radian $= \frac{21}{25}$ radian

EXERCISE - 3.2

Find the values of other five trigonometric functions in Q. 1 to 5

$cos x = - \frac{1}{2}$ , x lies in third quadrant.

Sol. $cos x = - \frac{1}{2}$

$∴ \Rightarrow \Rightarrow sec x = \frac{1}{c o s x} = \frac{1}{( - \frac{1}{2} )} = - 2 sin^{2} x + cos^{2} x = 1 sin^{2} x = 1 - cos^{2} x \Rightarrow sin^{2} x = 1 - (- \frac{1}{2})^{2} sin^{2} x = 1 - \frac{1}{4} = \frac{3}{4} \Rightarrow sin x = \pm \frac{3}{2}$

Since x lies in the $3^{rd}$ quadrant, the value of $sin$ x will be negative.

$∴ sin x = - \frac{3}{2} cosec x = \frac{1}{sin x} = \frac{1}{( - \frac{3}{2} )} = - \frac{2}{3} tan x = \frac{sin x}{cos x} = \frac{( - \frac{3}{2} )}{( - \frac{1}{2} )} = 3 cot x = \frac{1}{tan x} = \frac{1}{3}$

$sin x = \frac{3}{5}$ , x lies in second quadrant.

Sol. $sin x = \frac{3}{5}$

$cosec x = \frac{1}{s i n x} = \frac{1}{( \frac{3}{5} )} = \frac{5}{3}$

$\Rightarrow cos^{2} x = 1 - sin^{2} x \Rightarrow cos^{2} x = 1 - \frac{9}{25} \Rightarrow cos^{2} x = 1 - (\frac{3}{5})^{2} \Rightarrow cos^{2} x = \frac{16}{25}$

$\Rightarrow cos x = \pm \frac{4}{5}$ Since x lies in the $2^{nd}$ quadrant, the value of cos x will be negative

$∴ cos x sec x tan x cot x = - \frac{4}{5} = \frac{1}{cos x} = \frac{1}{( - \frac{4}{5} )} = - \frac{5}{4} = \frac{sin x}{cos x} = \frac{( \frac{3}{5} )}{( - \frac{4}{5} )} = - \frac{3}{4} = \frac{1}{tan x} = - \frac{4}{3}$

$cot x = \frac{3}{4}$ , x lies in third quadrant. Sol. $cot x = \frac{3}{4}$

$tan x = \frac{1}{c o t x} = \frac{1}{( \frac{3}{4} )} = \frac{4}{3}$

$1 + tan^{2} x = sec^{2} x$

$\Rightarrow \Rightarrow 1 + (\frac{4}{3})^{2} = sec^{2} x \frac{25}{9} = sec^{2} x \Rightarrow 1 + \frac{16}{9} = sec^{2} x = \pm \frac{5}{3}$

Since x lies in the $3^{rd}$ quadrant, the value of sec x will be negative.

$∴ \Rightarrow \Rightarrow sec x = - \frac{5}{3} cos x = \frac{1}{s e c x} = \frac{1}{( - \frac{5}{3} )} = - \frac{3}{5} tan x = \frac{s i n x}{c o s x} \Rightarrow \frac{4}{3} = \frac{s i n x}{( - \frac{3}{5} )} sin x = (\frac{4}{3}) \times (\frac{- 3}{5}) = - \frac{4}{5} cosec x = \frac{1}{s i n x} = - \frac{5}{4}$

$sec x = \frac{13}{5}$ , x lies in fourth quadrant. Sol. $sec x = \frac{13}{5}$

$cos x = \frac{1}{sec x} = \frac{1}{( \frac{13}{5} )} = \frac{5}{13} sin^{2} x + cos^{2} x = 1 \Rightarrow sin^{2} x = 1 - (\frac{5}{13})^{2}$

Since x lies in the $4^{th}$ quadrant, the value of $sin$ x will be negative. $∴ sin x = - \frac{12}{13}$

$cosec x = \frac{1}{sin x} = \frac{1}{( - \frac{12}{13} )} = - \frac{13}{12} tan x = \frac{sin x}{cos x} = \frac{( \frac{- 12}{13} )}{( \frac{5}{13} )} = - \frac{12}{5} cot x = \frac{1}{tan x} = \frac{1}{( - \frac{12}{5} )} = - \frac{5}{12}$

$tan x = - \frac{5}{12}$ , x lies in second quadrant. Sol. $tan x = - \frac{5}{12}$

$cot x = \frac{1}{t a n x} = \frac{1}{( - \frac{5}{12} )} = - \frac{12}{5}$

$\Rightarrow 1 + (- \frac{5}{12})^{2} = sec^{2} x \Rightarrow 1 + \frac{25}{144} = sec^{2} x$

$\Rightarrow \frac{169}{144} = sec^{2} x \Rightarrow sec x = \pm \frac{13}{12}$

Since x lies in the 2nd quadrant, the value of sec x will be negative. $∴ sec x = - \frac{13}{12}$

$cos x = \frac{1}{s e c x} = \frac{1}{( - \frac{13}{12} )} = - \frac{12}{13}$

$tan x = \frac{s i n x}{c o s x}$

$\Rightarrow - \frac{5}{12} = \frac{s i n x}{( - \frac{12}{13} )}$ $\Rightarrow sin x = (\frac{- 5}{12}) \times (- \frac{12}{13}) = \frac{5}{13}$ $\Rightarrow cosec x = \frac{1}{s i n x} = \frac{1}{( \frac{5}{13} )} = \frac{13}{5}$

Find the value of the trigonometric function in Q. 6 to 10

6. $sin 76 5^{\circ}$ .

Sol. It is known that the values of $sin x$ repeat after an interval of $2 π$ or 360°.

