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1 All In One Multi-Color Printing A way to 100% PLUS ONE MATHEMATICS REFERENCE BOOK-CHAPTER 1 f(x) = 5x + For CBSE Schools/ State Boards with NCERT Syllabus Theory explained with simple examples NCERT Exercises Fully Solved Solutions of Kerala and Karnataka Boards Question Papers Prof.Narayanan Gopi M.Sc, M.Phil Prof.Narayana Pillai Raju M.Sc, PhD

2 Visit for CBSE,State Board QP s, Plus 1 & Reference Books WELCOME Thank you for visiting our website and downloading Chapter : Trigonometric Functions of the book (Plus One Mathematics Reference Book). We are sure that you will really enjoy the way the contents are presented and students will find it useful.if you wish to have a copy of the book you can buy the full book (94 pages) from the same site at a cost of just 50. The book is designed in multi-color and all NCERT Textbook Exercises are solved in detail. The theory is explained with simple examples, so that students can cover the topic with little effort and time. As you know,about 0 State Boards in India including Kerala and Karnataka have adopted NCERT syllabus and Textbooks for Plus One and Plus Two courses. Since at present CBSE is not conducting annual examination for Plus one course, Kerala and Karnataka Board Question papers are included in our Reference book with an intention to help students to practice the topics effectively. Kerala Board Question papers are arranged chapter-wise and Karnataka Board Question papers and Model papers are given as it is in full, separately (and therefore not available in chapter ). Plus Two Reference Books contain CBSE Question papers as well as State Board Question Papers. You can visit for Plus One and Plus Two Question papers and Reference Books. Narayanan Gopi Narayana Pillai Raju

3 Visit for CBSE,State Board QP s, Plus 1 & Reference Books WARNING No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise. Any clarifications in this regard may be addressed to the authors. LICENSE AGREEMENT Your license is for a single user and please do not distribute the PDF document to any other Person or Institution. Also you are requested to not share the book through any Website or Face Book page or Whatsapp or any similar electronic media, but the free chapter (Chapter ) provided in the Website can be downloaded and shared with anybody as you wish. All pages of the document are stamped using an invisible watermark. It is your responsibility to honor the license agreement. Narayanan Gopi Narayana Pillai Raju

4 Visit for CBSE,State Board QP s, Plus 1 & Reference Books About the Authors Narayanan Gopi Narayanan Gopi has an extensive teaching experience of more than 0 years, working at different Government colleges in Kerala. He has a post graduate and an M.Phil degree from University of Kerala and was also been a member of Board of Studies, University of Calicut. To pursue his academic teaching interest after retirement, he was appointed as the Head of the Department of Mathematics at Nehru College of Engineering, Pampady, Kerala. Other than teaching, his passions lie in learning computer languages and developing Mathematical and Statistical packages. Dr.Narayana Pillai Raju Dr.Raju has more than 0 years of graduate and post-graduate teaching experience. He obtained his M.Sc from University of Kerala and Ph.D from University of Calicut. He has published a number of articles in National and International Journals and has supervised four graduate students for their Ph.D. As a member of various academic bodies in Universities, he has contributed significantly towards the development of syllabus geared to cater graduate as well as postgraduate programs. He has served as an integral liaison for a number of educational workshops, orientation programs and refresher courses for students and teachers alike.

5 Visit for CBSE,State Board QP s, Plus 1 & Reference Books Chapter TRIGONOMETRIC FUNCTIONS Trigonometry is an important branch of Mathematics and has tremendous applications in Astronomy, Engineering, Physics, Surveying,Navigation and various other branches of Mathematics. The initial works on trigonometry was started in India, perhaps around third and fourth century. The sine function was invented in India. The ancient Indian mathematicians, Aryabhatta(476 AD), Brahmagupta(598 AD), Bhaskara I(600 AD) and Bhaskara II (1114 AD) had made a lot of contributions to the literature. The word Trigonometry means measurements of a triangle. In its early stages, trigonometry was mainly concerned with the study of relationships among angles and sides of a right angled triangle. Later, its application has been extended beyond the study of measurements of right angled triangles. We shall begin our discussion with the study of trigonometrical ratios of acute angles of a right triangle(right angled triangle). These ratios are also called trigonometric functions of acute angles. Later we shall study trigonometric functions in a wider sense. Trigonometric ratios of an acute angle Let PQR be a right triangle ( PQR = 90 ) with PRQ = θ. PQ is the opposite side of θ and QR the adjacent side of θ. PR is the hypotenuse of PQR. Let PQ = x, QR = y and PR = r. Next we shall define the six trigonometric ratios namely sinθ, cosθ, tanθ, cotθ, secθ and cosec θ(cscθ). Q P x y r θ R These ratios are abbreviations of sine θ, cosine θ, tangent θ, cotangent θ, secant θ and cosecant θ, respectively

6 Visit for CBSE,State Board QP s, Plus 1 & Reference Books sinθ = x r, cosθ = y r, tanθ = x, (y 0) y cotθ = y x, (x 0), secθ = r y, (y 0), cosec θ = r, (x 0). x sinθ cosec θ = 1, cosθ secθ = 1, tanθ cotθ = 1. tanθ = x y = x/r y/r = sinθ cosθ sin θ + cos θ = x + y r = cotθ = cosθ sinθ. 1 + tan θ =1 + x y = x + y y = r y = sec θ. 1 + cot θ =1 + y x = x + y x = r x = cosec θ. Example If sinx = 1, find other trigonometric ratios. Draw the simplest right triangle for which = r r = 1( x + y = r by Pythagorean theorem) sinx = 1. PQ = 1, QR = 9 1 = 8, PR = r =. 8 By definition, cos x =, tanx = 1, 8 8 cotx = 1 = 8, secx =,cosec x = 8 1 =. Angles and Angle Measures Q P 1 8 x R An Angle may be generated by rotating two rays that share a fixed endpoint(point of rotation) at the origin. Usually one ray is fixed along the positive x axis and it is called the initial side of the angle and the other ray is called the terminal side of the angle. The point of rotation is called vertex (O). Rotations in the anticlockwise directions are considered as positive and in the clockwise directions as negative. Therefore measure of an angle is determined by the amount and direction of rotation from the initial side to the terminal side

7 Visit for CBSE,State Board QP s, Plus 1 & Reference Books Y Y Y O X O X O X blue line - initial side Positive angles are shown in the figure sky blue line - terminal side Negative angles, generated by the rotation in the clockwise direction, are shown in the next figure. Y Y Y O 45 X blue line - initial side 15 O Negative angles are shown in the figure X 15 O X sky blue line - terminal side If a wheel is rotated from the initial side to the terminal side then the terminal side coincides with the initial side. Then we say that the wheel completes one revolution. One revolution is usually used to measure large angles, for example, a spinning wheel is making an angle of 5(say)revolutions per second. An angle of 405 is shown in the figure given below

8 Visit for CBSE,State Board QP s, Plus 1 & Reference Books Y 405 O X Angle in standard position An angle is in standard position if its vertex is at the origin and its initial side is along the positive x axis. Angles in the figures shown above are in the standard position. Now we shall discuss the most commonly used measures of angle namely Degree Measure and Radian Measure and the relation between them. ( 1 60 )th of a revolution is called a degree and is usually denoted by 1. One complete rotation(counter-clockwise) = 1 revolution = 60. A degree is divided into minutes and a minute into seconds. 1 = 60 minutes (60 ), 1 = 60 seconds (60 ) Radian measure Measure of the angle subtended at the center of the circle of radius r by an arc of length r is called 1 radian. In the figure length of arc AB = r and radius = r = OA = OB. r O O 1 radian r A r B Since the circumference of a circle of radius 1 unit is π (1) = π, one complete revolution of the initial side subtends an angle of π radians at the center