$∴ sin 76 5^{\circ} = sin (2 \times 36 0^{\circ} + 4 5^{\circ})$

$= sin 4 5^{\circ} = \frac{1}{2}$

$cosec (- 141 0^{\circ})$ .

Sol. It is known that the values of cosec x repeat after an interval of $2 π$ or 360°.

$∴ cosec (- 141 0^{\circ})$

$= cosec (- 141 0^{\circ} + 4 \times 36 0^{\circ}) = cosec (- 141 0^{\circ} + 144 0^{\circ}) = cosec 3 0^{\circ} = 2$

$tan \frac{19 π}{3}$ .

Sol. It is known that the values of tan x repeat after an interval of $π$ or 180°

$∴ tan \frac{19 π}{3} = tan 6 \frac{1}{3} π = tan (6 π + \frac{π}{3})$

$= tan \frac{π}{3} = tan 6 0^{\circ} = 3$

$sin (- \frac{11 π}{3})$

Sol. It is known that the values of $sin x$ repeat after an interval of $2 π$ or 360°.

$∴ sin (- \frac{11 π}{3}) = sin (- \frac{11 π}{3} + 2 \times 2 π)$

$= sin (\frac{π}{3}) = sin \frac{3}{2}$

$cot (- \frac{15 π}{4})$ .

Sol. It is known that the values of $cot x$ repeat after an interval of $π$ or 180°. $∴ cot (- \frac{15 π}{4}) = cot (- \frac{15 π}{4} + 4 π) = cot \frac{π}{4} = 1$

EXERCISE - 3.3

Prove that :

$sin^{2} \frac{π}{6} + cos^{2} \frac{π}{3} - tan^{2} \frac{π}{4} = - \frac{1}{2}$

Sol. L.H.S. $= sin^{2} \frac{π}{6} + cos^{2} \frac{π}{3} - tan^{2} \frac{π}{4}$

$= (\frac{1}{2})^{2} + (\frac{1}{2})^{2} - (1)^{2} = \frac{1}{4} + \frac{1}{4} - 1 = - \frac{1}{2} = R.H.S.$

$2 sin^{2} \frac{π}{6} + cosec^{2} \frac{7 π}{6} cos^{2} \frac{π}{3} = \frac{3}{2}$

Sol. L.H.S. $= 2 sin^{2} \frac{π}{6} + cosec^{2} \frac{7 π}{6} cos^{2} \frac{π}{3}$

$= 2 (\frac{1}{2})^{2} + cosec^{2} (π + \frac{π}{6}) (\frac{1}{2})^{2} = 2 \times \frac{1}{4} + (- cosec \frac{π}{6})^{2} (\frac{1}{4}) = \frac{1}{2} + (- 2)^{2} (\frac{1}{4}) = \frac{1}{2} + \frac{4}{4} = \frac{1}{2} + 1 = \frac{3}{2} = R.H.S.$

$cot^{2} \frac{π}{6} + cosec \frac{5 π}{6} + 3 tan^{2} \frac{π}{6} = 6$

Sol. L.H.S. $= cot^{2} \frac{π}{6} + cosec \frac{5 π}{6} + 3 tan^{2} \frac{π}{6}$

$= (3)^{2} + cosec (π - \frac{π}{6}) + 3 (\frac{1}{3})^{2} = 3 + cosec \frac{π}{6} + 3 \times \frac{1}{3} = 3 + 2 + 1 = 6 = R.H.S.$

$2 sin^{2} \frac{3 π}{4} + 2 cos^{2} \frac{π}{4} + 2 sec^{2} \frac{π}{3} = 10$

Sol. L.H.S. $= 2 sin^{2} \frac{3 π}{4} + 2 cos^{2} \frac{π}{4} + 2 sec^{2} \frac{π}{3}$

$= 2 {sin (π - \frac{π}{4})}^{2} + 2 (\frac{1}{2})^{2} + 2 (2)^{2} = 2 {sin \frac{π}{4}}^{2} + 2 \times \frac{1}{2} + 8 = 2 (\frac{1}{2})^{2} + 1 + 8 = 1 + 1 + 8 = 10 = R.H.S.$

Find the value of: (i) $sin 7 5^{\circ}$ (ii) $tan 1 5^{\circ}$

Sol.

(i)

$sin 7 5^{\circ} = sin (4 5^{\circ} + 3 0^{\circ}) = sin 4 5^{\circ} cos 3 0^{\circ} + cos 4 5^{\circ} sin 3 0^{\circ} [sin (x + y) = sin x cos y + cos x sin y] = (\frac{1}{2}) (\frac{3}{2}) + (\frac{1}{2}) (\frac{1}{2}) = \frac{3}{2 2} + \frac{1}{2 2} = \frac{3 + 1}{2 2}$

(ii)

$tan 1 5^{\circ} = tan (4 5^{\circ} - 3 0^{\circ}) = \frac{tan 4 5 ^{\circ} - tan 3 0 ^{\circ}}{1 + tan 4 5 ^{\circ} tan 3 0 ^{\circ}} [tan (x - y) = \frac{tan x - tan y}{1 + tan x tan y}] = \frac{1 - \frac{1}{3}}{1 + 1 ( \frac{1}{3} )} = \frac{\frac{3 - 1}{3}}{\frac{3 + 1}{3}} = \frac{3 - 1}{3 + 1} = \frac{( 3 - 1 ) ^{2}}{( 3 + 1 ) ( 3 - 1 )} = \frac{3 + 1 - 2 3 ^{2}}{( 3 ) ^{2} - ( 1 ) ^{2}} = \frac{4 - 2 3}{3 - 1} = 2 - 3$

Prove the following :

$cos (\frac{π}{4} - x) cos (\frac{π}{4} - y) - sin (\frac{π}{4} - x) sin (\frac{π}{4} - y)$

$= sin (x + y)$

Sol. $cos (\frac{π}{4} - x) cos (\frac{π}{4} - y) - sin (\frac{π}{4} - x) sin (\frac{π}{4} - y)$

$= \frac{1}{2} [2 cos (\frac{π}{4} - x) cos (\frac{π}{4} - y)] + \frac{1}{2} [- 2 sin (\frac{π}{4} - x) sin (\frac{π}{4} - y)]$

$+ \frac{1}{2} [cos {(\frac{π}{4} - x) + (\frac{π}{4} - y)} - cos {(\frac{π}{4} - x) - (\frac{π}{4} - y)}]$

$∵ 2 cos A cos B = cos (A + B) + cos (A - B) - 2 sin A sin B = cos (A + B) - cos (A - B)]$

$= 2 \times \frac{1}{2} [cos {(\frac{π}{4} - x) + (\frac{π}{4} - y)}] = sin (x + y) = R.H.S.$

$\frac{t a n ( \frac{π}{4} + x )}{t a n ( \frac{π}{4} - x )} = (\frac{1 + t a n x}{1 - t a n x})^{2}$ Sol. It is known that

$tan (A + B) and tan (A - B) = \frac{tan A + tan B}{1 - tan A tan B} = \frac{tan A - tan B}{1 + tan A tan B}$

L.H.S.

$= \frac{tan ( \frac{π}{4} + x )}{tan ( \frac{π}{4} - x )} = \frac{( \frac{t a n \frac{π}{4} + t a n x}{1 - t a n \frac{π}{4} t a n x} )}{( \frac{t a n \frac{π}{4} - t a n x}{1 + t a n \frac{π}{4} t a n x} )} = \frac{( \frac{1 + t a n x}{1 - t a n x} )}{( \frac{1 - t a n x}{1 + t a n x} )} = (\frac{1 + tan x}{1 - tan x})^{2} = R.H.S.$

$\frac{c o s ( π + x ) c o s ( - x )}{s i n ( π - x ) c o s ( \frac{π}{2} + x )} = cot^{2} x$

Sol. L.H.S.

$= \frac{cos ( π + x ) cos ( - x )}{sin ( π - x ) cos ( \frac{π}{2} + x )} = \frac{[ - cos x ] [ cos x ]}{( sin x ) ( - sin x )} = \frac{- cos ^{2} x}{- sin ^{2} x} = cot^{2} x = R.H.S.$

$cos (\frac{3 π}{2} + x) cos (2 π + x)$

$[cot (\frac{3 π}{2} - x) + cot (2 π + x)] = 1$

Sol. L.H.S.

$= cos (\frac{3 π}{2} + x) cos (2 π + x) [cot (\frac{3 π}{2} - x) + cot (2 π + x)] = sin x cos x [tan x + cot x] = sin x cos x (\frac{sin x}{cos x} + \frac{cos x}{sin x}) = (sin x cos x) [\frac{sin ^{2} x + cos ^{2} x}{sin x cos x}] = 1 = R.H.S.$

$sin (n + 1) x sin (n + 2) x$

$+ cos (n + 1) x cos (n + 2) x = cos x$

Sol. L.H.S. $= sin (n + 1) x sin (n + 2) x$

$+ cos (n + 1) x cos (n + 2) x$

$= \frac{1}{2} [2 sin (n + 1) x sin (n + 2) x + 2 cos (n + 1) x cos (n + 2) x] = \frac{1}{2} [cos {(n + 1) x - (n + 2) x} - cos {(n + 1) x + (n + 2) x} + cos {(n + 1) x + (n + 2) x} + cos {(n + 1) x - (n + 2) x}]$

$[∵ - 2 sin A sin B = cos (A + B) - cos (A - B), 2 cos$

$A cos B = cos (A + B) + cos (A - B)]$

$cos (\frac{3 π}{4} + x) - cos (\frac{3 π}{4} - x) = - 2 sin x$

Sol. It is known that

$cos A - cos B = - 2 sin (\frac{A + B}{2}) \cdot sin (\frac{A - B}{2})$

L.H.S.

$= cos (\frac{3 π}{4} + x) - cos (\frac{3 π}{4} - x) = - 2 sin {\frac{( \frac{3 π}{4} + x ) + ( \frac{3 π}{4} - x )}{2}} \cdot sin {\frac{( \frac{3 π}{4} + x ) - ( \frac{3 π}{4} - x )}{2}} = - 2 sin (\frac{3 π}{4}) sin x = - 2 sin (π - \frac{π}{4}) sin x = - 2 sin \frac{π}{4} sin x = - 2 \times \frac{1}{2} \times sin x = - 2 sin x = R.H.S.$

$sin^{2} 6 x - sin^{2} 4 x = sin 2 x sin 10 x$

Sol. it is known that

$sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2}) sin A - sin B = 2 cos (\frac{A + B}{2}) sin (\frac{A - B}{2})$

L.H.S.

$= sin^{2} 6 x - sin^{2} 4 x = (sin 6 x + sin 4 x) (sin 6 x - sin 4 x) = [2 sin (\frac{6 x + 4 x}{2}) cos (\frac{6 x - 4 x}{2})] [2 cos (\frac{6 x + 4 x}{2}) \cdot sin (\frac{6 x - 4 x}{2})] = (2 sin 5 x cos x) (2 cos 5 x sin x) = (2 sin 5 x cos 5 x) (2 sin x cos x) = sin 10 x sin 2 x = R.H.S.$

$cos^{2} 2 x - cos^{2} 6 x = sin 4 x sin 8 x$

Sol. It is known that

$cos A + cos B = 2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2}) cos A - cos B = - 2 sin (\frac{A + B}{2}) sin (\frac{A - B}{2})$

L.H.S.