9 Visit for CBSE,State Board QP s, Plus 1 & Reference Books In a circle of radius r, an arc of length l, subtends an angle θ radian at the center. Then l = rθ. Relation between Degree measure and Radian measure π radian = 60 = π radian = 180. = 1 radian = 180 π Example and 1 = π 180 radian. Convert 50 into radian measure. 1 = π 180 radian = 50 = 50 π 180 = 5π 18 radian. Example Convert 5π 18 1 radian = 180 π radians into degree measure. = 5π 18 radians = (5π 18 )180 π = 50. Relation between Degree measure and radian measure for some common angles. Degree π π π π π π 5 π π Radian π π Unit Circle: A circle with centre at the origin (O(0,0)) and radius is equal to 1 unit is called the unit circle, defined in the two dimensional plane. The equation of the unit circle is x + y = 1 Relation between Radian measure and Real numbers 111

10 Visit for CBSE,State Board QP s, Plus 1 & Reference Books Associating Real Numbers with Points on the Unit Circle Think of the real line R, as an infinite string. Black dot represents the number 0, red line represents positive real numbers and green line represents negative real numbers. R Pick up this string and wrap it around the unit circle. Put the black dot at the point (1, 0) on the unit circle. Wrap the positive real numbers counter-clockwise. Wrap the negative real numbers clockwise. In this way every real number can be associated with a unique point on the unit circle. (1,0) Thus radian measures of angles and real numbers can be considered as one and the same. Trigonometric Functions So far we have defined trigonometric ratios(functions) for acute angles only(angles between 0 and π ); i.e we have defined trigonometric functions whose domains are the set of acute angles of a right triangle. Now we can extend the definition of trigonometric functions to include angles of any measure. Quadrantal Angles Angles in standard position having their terminal sides along the x axis or y axis are called Quadrantal angles. Angles 0, 90, 180, 70,60, 450 etc are Quadrantal angles. Coterminal Angles Angles like 90 and 0 are Coterminal angles. These angles have the same initial and terminal sides, but different amounts of rotation. Similarly 40 and 60 are Coterminal angles. In other words, when two angles in standard position have the same terminal sides, they are called coterminal angles. Cotermihttp:// 11

11 Visit for CBSE,State Board QP s, Plus 1 & Reference Books nal angles can be found by adding ( subtracting ) an integral multiple of π (or 60 ). Example Find a positive coterminal angle and a negative coterminal angle of 160. Positive coterminal angle = = 50. Negative coterminal angle = = 00. Find a positive coterminal angle and a negative coterminal angle of 11 π. Positive coterminal angle = 11 π 17 π + π =. Negative coterminal angle = 11 π (π) = π. Note: One complete rotation(counter-clockwise) = 60. Note: One complete rotation(clockwise) = 60. Trigonometric functions of any angle θ We shall divide the procedure into steps 1. θ is an acute angle in standard position (0 < θ < 90 ).. θ is a non-quadrantal angle in standard position (90 < θ < 60 ). θ is a quadrantal angle in standard position (θ = 0, 90, 180, 70,60 ). 4. θ < 0 or θ > Let P(x,y) be a point on the terminal side of θ, where θ is in standard position (x > 0,y > 0). Let OPQ be a right triangle with POQ = θ. PQ is the opposite side of θ and OQ is the adjacent side of θ. OP is the hypotenuse of OPQ. r P(x, y) x PQ = x, OQ = y, OP = r. We can define six trigonometric functions namely sinθ, cosθ, tanθ, cotθ, secθ and cosec θ(cscθ). O θ y Q 11

12 Visit for CBSE,State Board QP s, Plus 1 & Reference Books sinθ = x r, cosθ = y r, tanθ = x y, (y 0) cotθ = y, (x 0) x secθ = r y, (y 0) cosec θ = r, (x 0). x 1 + tan θ = 1 + x y = x + y y = r y = sec θ. 1 + cot θ = 1 + y x = x + y x = r x = cosec θ. When the terminal side of θ lies in the first quadrant, all trigonometric functions are positive.. To evaluate a non-quadrantal angle θ > 90, we need the concept of reference angles. The definition is given below. Reference angles Let θ be a non-quadrantal angle in standard position. Then its reference angle θ is an acute angle formed by the terminal side of θ and the x axis(see the diagrams shown below). θ θ θ = π θ θ θ θ = θ π Note: In the first quadrant θ = θ θ θ = π θ θ The trigonometric functions for any angle are determined using the following rule i) Find the reference angle θ. ii) Evaluate the function at the reference angle iii) The sign of the function is determined by the quadrant in which the terminal side of θ lies. The signs of trigonometric functions in all the 4 quadrants are shown in the 114

13 Visit for CBSE,State Board QP s, Plus 1 & Reference Books following table. Function\Quadrant I II III IV sin x + + cos x + + tan x + + cot x + + sec x + + cosec x + + Example Evaluate cos40. cos40 is negative ( cos function is negative in the rd quadrant). Reference angle θ = = 60. cosθ = cos60 = 1 = cos40 = 1. Evaluate sin 11π 6 sin 11π 6. is negative ( sin function is negative in the 4 th Reference angle θ = π 11π 6 = π 6. sin π 6 = 1 = sin 11π 6 = 1. quadrant)

14 Visit for CBSE,State Board QP s, Plus 1 & Reference Books. Consider a unit circle with centre at the origin O and radius = 1 unit. Let P(a, b) be any point on the unit circle. Let the length of arc PQ = x. Then QOP = x radian. Draw PM OQ. Then by Pythagorean theorem, we have a + b = 1 (1). Put a = cosx and b = sinx, in (1), we get cos x + sin x = 1. Hence P(cos x, sin x) is any point on the unit circle. S( 1, 0) T (0,1) O 1 x a R(0, 1) P(a, b) b M Q(1, 0) The coordinates of the point Q, T, S, R are (1, 0), (0, 1), ( 1, 0),(0, 1) respectively. Since the arc QT subtends an angle π at the centre, we have cos π = 0 and sin π = 1. The arc QS subtends an angle π at the centre. cos π = 1, sin π = 0. Similarly the arc QR subtends an angle cos π = 0, sin π = 1. π at the centre. Since one complete revolution subtends π at the centre, we have cos π = 1, sin π = If θ > 60 or θ < 0, the procedure for finding the trigonometric ratios are given below. i) Find the coterminal angle of θ. ii) Find the reference angle of coterminal angle. iii) Evaluate the function at the reference angle. iv) Determine the sign of the function,by looking at the quadrant in which the terminal side of θ lies. Example ( ) π Evaluate tan 116

15 Visit for CBSE,State Board QP s, Plus 1 & Reference Books θ = π lies in the rd quadrant and tangent function is positive in the rd ( ) π quadrant. Hence tan > 0. Coterminal angle = π + π = 4π. Coterminal angle lies in the third quadrant. Therefore reference angle θ = 4π π = π. ( ) π tan = tan π =. Evaluate cos40. θ = 40. Terminal side of θ lies in the first quadrant and therefore cos40 > 0. Coterminal angle = = 60, which is in the first quadrant and hence the coterminal angle and reference angle are the same. cos40 = cos60 = 1. Evaluate cos( 10 ). θ = 10. Terminal side of θ lies in the rd negative in the third quadrant. cos( 10 ) < 0. quadrant. Cosine function is Coterminal angle = = 40, which is in the third quadrant. Reference angle = = 60. cos( 10 ) = cos60 = 1. Trigonometric functions for some well known angles are given in the following table

16 Visit for CBSE,State Board QP s, Plus 1 & Reference Books function\θ 0 sinθ 0 cosθ 1 tanθ 0 π π π π π π (Not defined) π (Not defined) 0 Other trigonometric functions can be obtained,whenever they are defined, by using the relations given below. cosec θ = 1 sinθ, secθ = 1 cosθ, cotθ = 1 tanθ. Some important results sin( x) = sinx, cos( x) = cosx, tan( x) = tanx, x R and cot( x) = cotx, sec( x) = secx, cosec ( x) = cosec x, x R. Trigonometric functions of some related angles are given below π function\angle θ or (π θ) θ π + θ π θ π + θ π θ π + θ sin sinθ cosθ cosθ sinθ sinθ cosθ cosθ cos cosθ sinθ sinθ cosθ cosθ sinθ sinθ tan tanθ cotθ cotθ tanθ tanθ cotθ cotθ cot cotθ tanθ tanθ cotθ cotθ tanθ tanθ sec sec θ cosec θ cosec θ sec θ sec θ cosec θ cosec θ cosec cosec θ sec θ sec θ cosec θ cosec θ sec θ sec θ Example Evaluate cos40. cos40 = cos(40 60 ) = cos60 = 1. Example 118

17 Visit for CBSE,State Board QP s, Plus 1 & Reference Books ( ) π Evaluate tan ( ) ( ) ( π π tan = tan = tan π π ) = tan π =. Note If θ > 60, then reduce it to an angle in the interval 0 < θ < 60 by subtraction integral multiples of 60 from θ. Example Evaluate cot = lies in the first quadrant. Hence cot750 > 0. cot750 = cot( ) = cot0 =. (Co-terminal and reference angles are the same and each equal to 0 ) Example Evaluate tan040. tan040 = tan( ) = tan( ) = tan40 = tan( ) = tan60 =. Domain and Range of Trigonometric functions Both sinx and cosx are defined for all real numbers x. Also for each real number x, 1 sinx 1 and 1 cosx

18 Visit for CBSE,State Board QP s, Plus 1 & Reference Books y Graph of y = sinx 1 7π π 5π π π π π 1 π π π π 5π π 7π x Hence the domain, of both the functions, is the real line R closed interval [ 1,1]. y Graph of y = cosx and range is the 7π π 5π π π π π 1 1 π π π π 5π π 7π Also sin(nπ + x) = sinx, n Z, cos(nπ + x) = cosx, n Z. sinx = 0 = x = n π, n Z and In other words, sinx = 0, if x = 0, ±π, ±π,... and cosx = 0, if x = ± π, ± π, ±5 π,... cosx = 0 = x = (n + 1) π, n Z Now we define the other 4 trigonometric functions namely tanx, cotx, secx, cosec x. tanx = sinx cosx, x (n + 1) π, n Z. cotx = cosx sinx, x n π, n Z. secx = 1 cosx, x (n + 1) π, n Z. cosec x = 1 sinx, x n π, n Z. The Domains and Ranges of all trigonometric functions are listed in the following table. x 10

19 Visit for CBSE,State Board QP s, Plus 1 & Reference Books Function Domain Range sin R [ 1, 1] cos R [ 1, 1] tan R {(n + 1) π }, n Z R cot R {n π, n Z } R sec R {(n + 1) π }, n Z R ( 1,1) cosec R {n π, n Z } R ( 1,1) Graph of y = tanx, x ( π, π ) is shown in the right side. Since tanx is a periodic function (to be discussed in the next section) with period π, the graph (values of ) of tanx repeats on every interval of length π. From the graph, it is clear that tan0 = 0 and tanx is not defined at the points x = π and x = π. Periodic functions π y = tanx A function f (x) is said to be periodic with period θ, if f (x + θ) = f (x), x and the least positive value of θ is called the fundamental period of the function f (x). All trigonometric functions are periodic functions. We know that sin(nπ + x) = sinx, n Z and cos(nπ + x) = cosx, n Z. In both the functions, the shape of the graph begins repeating after π( see the graph of sinx and cosx). In other words, if we were to shift either graph horizontally by π, the resulting shape would be identical to the original function. Here θ = π is called the fundamental period of the sine and cosine function. Since tan(nπ+x) = tanx, n Z and cot(nπ+x) = cotx, n Z, the fundamental period of tangent and cotangent functions are π. Also we know, sec(nπ+x) = secx, n Z and cosec (nπ+x) = cosec x, n Z. Hence the fundamental period of secant and cosecant functions are π. y π x 11

20 Visit for CBSE,State Board QP s, Plus 1 & Reference Books Trigonometric Identities A trigonometric identity is an equation involving trigonometric functions that is true for all values for which every expression in the equation is defined. The following are trigonometric identities. i) tanx = sinx cosx, x (n + 1) π, where n Z. ii) sin x + cos x = 1, x R. sinx + cosx = 10 is not a trigonometric identity, but it is a trigonometric equation, which has no solution. Next we shall list some more important trigonometric identities(without proof) 1. cos(x + y) = cosxcosy sinxsiny. cos(x y) = cosxcosy + sinxsiny. sin(x + y) = sinxcosy + cosxsiny 4. sin(x y) = sinxcosy cosxsiny 5. If none of the angles x,y and (x + y) is an odd multiple of π, then tanx + tany tan(x + y) = 1 tanxtany. 6. If none of the angles x,y and (x y) is an odd multiple of π, then tanx tany tan(x y) = 1 + tanxtany. 7. If none of the angles x,y and (x + y) is a multiple of π, then cot(x + y) = cotxcoty 1 coty + cotx. 8. If none of the angles x,y and (x y) is a multiple of π, then cot(x y) = cotxcoty + 1 coty cotx. 9. cosx = cos x sin x = cos x 1 = 1 sin x = 1 tan x 1 + tan x. 1

21 Visit for CBSE,State Board QP s, Plus 1 & Reference Books 10. sinx = sinxcosx = tanx 1 + tan x. 11. tanx = tanx 1 tan x. 1. sinx = sinx 4sin x. 1. cosx = 4cos x cosx. 14. tanx = tanx tan x 1 tan x. 15. (i) cosx + cosy = cos x + y cos x y (ii) cosx cosy = sin x + y sin x y (iii) sinx + siny = sin x + y cos x y (iv) sinx siny = cos x + y sin x y 16. (i) cosxcosy = cos(x + y) + cos(x y) (ii) sinxsiny = cos(x + y) cos(x y) (iii) sinxcosy = sin(x + y) + sin(x y) (iv) cosxsiny = sin(x + y) sin(x y) 17. cosx = cos x 1 = 1 sin x = 1 tan x 1 + tan x 18. sinx = sin x cos x = tan x 1 tan x Trigonometric Equations Equations involving trigonometric functions such as sin x, cos x etc are called Trigonometric Equations. Examples 1