$= cos^{2} 2 x - cos^{2} 6 x = (cos 2 x + cos 6 x) (cos 2 x - 6 x) = [2 cos (\frac{2 x + 6 x}{2}) cos (\frac{2 x - 6 x}{2})] [- 2 sin (\frac{2 x + 6 x}{2}) sin (\frac{2 x - 6 x}{2})] = [2 cos 4 x cos (- 2 x)] [- 2 sin 4 x sin (- 2 x)] = [2 cos 4 x cos 2 x] [- 2 sin 4 x (- sin 2 x)] = (2 sin 4 x cos 4 x) (2 sin 2 x cos 2 x) = sin 8 x sin 4 x = R.H.S.$

$sin 2 x + 2 sin 4 x + sin 6 x = 4 cos^{2} x sin 4 x$

Sol. L.H.S.

$= sin 2 x + 2 sin 4 x + sin 6 x = [sin 2 x + sin 6 x] + 2 sin 4 x = [2 sin (\frac{2 x + 6 x}{2}) cos (\frac{2 x - 6 x}{2})] + 2 sin 4 x [∵ sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})] = 2 sin 4 x cos (- 2 x) + 2 sin 4 x = 2 sin 4 x cos 2 x + 2 sin 4 x = 2 sin 4 x (cos 2 x + 1) = 2 sin 4 x (2 cos^{2} x - 1 + 1) = 2 sin 4 x (2 cos^{2} x) = 4 cos^{2} x sin 4 x = R.H.S.$

$cot 4 x (sin 5 x + sin 3 x) = cot x (sin 5 x - sin 3 x)$

Sol. L.H.S.

$= cot 4 x (sin 5 x + sin 3 x) = \frac{cos 4 x}{sin 4 x} [2 sin (\frac{5 x + 3 x}{2}) cos (\frac{5 x - 3 x}{2})] ∵ sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})] = (\frac{cos 4 x}{sin 4 x}) [2 sin 4 x cos x] = 2 cos 4 x cos x R.H.S. = cot x (sin 5 x - sin 3 x) = \frac{cos x}{sin x} [2 cos (\frac{5 x + 3 x}{2}) sin (\frac{5 x - 3 x}{2})] [∵ sin A - sin B = 2 cos (\frac{A + B}{2}) sin (\frac{A - B}{2})] = 2 cos 4 x . cos x L.H.S. = R.H.S.$

$\frac{c o s 9 x - c o s 5 x}{s i n 17 x - s i n 3 x} = - \frac{s i n 2 x}{c o s 10 x}$ Sol. It is known that

$cos A - cos B = - 2 sin (\frac{A + B}{2}) sin (\frac{A - B}{2}) sin A - sin B = 2 cos (\frac{A + B}{2}) sin (\frac{A - B}{2})$

L.H.S.

$= \frac{cos 9 x - cos 5 x}{sin 17 x - sin 3 x} = \frac{- 2 sin ( \frac{9 x + 5 x}{2} ) \cdot sin ( \frac{9 x - 5 x}{2} )}{2 cos ( \frac{17 x + 3 x}{2} ) \cdot sin ( \frac{17 x - 3 x}{2} )} = \frac{- 2 sin 7 x \cdot sin 2 x}{2 cos 10 x \cdot sin 7 x} = - \frac{sin 2 x}{cos 10 x} = R.H.S.$

$\frac{s i n 5 x + s i n 3 x}{c o s 5 x + c o s 3 x} = tan 4 x$ Sol. It is known that

$sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2}) cos A + cos B = 2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2})$

L. H.S.

$= \frac{sin 5 x + sin 3 x}{cos 5 x + cos 3 x} = \frac{2 sin ( \frac{5 x + 3 x}{2} ) \cdot cos ( \frac{5 x - 3 x}{2} )}{2 cos ( \frac{5 x + 3 x}{2} ) \cdot cos ( \frac{5 x - 3 x}{2} )} = \frac{2 sin 4 x \cdot cos x}{2 cos 4 x \cdot cos x} = \frac{sin 4 x}{cos 4 x} = tan 4 x = R.H.S.$

$\frac{s i n x - s i n y}{c o s x + c o s y} = tan \frac{x - y}{2}$ Sol. It is known that

$sin A - sin B = 2 cos (\frac{A + B}{2}) sin (\frac{A - B}{2}) cos A + cos B = 2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2})$

L. H.S.

$= \frac{sin x - sin y}{cos x + cos y} = \frac{2 cos ( \frac{x + y}{2} ) \cdot sin ( \frac{x - y}{2} )}{2 cos ( \frac{x + y}{2} ) \cdot cos ( \frac{x - y}{2} )} = \frac{sin ( \frac{x - y}{2} )}{cos ( \frac{x - y}{2} )} = tan (\frac{x - y}{2}) = R.H.S.$

$\frac{s i n x + s i n 3 x}{c o s x + c o s 3 x} = tan 2 x$ Sol. It is known that

$sin A sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2}) cos A cos B = 2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2})$

L. H.S.

$= \frac{sin x + sin 3 x}{cos x + cos 3 x} = \frac{2 sin ( \frac{x + 3 x}{2} ) cos ( \frac{x - 3 x}{2} )}{2 cos ( \frac{x + 3 x}{2} ) cos ( \frac{x - 3 x}{2} )} = \frac{sin 2 x}{cos 2 x} = tan 2 x = R.H.S.$

$\frac{s i n x - s i n 3 x}{s i n ^{2} x - c o s ^{2} x} = 2 sin x$ Sol. It is known that

$sin A - sin B = 2 cos (\frac{A + B}{2}) sin (\frac{A - B}{2}) cos^{2} A - sin^{2} A = cos 2 A$

$L.H.S. = \frac{sin x - sin 3 x}{sin ^{2} x - cos ^{2} x} = \frac{2 cos ( \frac{x + 3 x}{2} ) sin ( \frac{x - 3 x}{2} )}{- cos 2 x} = \frac{2 cos 2 x sin ( - x )}{- cos 2 x} = - 2 \times (- sin x) = 2 sin x = R.H.S.$

$\frac{c o s 4 x + c o s 3 x + c o s 2 x}{s i n 4 x + s i n 3 x + s i n 2 x} = cot 3 x$

Sol. L.H.S.