22 Visit for CBSE,State Board QP s, Plus 1 & Reference Books 1. sinx + sinx + sin5x = 0.. cosx = 1.. cos x + sinx = sinx =. Solutions of a trigonometric equations A solution of a trigonometric equation is the unknown angle (x ) that satisfies the equation. For example x = π 4 is a solution of the equation tanx = 1. Trigonometric equations may have infinite number of solutions or have no solution. Trigonometric equation in example 4 has no solution. Principal solutions and General solutions Since the trigonometric functions are periodic functions, they have many solutions.the equations are solved initially to get the principle solutions falling in one period interval of the function. There are solutions outside this interval. These other solutions differ by integral multiples of the period of the function. The periods of both sine and cosine functions are π. Hence the principle solutions of equations involving sinx (or cosx) are in the interval [0,π). The period of tangent(or cotangent) function is π. Hence the principal solutions are the solutions in the interval [0,π). The periods of both secant and cosecant functions are π. Hence the principle solutions of equations involving sec x (or cosec x) are in the interval [0, π). The set of all solutions of the trigonometric equation is called its general solution. General solution contains expressions involving integer n. For example, the trigonometric equation sin x = 0 has principal solutions x = 0, π and has general solutions x = n π, n Z. Note: Some authors define the principal solution as the solution in which the absolute value of the angle is the least. For x, y R, sinx = siny = x = nπ + ( 1) n y, n Z 14

23 Visit for CBSE,State Board QP s, Plus 1 & Reference Books For x, y R, cosx = cosy = x = nπ ± y, n Z If x and y are not odd multiples of π, then tanx = tany = x = nπ + y, n Z Example Find all solutions of sinx = cosx, x [0,π). sinx = cosx = sinx cosx = cosx = cosx (sinx 1) = 0 = cosx = 0 or sinx = 1. cosx = 0 = x = π, π ( x [0,π)). sinx = 1 = x = π ( x [0,π)). Hence the solutions are x = π, π. Example Find the principal and general solutions of the equation: tanx =. We have tan π =. and π [0,π). Hence the principal solution is x = π General solutions are x = nπ + π, n Z. Example Find the principal and general solutions of the equation: sinx =. We have sin( π ) = sin(π ) = and 15

24 Visit for CBSE,State Board QP s, Plus 1 & Reference Books sin( 4π ) = sin(π + π ) = sin(π ) = Hence the given equation has two solutions in the interval 0 x < π. Therefore the principal solutions are x = π and x = 4π. For writing the general solution, we can use either of the two solutions, both gives the same result, although they may look different. ( General solution is x = nπ + ( 1) n π ), n Z. Example Solve the equation: cos x + sinx = 0. cos x + sinx = (1 sin x) + sinx = sin x + sinx + cos x + sinx = 0 = sin x sinx = 0 = (sinx + 1)(sinx ) = 0 = sinx = 1 or sinx =, which is not possible. sinx = 1 = sin π π = sin 6 6. = x = π 6. general solution is x = nπ + ( 1) n ( π 6 ), n Z. EXERCISE.1 1. Find the radian measures corresponding to the following degree measures: (i) 5 (ii) 47 0 (iii) 40 (iv)

25 Visit for CBSE,State Board QP s, Plus 1 & Reference Books 1 = π 180 radian (i) 5 = 5 π 180 radian = 5 π 6 radian. (ii) 47 0 = 47.5 = 95. or 47 0 = 95 ( π 180 radian ) = 19 π 7 radian (iii) 40 = 40 ( π 180 radian ) = 4 π radian. (iv) 50 = 50 ( π 180 radian ) = 6 π 9 radian.. Find the degree measures corresponding to the following radian measures (Use π = 7 ). (i) (ii) 4 1 radian = 180 π (i) radian (iii) 5 π = ( )180 π = π = 495 4() 7 (ii) 4 radian = ( 4) 180 π (iii) 5 π (iv) 7 π 6 (iv) 7 π 6 = 9.75 = 9.5 = 9 0 = ( 4) π radian = (5 )180 π = 00. radian = (7 π 6 )180 π = 10. = = = A wheel makes 60 revolutions in one minute. Through how many radians does it turn in one second? 17

26 Visit for CBSE,State Board QP s, Plus 1 & Reference Books Since the wheel makes 60 revolutions in one minute, in one second it makes = 6 revolutions. But 1 revolution = π radians. Hence the angle turned in 1 second = 6( π) = 1 π radians. 4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length cm (Use π = 7 ). Let θ radian be the angle subtended at the centre and radius r = 100 cm and arc length = l = cm. Then l = r θ = θ = l r = = 0. radian = 0.( π ) = 0.(180 7 = θ = 1.6 = In a circle of diameter 40 cm, the length of a chord is 0 cm. Find the length of minor arc of the chord. In the figure length of arc AB = l and radius = r = OA = OB = 0 cm and length of chord = AB = 0 cm. Since OA = OB = AB = 0, θ = π radian. l = r θ = 0( π ) = 0 π cm ) O O π A B l 6. If in two circles, arcs of the same length subtend angles 60 and 75 at the centre. Find the ratio of their radii. 18

27 Visit for CBSE,State Board QP s, Plus 1 & Reference Books Let r 1 and r be the radii of the two circles. Given that θ 1 = 60 = 60 ( π 180 ) = π radian and θ = 75 = 75 ( π 180 ) = 15 π 6 radian = 5 π 1 radian. Let l be the length of each arc. Then l = r 1 θ 1 = r θ = π r 1 = 5 π 1 r = r 1 = 5 4 r = r 1 r = 5 4 = r 1 : r = 5 : Find the angle in radian through 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) 1 cm l = r θ = θ = l r (i) Here r = 75 and l = 10 cm. Hence θ = l r = = 15 radian. (ii) Here r = 75 and l = 15 cm. Hence θ = l r = = 1 5 radian. (iii) Here r = 75 and l = 1 cm. Hence θ = l r = 1 75 = 7 5 radian. EXERCISE. Find the values of other five trigonometric functions in Exercises 1 to cosx = 1, x lies in third quadrant. secx = 1 cosx =, sin x = 1 cos x = = = sinx = ± 4 But sinx < 0 in the third quadrant. sinx =

28 Visit for CBSE,State Board QP s, Plus 1 & Reference Books tanx = sinx cosx = 1 =. cotx = 1 tanx = 1, cosec x = 1 sinx =.. sinx =, x lies in second quadrant. 5 cos x = 1 sin x = = 16 5 = cosx = ±4 5. But cosx < 0 in the second quadrant. cosx = 4 5. tanx = sinx cosx = 5 = cotx = 1 tanx = 4, secx = 1 cosx = 5 4 cosec x = 1 sinx = 5.. cotx =, x lies in third quadrant. 4 tanx = 1 cotx = 4, sec x = 1 + tan x = = 5 9 = secx = ± 5. But secx < 0 in the third quadrant. secx = 5, cosx = 1 secx = 5. sinx = tanx.cosx = 4 ( 5 ) = 4 5. cosec x = 1 sinx = secx = 1, x lies in fourth quadrant

29 Visit for CBSE,State Board QP s, Plus 1 & Reference Books cosx = 1 secx = 5 1. sin x = 1 cos x = = = sinx = ±1 1. But sinx < 0 in the fourth quadrant. sinx = 1 1. tanx = sinx cosx = = 1 5 cotx = 1 tanx = 5 1, cosec x = 1 sinx = tanx = 5, x lies in second quadrant. 1 cotx = 1 tanx = 1 5, cosec x = 1 + cot x = = = cosec x = ± 1. But cosec x > 0 in the second quadrant. 5 cosec x = 1 5, sinx = 1 cosec x = 5 1. cosx = cotx.sinx = 1 5.( 5 1 ) = 1 1. secx = 1 cosx = 1 1. Find the values of the trigonometric functions in Exercises 6 to sin lies in the first quadrant. Hence sin765 > 0. sin765 = sin((60 ) + 45 ) = sin45 ( sinx repeats after an interval of 60 or we can say that coterminal and reference angles are the same, each equal to 45 ) 11

30 Visit for CBSE,State Board QP s, Plus 1 & Reference Books = sin765 = cosec ( 1410 ) cosec ( 1410 ) > 0 ( ( 1410) lies in the first quadrant). cosec ( 1410 ) = cosec ( (4 60) ) = cosec (0 ) =. (coterminal and reference angles are the same,each equal to 0 ) 8. tan 19 π 19 π lies in the first quadrant. Hence tan 19 π > 0. tan 19 π = tan(6 π + π ) = tan(π ) =. (coterminal and reference angles are the same,each equal to 60 ) 9. sin( 11 π ) 11 π lies in the first quadrant. Hence sin( 11 π ) > 0. sin( 11 π 11 π ) = sin(4π ) = sin(π ) = (coterminal and reference angles are the same,each equal to 60 ) 10. cot( 15 π 4 ) cot( 15 π 4 π ) > 0( ( 15 ) lies in the first quadrant). 4 cot( 15 π 15 π ) = cot(4π 4 4 ) = cot(π 4 ) =

31 Visit for CBSE,State Board QP s, Plus 1 & Reference Books Prove that: (coterminal and reference angles are the same, each equal to 45 ) EXERCISE. 1. sin π 6 + cos π tan π 4 = 1 sin π 6 = 1, cos π = 1, tan π 4 = 1 L.H.S = sin π 6 + π cos π tan 4 = = 1 1 = 1 = R.H.S. sin π 6 + cosec 7π π 6 cos = sin π 6 = 1, cos π = 1, cosec 7π 6 = cosec (π + π 6 ) = cosec π 6 = L.H.S = sin π 6 + cosec 7π π 6 cos = (1 4 ) + 1 ( ) 4 = = =R.H.S. cot π 6 + cosec 5π 6 + π tan 6 = 6 cot π 6 =, cosec 5π 6 = cosec (π π 6 ) = cosec π 6 =, tan π 6 = 1 L.H.S = cot π 6 + cosec 5π 6 + π tan 6 = + + (1 ) = = 6 = R.H.S 4. sin π 4 + cos π 4 + sec π = 10 sin π 4 = sin(π π 4 ) = sin π 4 = 1, cos π 4 = 1, sec π = L.H.S = sin π 4 + π cos 4 + π sec = (1 ) + (1 ) + (4) = 10 =R.H.S 1

32 Visit for CBSE,State Board QP s, Plus 1 & Reference Books 5. Find the value of (i) sin75 (ii) tan15 sin(x + y) = sinxcosy + cosxsiny, tan(x y) = tanx tany 1 + tanxtany (i) sin75 = sin( ) = sin45 cos0 + cos45 sin0 = 1 ( ) + 1 ( ) = (ii) tan15 = tan(45 0 ) = tan45 tan0 1 + tan45 tan0 = (1)( 1 ) = = ( 1)( 1) ( + 1)( 1) = + 1 =. 1 Prove the following: 6. cos( π 4 x)cos(π 4 y) sin(π 4 x)sin(π y) = sin(x + y) cos(x + y) = cosxcosy sinxsiny, {}}{ π L.H.S = cos( 4 x+ π 4 y ) = cos( π }{{} tan( π 4 + x) tan( π 4 x) = ( 1 + tanx 1 tanx ) cos( π θ) = sinθ (x + y)) = sin(x + y) = R.H.S tanx + tany tanx tany tan(x + y) =, tan(x y) = 1 tanxtany 1 + tanxtany, tan π 4 = 1 L.H.S = tan π 4 + tanx 1 tan π 4 tanx tan π 4 tanx 1 + tan π 4 tanx = cos(π + x)cos( x) sin(π x)cos( π + x) = cot x 1 + tanx 1 tanx 1 tanx 1 + tanx = ( ) 1 + tanx = R.H.S 1 tanx

33 Visit for CBSE,State Board QP s, Plus 1 & Reference Books L.H.S = ( cosx)cosx sinx( sinx) = cos x sin x = cot x = R.H.S 9. cos( π + x)cos(π + x) [ cot( π x) + cot(π + x) ] = 1 [ sinx L.H.S = (sinx)cosx[tanx + cotx] = sinxcosx cosx + cosx ] sinx [ sin x + cos ] x = sinxcosx = sin x + cos x = 1 = R.H.S sinxcosx 10. sin(n + 1)x sin(n + )x + cos(n + 1)x cos(n + )x = cosx cos(x y) = cosxcosy + sinxsiny L.H.S = sin(n + 1)x sin(n + )x + cos(n + 1)x cos(n + )x = cos(n + 1)x cos(n + )x + sin(n + 1)x sin(n + )x = cos[(n + 1)x (n + )x] = cos( x) = cosx = R.H.S 11. cos( π 4 + x) cos(π 4 x) = sinx cosx cosy = sin x + y sin x y L.H.S = sin ( π 4 + x) + ( π 4 x) sin ( π 4 + x) ( π 4 x) = sin π 4 sinx = sin(π π 4 )sinx = sin π 4 sinx = ( )( 1 )sinx = sinx = R.H.S 15

34 Visit for CBSE,State Board QP s, Plus 1 & Reference Books 1. sin 6x sin 4x = sinx sin10x sinx + siny = sin x + y cos x y x + y, sinx siny = cos sin x y, sinx = sinx cosx L.H.S = sin 6x sin 4x = (sin6x + sin4x)(sin6x sin4x) = (sin 6x + 4x cos 6x 4x 6x + 4x )(cos sin 6x 4x ) = sin5x cosx(cos5xsinx) = (sinxcosx) (sin5xcos5x) = sinx(sin10x) = R.H.S 1. cos x cos 6x = sin4x sin8x cosx cosy = sin x + y sin x y x + y, cosx + cosy = cos cos x y L.H.S = = cos cos x cos 6x = (cosx + cos6x)(cosx cos6x) x + 6x x 6x x + 6x x 6x cos ( sin sin ) = cos4x cos( x)[ sin4x sin( x)] = cos4x cos(x)[sin4x sin(x)] = (sinx cosx)(sin4xcos4x) = sin4x sin8x = R.H.S 14. sinx + sin4x + sin6x = 4cos x sin4x sinx + siny = sin x + y cos x y L.H.S = sinx + sin4x + sin6x = (sinx + sin4x) + (sin4x + sin6x) 16

35 Visit for CBSE,State Board QP s, Plus 1 & Reference Books = sin x + 4x cos x 4x + sin 4x + 6x cos 4x 6x = sinx cos( x) + sin5x cos( x) = sinx cos(x) + sin5x cos(x) = cosx[sinx + sin5x] = cosx[sin x + 5x cos = 4cos x sin4x = R.H.S x 5x ] = cosx [sin4xcos( x)] 15. cot4x(sin5x + sinx) = cotx (sin5x sinx) sinx + siny = sin x + y cos x y x + y, sinx siny = cos sin x y 5x + x 5x x L.H.S = cot4x(sin5x + sinx) = cot4x(sin cos ) = cot4x(sin4x cosx) = cos4x (sin4x cosx) = cosx cos4x (1) sin4x R.H.S = = cosx sinx 5x + x cotx (sin5x sinx) = cotx (cos sin (cos4xsinx) = cosx cos4x () From (1) and (), we get cos9x cos5x sin17x sinx = sinx cos 10x L.H.S = R.H.S 5x x ) cosx cosy = sin x + y sin x y x + y, sinx siny = cos sin x y 9x + 5x 9x 5x sin sin L.H.S = 17x + x 17x x cos sin sin7x sinx = cos10x sin7x = sinx cos10x = R.H.S sin5x + sinx cos5x + cosx = tan4x 17