$= \frac{cos 4 x + cos 3 x + cos 2 x}{sin 4 x + sin 3 x + sin 2 x} = \frac{( cos 4 x + cos 2 x ) + cos 3 x}{( sin 4 x + sin 2 x ) + sin 3 x} = \frac{2 cos ( \frac{4 x + 2 x}{2} ) cos ( \frac{4 x - 2 x}{2} ) + cos 3 x}{2 sin ( \frac{4 x + 2 x}{2} ) cos ( \frac{4 x - 2 x}{2} ) + sin 3 x} [∵ cos A + cos B = 2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2}), sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})] = \frac{2 cos 3 x cos x + cos 3 x}{2 sin 3 x cos x + sin 3 x} = \frac{cos 3 x ( 2 cos x + 1 )}{sin 3 x ( 2 cos x + 1 )} = cot 3 x = R.H.S.$

$cot x cot 2 x - cot 2 x cot 3 x - cot 3 x cot x = 1$

Sol. L.H.S.

$= cot x cot 2 x - cot 2 x cot 3 x - cot 3 x cot x = cot x cot 2 x - cot 3 x (cot 2 x + cot x) = cot x cot 2 x - cot (2 x + x) (cot 2 x + cot x) = cot x cos 2 x - [\frac{cot 2 x cot x - 1}{cot x + cot 2 x}] (cot 2 x + cot x) [∵ cot (A + B) = \frac{cot A cot B - 1}{cot A + cot B}] = cot x cot 2 x - (cot 2 x cot x - 1) = 1 = R.H.S.$

$tan 4 x = \frac{4 t a n x ( 1 - t a n ^{2} x )}{1 - 6 t a n ^{2} x + t a n ^{4} x}$

Sol. It is known that $tan 2 A = \frac{2 t a n A}{1 - t a n ^{2} A}$ L.H.S.

$= tan 4 x = tan 2 (2 x) = \frac{2 tan 2 x}{1 - tan ^{2} ( 2 x )} = \frac{2 ( \frac{2 t a n x}{1 - t a n ^{2} x} )}{1 - ( \frac{2 t a n x}{1 - t a n ^{2} x} ) ^{2}} = \frac{( \frac{4 t a n x}{1 - t a n ^{2} x} )}{[ 1 - \frac{4 t a n ^{2} x}{( 1 - t a n ^{2} x ) ^{2}} ]} = \frac{( \frac{4 t a n x}{1 - t a n ^{2} x} )}{[ \frac{( 1 - t a n ^{2} x ) ^{2} - 4 t a n ^{2} x}{( 1 - t a n ^{2} x ) ^{2}} ]} = \frac{4 tan x ( 1 - tan ^{2} x )}{( 1 - tan ^{2} x ) ^{2} - 4 tan ^{2} x} = \frac{4 tan x ( 1 - tan ^{2} x )}{1 + tan ^{4} x - 2 tan ^{2} x - 4 tan ^{2} x} = \frac{4 tan x ( 1 - tan ^{2} x )}{1 - 6 tan ^{2} x + tan ^{4} x} = R.H.S.$

$cos 4 x = 1 - 8 sin^{2} x cos^{2} x$

Sol. L.H.S.

$= cos 4 x = cos 2 (2 x) = 1 - 2 sin^{2} 2 x [cos 2 A = 1 - 2 sin^{2} A] = 1 - 2 (2 sin x cos x)^{2} [sin 2 A = 2 sin A cos A] = 1 - 8 sin^{2} x cos^{2} x = R H S$

$cos 6 x = 32 cos^{6} x - 48 cos^{4} x + 18 cos^{2} x - 1$

Sol. L.H.S.

$= cos 6 x = cos 3 (2 x) = 4 cos^{3} 2 x - 3 cos 2 x [∵ cos 3 A = 4 cos^{3} A - 3 cos A] = 4 [(2 cos^{2} x - 1)^{3} - 3 (2 cos^{2} x - 1) [∵ cos 2 x = 2 cos^{2} x - 1] = 4 [(2 cos^{2} x)^{3} - (1)^{3} - 3 (2 cos^{2} x)^{2} + 3 (2 cos^{2} x)] - 6 cos^{2} x + 3 = 4 [8 cos^{6} x - 1 - 12 cos^{4} x + 6 cos^{2} x] - 6 cos^{2} x + 3 = 32 cos^{6} x - 4 - 48 cos^{4} x + 24 cos^{2} x - 6 cos^{2} x + 3 = 32 cos^{6} x - 48 cos^{4} x + 18 cos^{2} x - 1 = R.H.S.$

MISCELLANEOUS EXERCISE

Prove that :

$2 cos \frac{π}{13} cos \frac{9 π}{13} + cos \frac{3 π}{13} + cos \frac{5 π}{13} = 0$

Sol. L.H.S.

$= 2 cos \frac{π}{13} cos \frac{9 π}{13} + cos \frac{3 π}{13} + cos \frac{5 π}{13} = 2 cos \frac{π}{13} cos \frac{9 π}{13} + 2 cos (\frac{3 π}{13} + \frac{5 π}{13}) cos (\frac{\frac{3 π}{13} - \frac{5 π}{13}}{2}) [cos x + cos y = 2 cos (\frac{x + y}{2}) cos (\frac{x - y}{2})] = 2 cos \frac{π}{13} cos \frac{9 π}{13} + 2 cos \frac{4 π}{13} cos (\frac{- π}{13}) = 2 cos \frac{π}{13} cos \frac{9 π}{13} + 2 cos \frac{4 π}{13} cos \frac{π}{13} = 2 cos \frac{π}{13} [cos \frac{9 π}{13} + cos \frac{4 π}{13}] = 2 cos \frac{π}{13} [2 cos (\frac{9 π}{13} + \frac{4 π}{13}) cos (\frac{\frac{9 π}{13} - \frac{4 π}{13}}{2})] = 2 cos \frac{π}{13} [2 cos \frac{π}{2} cos \frac{5 π}{26}] = 2 cos \frac{π}{13} \times 2 \times 0 \times cos \frac{5 π}{26} = 0 = R.H.S.$

$(sin 3 x + sin x) sin x + (cos 3 x - cos x) cos x$

$= 0$

Sol. L.H.S.

$= (sin 3 x + sin x) sin x + (cos 3 x - cos x) cos x = sin 3 x sin x + sin^{2} x + cos 3 x cos^{2} x - cos^{2} x = cos 3 x cos x + sin 3 x sin x - (cos^{2} x - sin^{2} x) = cos (3 x - x) - cos 2 x [cos (A - B) = cos A cos B + sin A sin B] = cos 2 x - cos 2 x = 0 = RH.S.$

$(cos x + cos y)^{2} + (sin x - sin y)^{2} = 4 cos^{2} \frac{x + y}{2}$ Sol. L.H.S.

$= = = = = = = = = (cos x + cos y)^{2} + (sin x - sin y)^{2} cos^{2} x + cos^{2} y + 2 cos x cos y + sin^{2} x + sin^{2} y - 2 sin x sin y (cos^{2} x + sin^{2} x) + (cos^{2} y + sin^{2} y) + 2 (cos x cos y - sin x sin y) 1 + 1 + 2 cos (x + y) [cos (A + B) = (cos A cos B - sin A sin B)] 2 + 2 cos (x + y) 2 [1 + cos (x + y)] 2 [1 + 2 cos^{2} (\frac{x + y}{2}) - 1] [cos 2 A = 2 cos^{2} A - 1] 4 cos^{2} (\frac{x + y}{2}) R.H.S.$

$(cos x - cos y)^{2} + (sin x - sin y)^{2} = 4 sin^{2} \frac{x - y}{2}$ Sol. L.H.S.

$= = = = = = = = (cos x - cos y)^{2} + (sin x - sin y)^{2} cos^{2} x + cos^{2} y - 2 cos x cos y + sin^{2} x + sin^{2} y - 2 sin x sin y (cos^{2} x + sin^{2} x) + (cos^{2} y + sin^{2} y) - 2 [cos x cos y + sin x sin y] 1 + 1 - 2 [cos (x - y)] [cos (A - B) = cos A cos B + sin A sin B] 2 [1 - cos (x - y)] 2 [1 - {1 - 2 sin^{2} (\frac{x - y}{2})}] [cos 2 A = 1 - 2 sin^{2} A] 4 sin^{2} (\frac{x - y}{2}) R.H.S.$

$sin x + sin 3 x + sin 5 x + sin 7 x = 4 cos x cos 2 x sin 4 x$

Sol. It is known that

$sin A + sin B = 2 sin (\frac{A + B}{2}) \cdot cos (\frac{A - B}{2})$

L.H.S.

$= sin x + sin 3 x + sin 5 x + sin 7 x = (sin x + sin 5 x) + (sin 3 x + sin 7 x) = 2 sin (\frac{x + 5 x}{2}) \cdot cos (\frac{x - 5 x}{2}) + sin (\frac{3 x + 7 x}{2}) cos (\frac{3 x - 7 x}{2}) = 2 sin 3 x cos (- 2 x) + 2 sin 5 x cos (- 2 x) = 2 sin 3 x cos 2 x + 2 sin 5 x cos 2 x = 2 cos 2 x [sin 3 x + sin 5 x] = 2 cos 2 x [2 sin (\frac{3 x + 5 x}{2}) \cdot cos (\frac{3 x + 5 x}{2})] = 2 cos 2 x [2 sin 4 x \cdot cos (- x)] = 4 cos 2 x sin 4 x cos x = R.H.S.$

$\frac{( s i n 7 x + s i n 5 x ) + ( s i n 9 x + s i n 3 x )}{( c o s 7 x + c o s 5 x ) + ( c o s 9 x + c o s 3 x )} = tan 6 x$ Sol. It is known that

$sin A + sin B = 2 sin (\frac{A + B}{2}) \cdot cos (\frac{A - B}{2}) cos A + cos B = 2 cos (\frac{A + B}{2}) \cdot cos (\frac{A - B}{2})$

L.H.S.