36 Visit for CBSE,State Board QP s, Plus 1 & Reference Books sinx + siny = sin x + y cos x y 5x + x 5x x sin cos L.H.S = 5x + x 5x x cos cos sinx siny cosx + cosy = tan x y x + y, cosx + cosy = cos cos x y = sin4x = tan4x = R.H.S cos4x sinx siny = cos x + y sin x y x + y, cosx + cosy = cos cos x y cos x + y L.H.S = sin x y sin x y cos x + y cos x y = cos x y = tan x y = R.H.S sinx + sinx cosx + cosx = tanx sinx + siny = sin x + y cos x y x + y, cosx + cosy = cos cos x y sin x + x cos x x L.H.S = (cos x + x cos x x = sinx = tanx = R.H.S cosx ) 0. sinx sinx sin x cos x = sinx sinx + siny = sin x + y cos x y x + y, sinx siny = cos sin x y 18

37 Visit for CBSE,State Board QP s, Plus 1 & Reference Books 1. L.H.S = cos x + x sin x x cosx sin( x) sin x cos = x (cos x sin x) = cosxsinx cosx = sinx = R.H.S cos4x + cosx + cosx sin4x + sinx + sinx = cotx sinx + siny = sin x + y cos x y x + y, cosx + cosy = cos cos x y 4x + x 4x x L.H.S = = cos ) + cosx 4x + x 4x x (sin cos ) + sinx cosx( cosx + 1) = sinx (cosx + 1) = cosx = cotx = R.H.S sinx (cos4x + cosx) + cosx (cos (sin4x + sinx) + sinx = (cosx cosx) + cosx (sinx cosx) + sinx. cotx cotx cotx cotx cotx cotx = 1 cotx coty 1 cot(x + y) = coty + cotx cotx cot(x) 1 cotx = cot(x + x) = cot(x) + cotx = cotx (cotx + cotx) = cotx cotx 1 = cotx cotx cotx cotx cotx cotx = 1. tan4x = 4tanx (1 tan x) 1 6tan x + tan 4 x tanx = tanx 1 tan x tanx tan4x = tan(x) = tanx 1 tan x = 1 tan x ( ) tanx = 4tanx(1 tan x) (1 tan x) 4tan x 1 1 tan x 19

38 Visit for CBSE,State Board QP s, Plus 1 & Reference Books = 4tanx(1 tan x) (1 + tan 4 x tan x) 4tan x = 4tanx (1 tan x) 1 6tan x + tan 4 x. 4. cos4x = 1 8sin xcos x cosx = 1 sin x cos4x = cos(x) = 1 sin x = 1 (sinx) = 1 (sinxcosx) = 1 (4)sin x cos x = 1 8sin xcos x 5. cos6x = cos 6 x 48 cos 4 x + 18 cos x 1 cosx = 4cos x cosx, cosx = cos x 1 cos6x = cos(x) = cos x 1 = (cosx) 1 = (4cos x cosx) 1 = (16 cos 6 x + 9cos x 4cos 4 x) 1 = cos 6 x 48 cos 4 x + 18 cos x 1 EXERCISE.4 Find the principal and general solutions of the following equations: 1. tanx = We have tan π = and tan(π + π ) = tan 4π =. Hence the principal solutions are x = π and x = 4π. General solutions are x = nπ + π, n Z. Note: Solved in accordance with the theory given in the NCERT book

39 Visit for CBSE,State Board QP s, Plus 1 & Reference Books. secx = We have sec π = and sec(π π ) = sec 5π =. Hence the principal solutions are x = π and x = 5π. General solutions are x = nπ ± π, n Z.. cotx = We have cot(π π 6 ) = cot(5π 6 ) = and cot(π π 11π ) = cot 6 6 =. Hence the principal solutions are x = 5π 6 General solutions are x = nπ + 5π 6, n Z. 11π and x = 6. Note: Solved in accordance with the theory given in the NCERT book. 4. cosec x = We have cosec (π + π 6 ) = cosec (7π ) = and 6 cosec (π π 6 ) = cosec 11π 6 =. Hence the principal solutions are x = 7π 6 and x = 11π 6. General solutions are x = nπ + ( 1) n 7π 6, n Z. Find the general solution for each of the following equations: 5. cos4x = cosx 141

40 Visit for CBSE,State Board QP s, Plus 1 & Reference Books cos4x = cosx = 4x = nπ ± x = x = nπ ± x = x = nπ x or x = nπ + x = x = nπ or x = nπ, n Z. 6. cosx + cosx cosx = 0 cosx + cosx cosx = 0 = (cosx + cosx) cosx = 0 = cos x + x cos x x = cosx cosx cosx = 0 = cosx (cosx 1) = 0 cosx = 0 = cosx = 0 or cosx = 1 = cos π. = x = (n + 1) π or x = nπ ± π. = x = (n + 1) π 4 or x = nπ ± π, n Z. 7. sinx + cosx = 0 sinx + cosx = 0 = sinx cosx + cosx = 0 = cosx (sinx + 1) = 0 = cosx = 0 or sinx = 1 = sin(π + π 6 ) = sin 7π 6 = x = (n + 1) π or x = nπ + 7π ( 1)n 6, n Z. 8. sec x = 1 tanx 14

41 Visit for CBSE,State Board QP s, Plus 1 & Reference Books sec x = 1 tanx = 1 + tan x = 1 tanx = tanx (1 + tanx) = 0 = tanx = 0 or tanx = 1 = tan(π π 4 ) = tan(π 4 ) = x = nπ or x = nπ + π 4, n Z = x = nπ or x = nπ + π 8, n Z. 9. sinx + sinx + sin5x = 0 (sinx + sin5x) + sinx = 0 = sin x + 5x cos x 5x = sinx cosx + sinx = 0 ( cos( x) = cosx) = sinx (cosx + 1) = 0 = sinx = 0 or cosx = 1 = cos(π π ) = cos π = x = nπ or x = nπ ± π = x = nπ or x = nπ ± π, n Z. Prove that : Miscellaneous Exercise on Chapter + sinx = 0 1. cos π 9π π cos + cos cos 5 π 1 = 0 cosx cosy = cos(x + y) + cos(x y), cos π 9π cos 1 1 = cos( π 1 + 9π π + cos 1 + cos 5 π 1 1 ) + cos( π 1 9π 1 π ) + cos 1 + cos 5 π 1 cosx + cosy = cos x + y cos x y 14

42 Visit for CBSE,State Board QP s, Plus 1 & Reference Books = cos 10 π 1 + cos 8 π π + cos cos 5 π 1 = (cos 10 π π + cos 1 1 ) + (cos 8 π 1 + cos 5 π 1 ) 10 π = cos 1 + π 10 π 1 cos 1 π 8 π 1 + cos π 8 π 1 cos 1 5 π 1 = cos π cos 7π 6 + cos π cos π 6 = 0 ( cos π = 0).. (sinx + sinx)sinx + (cosx cosx)cosx = 0 sinx + siny = sin x + y cos x y L.H.S = (sin x + x cos x x ) sinx + ( sin x + x x + y, cosx cosy = sin sin x y = sinx cosx sinx sinx sinx cosx = 0 = R.H.S. (cosx + cosy) + (sinx siny) = 4 cos x + y cosx = cos x 1 = 1 + cosx = cos x sin x x ) cosx L.H.S = cos x + cos y + cosx cosy + sin x + sin y sinx siny = (cos x + sin x) + (cos y + sin y) + ( cosx cosy sinx siny) = cos(x + y) = (1 + cos(x + y)) = ( cos x + y ) = 4 cos x + y = R.H.S 4. (cosx cosy) + (sinx siny) = 4 sin x y 144