$= \frac{( sin 7 x + sin 5 x ) + ( sin 9 x + sin 3 x )}{( cos 7 x + cos 5 x ) + ( cos 9 x + cos 3 x )} = \frac{[ 2 sin ( \frac{7 x + 5 x}{2} ) \cdot cos ( \frac{7 x - 5 x}{2} ) ] + [ 2 sin ( \frac{9 x + 3 x}{2} ) \cdot cos ( \frac{9 x - 3 x}{2} ) ]}{[ 2 cos ( \frac{7 x + 5 x}{2} ) \cdot cos ( \frac{7 x - 5 x}{2} ) ] + [ 2 cos ( \frac{9 x + 3 x}{2} ) \cdot cos ( \frac{9 x - 3 x}{2} ) ]} = \frac{[ 2 sin 6 x \cdot cos x ] + [ 2 sin 6 x \cdot cos x ]}{[ 2 cos 6 x \cdot cos x ] + [ 2 cos 6 x \cdot cos 3 x ]} = \frac{2 sin 6 x [ cos x + cos 3 x ]}{2 cos 6 x [ cos x + cos 3 x ]} = tan 6 x = R.H.S.$

$sin 3 x + sin 2 x - sin x = 4 sin x cos \frac{x}{2} cos \frac{3 x}{2}$ Sol. L.H.S. $= sin 3 x + sin 2 x - sin x$

$= sin 3 x + (sin 2 x - sin x)$

$= 2 sin \frac{3 x}{2} \cdot cos \frac{3 x}{2} + 2 cos \frac{3 x}{2} sin \frac{x}{2}$

$[sin 2 A = 2 sin A \cdot cos B] = 2 cos (\frac{3 x}{2}) [sin (\frac{3 x}{2}) + sin (\frac{x}{2})] = 2 cos (\frac{3 x}{2}) [2 sin {\frac{( \frac{3 x}{2} ) + ( \frac{x}{2} )}{2}} cos {\frac{( \frac{3 x}{2} ) - ( \frac{x}{2} )}{2}}] [sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})] = 2 cos (\frac{3 x}{2}) \cdot 2 sin x cos (\frac{x}{2}) = 4 sin x cos (\frac{x}{2}) cos (\frac{3 x}{2}) = R.H.S.$

Find $sin \frac{x}{2}, cos \frac{x}{2}$ and $tan \frac{x}{2}$ , in Q. 8 to 10 of the following : 8. $tan x = - \frac{4}{3}$ , x in quadrant II. Sol. Here, x is in quadrant II. i.e. $\frac{π}{2} < x < π$ $\Rightarrow \frac{π}{4} < \frac{π}{2} < \frac{π}{2}$

Therefore, $sin \frac{x}{2}, cos \frac{x}{2}$ and $tan \frac{x}{2}$ are lines in first quadrant. It is given that $tan x = - \frac{4}{3}$

$sec^{2} x = 1 + tan^{2} x = 1 + (\frac{- 4}{3})^{2} = 1 + \frac{16}{9} = \frac{25}{9}$

$∴ cos^{2} x = \frac{9}{25} \Rightarrow cos x = \pm \frac{3}{5}$

As x is in quadrant II, cosx is negative.

$cos x = \frac{- 3}{5}$

Now, $cos x = 2 cos^{2} \frac{x}{2} - 1$

$\Rightarrow \frac{- 3}{5} = 2 cos^{2} \frac{x}{2} - 1 \Rightarrow 2 cos^{2} \frac{x}{2} = 1 - \frac{3}{5}$

$\Rightarrow cos \frac{x}{2} = \frac{1}{5} [∵ cos \frac{x}{2} is positive]$

$\Rightarrow sin^{2} \frac{x}{2} + (\frac{1}{5})^{2} = 1$

$\Rightarrow sin^{2} \frac{x}{2} = 1 - \frac{1}{5} = \frac{4}{5}$

$\Rightarrow sin \frac{x}{2} = \frac{2}{5} [∵ sin \frac{x}{2} is positive]$

$∴ sin \frac{x}{2} = \frac{2 5}{5} tan \frac{x}{2} = \frac{sin \frac{x}{2}}{cos \frac{x}{2}} = \frac{( \frac{2}{5} )}{( \frac{1}{5} )} = 2$

Thus, the respective values of $sin \frac{x}{2}, cos \frac{x}{2}$ and $tan \frac{x}{2}$ are $\frac{2 5}{5}, \frac{5}{5}$ and 2.

9. $cos x = - \frac{1}{3}$ , x in quadrant III.

Sol. Here, x is in quadrant III.

$i.e., π < x < \frac{3 π}{2} \Rightarrow \frac{π}{2} < \frac{x}{2} < \frac{3 π}{4}$

Therefore, $cos \frac{x}{2}$ and $tan \frac{x}{2}$ are negative, where $sin \frac{x}{2}$ as is positive. It is given that $cos x = - \frac{1}{3}$

$cos x = 1 - 2 sin^{2} \frac{x}{2}$

$\Rightarrow sin^{2} \frac{x}{2} = \frac{1 - cos x}{2} \Rightarrow sin^{2} \frac{x}{2} = \frac{1 - ( - \frac{1}{3} )}{2} = \frac{( 1 + \frac{1}{3} )}{2} = \frac{\frac{4}{3}}{2} = \frac{2}{3}$

$\Rightarrow sin \frac{x}{2} = \frac{2}{3} [∵ sin \frac{x}{2} is positive]$

$∴ sin \frac{x}{2} = \frac{2}{3} \times \frac{3}{3} = \frac{6}{3}$

Now, $cos x = 2 cos^{2} \frac{x}{2} - 1$

$\Rightarrow cos^{2} \frac{x}{2} = \frac{1 + cos x}{2} = \frac{1 + ( - \frac{1}{3} )}{2} = \frac{( \frac{3 - 1}{3} )}{2} = \frac{( \frac{2}{3} )}{2} = \frac{1}{3}$

$\Rightarrow cos \frac{x}{2} = - \frac{1}{3} [∵ cos \frac{x}{2} is negative]$

$∴ cos \frac{x}{2} = - \frac{1}{3} \times \frac{3}{3} = \frac{- 3}{3} tan \frac{x}{2} = \frac{sin \frac{x}{2}}{cos \frac{x}{2}} = \frac{( \frac{2}{3} )}{( \frac{- 1}{3} )} = - 2$

Thus, the respective values of $sin \frac{x}{2}, cos \frac{x}{2}$ , $tan \frac{x}{2}$ are $\frac{6}{3}, \frac{- 3}{3}$ and $- 2$

10. $sin x = \frac{1}{4}$ , x in quadrant II.

Sol. Here, x is in quadrant II.

$i.e., \Rightarrow \frac{3 π}{2} < x < π \frac{π}{4} < \frac{x}{2} < \frac{π}{2}$

Therefore, $sin \frac{x}{2}, cos \frac{x}{2}$ and $tan \frac{x}{2}$ are all positive.

$It is given that sin x = \frac{1}{4} cos^{2} x = = 1 - sin^{2} x = 1 - (\frac{1}{4})^{2} 1 - \frac{1}{16} = \frac{15}{16}$

$\Rightarrow cos x = - \frac{15}{4}$ [cos x is negative in quadrant II]

$sin^{2} \frac{x}{2} = \frac{1 - cos x}{2} = \frac{1 - ( - \frac{15}{4} )}{2} = \frac{4 + 15}{8}$

$\Rightarrow sin \frac{x}{2} = \frac{4 + 15}{8} [∵ sin \frac{x}{2}$ is positive $]$

$sin \frac{x}{2} cos^{2} \frac{x}{2} = \frac{4 + 15}{8} \times \frac{2}{2} = \frac{8 + 2 15}{16} = \frac{8 + 2 15}{4} = \frac{1 + cos x}{2} = \frac{1 + ( - \frac{15}{4} )}{2} = \frac{4 - 15}{8}$

$\Rightarrow cos \frac{x}{2} = \frac{4 - 15}{8} [∵ cos \frac{x}{2}$ is positive $]$

$cos \frac{x}{2} = \frac{4 - 15}{8} \times \frac{2}{2} = \frac{8 - 2 15}{16} = \frac{8 - 2 15}{4}$

$tan \frac{x}{2} = \frac{s i n \frac{x}{2}}{c o s \frac{x}{2}} = \frac{( \frac{8 + 2 15}{4} )}{( \frac{8 - 2 15}{4} )}$

$= \frac{8 + 2 15}{8 - 2 15}$

Thus, the respective values of $sin \frac{x}{2}, cos \frac{x}{2}$ and $tan \frac{x}{2}$ are $\frac{8 + 2 15}{4}, \frac{8 - 2 15}{4}$ and

$4 + 15$

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

Find NCERT Solutions for every Class 11 Maths chapter, with answers to textbook exercises, key formulas, and clear explanations for solving questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sets
Chapter 2	Relations and Functions
Chapter 4	Complex Numbers and Quadratic Equations
Chapter 5	Linear Inequalities
Chapter 6	Permutations and Combinations
Chapter 7	Binomial Theorem
Chapter 8	Sequence and Series
Chapter 9	Straight Lines
Chapter 10	Conic Sections
Chapter 11	Introduction to Three-dimensional Geometry
Chapter 12	Limits and Derivatives
Chapter 13	Statistics
Chapter 14	Probability

4.0Class 11 Maths Chapter 3 Trigonometric Functions: Exercise-wise Questions and Topics

Exercise	Number of Questions	Important Topics Covered
Exercise 3.1	7 Questions and Solutions	Trigonometric ratios, Pythagoras theorem, and complementary angles
Exercise 3.2	10 Questions and Solutions	Trigonometric ratios of special angles: 0°, 30°, 45°, 60°, and 90°
Exercise 3.3	25 Questions and Solutions	Trigonometric ratios, Pythagoras theorem, double-angle formula, and half-angle formula

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 3

Clear Understanding of the Unit Circle: The solutions explain how the coordinates of points on the unit circle are related to the values of sin θ and cos θ. This helps students understand trigonometric functions and their values.

Step-by-Step Explanation of Trigonometric Formulas: Important addition and subtraction formulas are explained with clear steps. Students can learn how to apply these formulas while solving questions.

Easy Explanation of Important Angles: The solutions explain trigonometric values at important angles, including angles of the form nπ and (2n + 1)π/2. Students can understand when trigonometric functions are zero or undefined.

Simple Methods for Trigonometric Identities: The solutions show how to prove trigonometric identities step by step. Students can learn how to choose suitable identities and simplify expressions correctly.

Prepared by ALLEN Subject Experts: These NCERT Solutions are prepared by ALLEN subject experts and follow the NCERT syllabus. The solutions use standard mathematical formulas, symbols, and notation.

Easy-to-Understand Solutions: The concepts are explained in simple language with clear steps. This helps students understand trigonometric functions, learn important formulas, and solve questions with greater ease.

Coverage of NCERT Exercises and Examples: The solutions cover the exercises and examples from NCERT Class 11 Maths Chapter 3 – Trigonometric Functions. Students can use them to understand concepts, check their answers, and revise the chapter.

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Frequently Asked Questions

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NCERT Solutions Class 11 Maths Chapter 3 – Trigonometric Functions

1.0Class 11 Maths Chapter 3 Trigonometric Functions: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 3

EXERCISE - 3.1

EXERCISE - 3.2

Find the value of the trigonometric function in Q. 6 to 10

EXERCISE - 3.3

Prove the following :

MISCELLANEOUS EXERCISE

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

4.0Class 11 Maths Chapter 3 Trigonometric Functions: Exercise-wise Questions and Topics

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 3

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