43 Visit for CBSE,State Board QP s, Plus 1 & Reference Books cosx = 1 sin x = 1 cosx = sin x L.H.S = cos x + cos y cosx cosy + sin x + sin y sinx siny = (cos x + sin x) + (cos y + sin y) ( cosx cosy + sinx siny) = cos(x y) = (1 cos(x y)) = ( sin x y ) = 4 sin x y = R.H.S 5. sinx + sinx + sin5x + sin7x = 4cosx cosx sin4x 6. L.H.S = = (sin x + 7x (sinx + sin7x) + (sinx + sin5x) cos x 7x x + 5x x 5x ) + (sin cos ) = sin4x cosx + sin4x cosx = sin4x (cosx + cosx) = sin4x (cos x + x cos x x ) = sin4x(cosx cosx) = 4cosx cosx sin4x = R.H.S (sin7x + sin5x) + (sin9x + sinx) (cos7x + cos5x) + (cos9x + cosx) = tan6x sinx + siny = sin x + y cos x y x + y, cosx + cosy = cos cos x y 7x + 5x 7x 5x 9x + x 9x x (sin cos ) + (sin cos ) L.H.S = 7x + 5x 7x 5x 9x + x 9x x (cos cos ) + (cos cos ) sin6x cosx + sin6x cosx = cos6x cosx + cos6x cosx sin6x ( cosx + cosx) = cos6x ( cosx + cosx) = sin6x cos6x = tan6x. 7. sinx + sinx sinx = 4 sinx cos x cos x 145

44 Visit for CBSE,State Board QP s, Plus 1 & Reference Books sinx siny = cos x + y sin x y x + y, cosx + cosy = cos cos x y L.H.S = (sinx sinx) + sinx = cos x + x sin x x = cosx sinx + sinx cosx = sinx (cosx + cosx) + sinx cosx = sinx (cos x + x cos x x ) = sinx (cos x cos x ) = 4 sinx cos x Find sin x, cos x and tan x 8. tanx = 4, x in quadrant II tanθ = tanθ 1 tan θ cos x = R.H.S in each of the following: x in quadrant II = π < x < π = π 4 < x < π are posi- = x lies in the first quadrant. Hence sin x, cos x and tan x tive. 4 = tanx = tan( x tan x ) = 1 tan x = = tan x 1 tan x = + tan x = tan x = tan x tan x = 0 = (tan x + 1)(tan x ) = 0 = tan x =, 1 But tan x > 0 = tan x =. 1 + tan x = sec x = = 5 = cos x = 1 5 = 5 5 = cos x =

45 Visit for CBSE,State Board QP s, Plus 1 & Reference Books sin x = tan x cos x 5 = ( 5 ) = cosx = 1, x in quadrant III. cosx = cos x 1 = 1 sin x 1 cosx = = 4, 1 + cosx = 1 1 = x in III quadrant = π < x < π = π < x < π 4. sin x > 0, cos x < 0 and tan x < 0. sin x = 1 cosx cos x = 1 + cosx tan x = sin x cos x = = = 6 9 = sin x 6 = = 1 = 9 = cos x = 6 =. 10. sinx = 1, x in quadrant II 4 x in quadrant II = π < x < π = π 4 < x < π are posi- = x lies in the first quadrant. Hence sin x, cos x and tan x tive. x 1 4 = sinx = sin( x ) = sin x cos x tan = 1 + tan x = 1 + tan x = 8 tan x (1) and 1 = 8sin x cos x () From (1), we get tan x 8 tan x + 1 =

46 Visit for CBSE,State Board QP s, Plus 1 & Reference Books = tan x = = Dividing () by cos x, we get = sec x = 8tan x = sec x = 8(4 + 15) = = = cos x = 1 8(4 + 15) = (4 15) 8(4 + 15)(4 15) = (4 15) 8(16 15) = = 8 15 = cos x 16 = sin x = tan x cos x = (4 + 15) = ( ( ) (8 15) ) = (4 + 15) (8 15) Now ( ) (8 15) = ( ) (8 15) = = sin x = KERALA : Questions from Previous years Question Papers a) The degree measure of 7π radians is... 6 i) 10 ii) 10 iii) 01 iv) 10 cos7x + cos5x b) Prove that sin7x sin5x = cotx. c) A lamp post is situated at the middle point M of the side AC of a triangular plot ABC with BC=7m, CA=8m, AB = 9 m. Lamp post subtends an angle 15 at the point B. Determine the height of the lamp post. 4 a) 7 π 6 radian = (7 π 6 )180 π =

47 Visit for CBSE,State Board QP s, Plus 1 & Reference Books cos7x + cos5x b) sin7x sin5x = cosx sinx = cotx. c) 7x+5x cos cos 7x 5x = cos 7x+5x sin 7x 5x A P c = 9 h M B 15 a = 7 b = 8 In the figure BC = a = 7 m, CA = b = 8 m, AB = c = 9 m. M is the midpoint of AC. Therefore AM = MC = 4 m. PM is the lamp post and its height is h m. PM BM. Hence BMP is a right angled triangle. The lamp subtends an angle 15 at B. Now apply cosine formula for ABC, we get cosc = a + b c = = (1) ab Applying cosine formula in BMC, we get cosc = BC +CM BM BC CM Using (1), we get = BM = BC +CM BC CM cosc BM = = 49 = BM = 7 m. 7 C From the right angled BMP, tan15 = h BM = h 7 = tan15 = tan(45 0 ) = tan45 tan0 1 + tan45.tan0 = = () 149

48 Visit for CBSE,State Board QP s, Plus 1 & Reference Books 015 = ( 1) = 4 = h = 7( ) m. = = h, (using ()) 7 1. a) Which of the following values of sinx is incorrect? (i) 0 (ii) 1 (iii) (iv) 1 b) Prove that cos( π 4 + x) + cos(π 4 x) = cosx c) A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle 0 with it. The distance between the foot of the tree to the point where top touches the ground is 8 m. Find the height of the tree. a) Answer: (iii) ( 1 sinx 1). cos(x + y) + cos(x y) = cosxcosy b) cos( π 4 + x) + cos(π 4 x) = cos π 4 cosx = ( 1 ) cosx = cosx c) The broken tree is shown in red color. Length of tree = AB + BC (1) tan0 = 1 = AB 8 = AB = 8 We have, AB + AC = BC = = BC 0 B C 8m A = BC = = 56 = BC = 16 = 16. Hence from (1), Height of the tree = = 4 = 8 m

49 Visit for CBSE,State Board QP s, Plus 1 & Reference Books. a) sin5 = i) 1 ii) iii) 1 iv) 1 b) Find the principal and general solution of sinx = ( ) A B c) Prove that tan = a b a + b cot C. a) sin5 = sin( ) = sin45 = 1.. b) For any real numbers x and y, sinx = siny = x = nπ + ( 1) n y, n Z. We have, sin π =. sinx = = sinx = sin π = sin(π + π ) = sin 4π. Also sin( π ) =. sinx = sin 4π = x = nπ + ( 1)n 4π, n Z (1) (1) is the general solution of the given equation. Principal solution is x = 4π π and x =. c) In ABC the sides opposite to the angles A, B, C are denoted by the letters a, b, c respectively. Thus BC = a, AC = b, AB = c. B Sine formula c a A b C a In any ABC, sina = circum circle of ABC. Napier s formula b sinb = c sinc = R, where R is the radius of the 151

50 Visit for CBSE,State Board QP s, Plus 1 & Reference Books ( ) A B In any ABC, tan = a b a + b cot C. From sine formula, we have a = RsinA and b = RsinB. Now = a b a + b cot C cos A + B sin A + B sin A B cos A B = RsinA RsinB RsinA + RsinB cot C cot C = cot A + B = cot( π C ) tan A B cot C = tan C tan A B cot C = tan A B ( ) A B = tan = a b a + b cot C. = sina sinb sina + sinb cot C tan A B cot C ( A + B +C = π = A + B. a) Which of the following is equal to 50? i) 6π ii) 9π iii) 6π iv) 9π 9 6 b) Solve: sinx sin4x + sin6x = 0 c) In any triangle ABC, prove that tan B C = b c b + c cot A. a) 50 = (180 ) 0 = π π 9 = 6π 9. = π = C ) b) The given equation can be written as sin 6x + sin x sin 4x = 0 = sin4xcosx sin4x = 0 = sin4x(cosx 1) = 0 = sin4x = 0 or cosx = 1 = sin4x = 0 or cosx = cos π = 4x = nπ or x = nπ ± π, n Z 15

51 Visit for CBSE,State Board QP s, Plus 1 & Reference Books = x = nπ 4 or x = nπ ± π 6, n Z. c) From sine formula, we have b = R sin B and c = R sinc. 014 Now = b c b + c cot A cos B +C sin B +C sin B C cos B C = cot( π A B C ) tan = tan A B C tan cot A = RsinB RsinC RsinB + RsinC cot A cot A cot A = tan B C ( ) B C = tan = b c b + c cot A. = cot B +C = sinb sinc sinb + sinc cot A tan B C cot A ( A + B +C = π = B +C 1. a) The value of sin(π x) = b) Find the principal and general solution of the equation sinx =. a) sin(π x) = sinx. b) sinx = = sin π = sin(π π ) = sin π. Hence the principle solutions are π, π. = π A ) For writing the general solution, we prefer the principal solution which has the least absolute value. Hence we take, sinx = sin π = the general solution is x = n π + ( 1) n π, n Z.. Prove that cos4x + cosx + cosx sin4x + sinx + sinx = cotx 15

52 Visit for CBSE,State Board QP s, Plus 1 & Reference Books cosx + cosy = cos x + y cos x y x + y, sinx + siny = sin cos x y cos4x + cosx + cosx (cos4x + cosx) + cosx = sin4x + sinx + sinx (sin4x + sinx) + sinx cosx cosx + cosx cosx(cosx + 1) = = sinx cosx + sinx sinx (cosx + 1) = cosx sinx = cotx.. a) π radian = degree b) cos(π x) = c) Find the general solution of sin x sin 4x + sin 6x = 0. 1 radian = 180 π a) π radian = 180 π ( π ) = 10. b) cos(π x) = cos( x) = cosx. sinx + siny = sin x + y cos x y c) sinx sin4x + sin6x = 0 = (sinx + sin6x) sin4x = 0 = (sin x+6x cos x 6x ) sin4x = 0 = sin4x cos( x) sin4x = 0 = sin4x cos(x) sin4x = 0 = sin4x (cosx 1) = 0 sin4x = 0 or cosx = 1. sin4x = 0 or cosx = cos π. = 4x = nπ or x = nπ ± π, n Z

53 Visit for CBSE,State Board QP s, Plus 1 & Reference Books 01 i.e. x = nπ 4 or x = nπ ± π 6, n Z. 1. a) Prove that sin5x + sinx cos5x + cosx = tan4x. b) Prove that sin π 6 sec π 4sin 5π 6 cot π 4 = 1. a) Refer problem 17 page 17 b) sin π 6 sec π 4sin 5π 6 cot π 4 = ( 1 )() 4sin(π π 6 )(1) = 4(sin π 6 ) = 4(1 ) = = 1.. Prove that tan( π 4 + x) tan( π 4 x) = Refer problem 7 page 14 ( 1 + tanx 1 tanx. a) Show that tan15 =. b) tan15 + cot15 = 4. cotx coty + 1 cot(x y) = coty cotx ) a) Refer problem 5(ii) page 14 b) cot15 = cot(45 0 ) = cot45 cot0 + 1 cot0 cot45 = (1) ( + 1) = = 1 1 ( 1)( + 1) = ( + 1) 1 = + = tan15 + cot15 = + + =

54 Visit for CBSE,State Board QP s, Plus 1 & Reference Books Consider the trigonometric equation tanx = a) Write the general solution b) Write the principal solutions. Refer problem 1 page a) Evaluate tan 1π 6 b) If tanx = 1 and x is in the third quadrant, find sinx and cosx. a) tan 1π 6 = tan(π + π 6 ) = tan π 6 = 1. b) sec x = 1 + tan x = = = secx = ± 5 But secx < 0 in the third quadrant. Hence secx =. cosx = 1 secx = 5. sinx = tanx. cosx = 1 ( 5 ) = Prove that cosx + cos7x cosx sin7x sinx sinx = cotx. cosx + cos7x cosx (cosx + cos7x) cosx = sin7x sinx sinx (sin7x sinx) sinx (cos5x cosx) cosx cosx (cos5x 1) = = (cos5x sinx) sinx sinx (cos5x 1) = cotx.. Show that tanx tanx tanx = tanx tanx tanx. tanx = tan(x + x) = tanx + tanx 1 tanx tanx 156

55 Visit for CBSE,State Board QP s, Plus 1 & Reference Books = tanx tanx tanx tanx = tanx + tanx = tanx tanx tanx = tanx tanx tanx. 4. Solve sinx sin4x + sin6x = Refer problem (c) page 154 ( ) 1π 1. i) Find the value of sin. ii) Find the principal and general solutions of the equation cosx =. ( ) 1π ( i) sin = sin 10π + π ) = sin π =. ii) cosx = = cos π 6 = cos(π π 6 ) = cos 5 π 6. cosx = = cos π 6 = cos(π + π 6 ) = cos 7 π 6. Hence the principle solutions are 5π 6, 7 π 6. For writing the general solution, we prefer the principal solution which has the least absolute value. Hence we take, cosx = cos 5π 6 = the general solution is x = n π ± 5π 6, n Z. ( ) x y. Show that (cosx + cosy) + (sinx + siny) = 4 cos

56 Visit for CBSE,State Board QP s, Plus 1 & Reference Books cosx + cosy = cos x + y cos x y x + y, sinx + siny = sin cos x y (cosx+cosy) +(sinx+siny) = (cos x + y cos x y ) +(sin x + y cos x y ( ) x y = 4cos [cos x + y + x + y ( ) x y sin ] = 4 cos.. If x is in the third quadrant, then a) Choose the possible value of cosec x from the bracket. [ 5, 5, 5, 5 ]. b) Evaluate tan x sec x for the x in part (a). ) a) In the third quadrant, cosec x < 0 and its value increases from to 1. i.e, cosec x (, 1). cosec x = 5. b) sinx = 5 and cosx = 1 sin x = But cosx < 0 in the third quadrant. l tanx = sinx cosx = 4, secx = 1 cosx = 5 4. tanx secx = = 8 4 = = cosx = Find the general solution of sin6x sin4x + sinx = = ± Refer problem (c) page i) Find the degree measure corresponding to radians (use π = 7 )