Trigonometric ratios of multiple and submultiple angles pdf

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Trigonometric ratios of multiple and submultiple angles pdf

Hi students, welcome to Amans Maths Blogs (AMB). On this post, you will get the Multiple And Submultiple Angles Question and Answer Set 1 is the collection of some important questions. Practice these questions for SSC CGL CAT exams etc. It will help you to practice the questions on the topics of multiple and submultiple angles trigonometric

functions based questions with answers.Multiple And Submultiple Angles Question and Answer Set 1: Ques No 1The value of sin (67.50) isOptions:A. 1/22B. 1/3C. (2 ? 2) / 2D. (2 + 2) / 2Answer: CMultiple And Submultiple Angles Question and Answer Set 1: Ques No 2The value of sin120sin480sin 540 isOptions:A. 1/22B. 1/4C. 1/23D.

1/8Answer: DMultiple And Submultiple Angles Question and Answer Set 1: Ques No 3The value of (sin224? ? sin226?)(sin242? ? sin212?) isOptions:A. 1/16B. 1/4C. 1/12D. 1/8Answer: AMultiple And Submultiple Angles Question and Answer Set 1: Ques No 4The value of tan (7.5)0 isOptions:A. 6 ? 4 ? 3 + 2B. 6 ? 4 + 3 ? 2C. 6 + 4 ? 3 ? 2D.

6 + 4 + 3 + 2Answer: AMultiple And Submultiple Angles Question and Answer Set 1: Ques No 5The value of 1 + tan(x)tan(x/2) isOptions:A. sinxB. cosxC. cosecxD. secxAnswer: DMultiple And Submultiple Angles Questions and Answer Set 1: Ques No 5The value of 1 + tan(x)tan(x/2) isOptions:A. sinxB. cosxC. cosecxD. secxAnswer: DMultiple And

Submultiple Angles Questions and Answer Set 1: Ques No 6If 4 sin3(x)cos(3x) + 4cos3(x) sin(3x) = msin(kx), then the value of (m + k) is Options:A. 0B. 5C. 7D. -3Answer: CMultiple And Submultiple Angles Questions and Answer Set 1: Ques No 7The value of cos2(Pi/8) + cos2(3Pi/8) + cos2(5Pi/8) + cos2(7Pi/8) is Options:A. 2B. 1C. 0D. 3Answer:

AMultiple And Submultiple Angles Questions and Answer Set 1: Ques No 8The value of cos200cos400cos800 isOptions:A. 1/16B. 1/4C. 1/12D. 1/8Answer: DMultiple And Submultiple Angles Questions and Answer Set 1: Ques No 9If cos(x) = 1/3 and 3Pi/2 < x < 2Pi in quadrant IV, then the value of sin(4x) is Options:A. 8 2 / 3 B. -8 2 / 3 C. ? 56 2 /

243D. 56 2 / 81Answer: DMultiple And Submultiple Angles Questions and Answer Set 1: Ques No 10If (a + b + c)(tanA/2 + tanB/2) isOptions:A. 2c cot C/2B. c cot C/2C. 3c cot C/2D. 4c cot C/2Answer: AMultiple And Submultiple Angles Questions and Answer Set 1: Ques No 11In a triangle ABC, if A = 60 degree, then the value of (1 + a/c + b/c)(1 +

c/b ? a/b)Options:A. 1B. 2C. 3D. 4Answer: CMultiple And Submultiple Angles Questions and Answer Set 1: Ques No 12The value of cotA/2 + cotB/2 + cotC/2 isOptions:A. /s2B. s3/C. s/D. s2/Answer: DMultiple And Submultiple Angles Questions and Answer Set 1: Ques No 13The value of (a2/sinA + b2/sinB + c2/sinC)/(sinAsinBsinC) isOptions:A.

B. sC. rD. RAnswer: AMultiple And Submultiple Angles Questions and Answer Set 1: Ques No 14In triangle ABC, tanA/2 = 5/6, tanC/2 = 2/5, then the relation between a, b, and c isOptions:A. b2 = acB. 2b = a + cC. 2ac = b(a + c)D. a + b = cAnswer: BMultiple And Submultiple Angles Questions and Answer Set 1: Ques No 15In triangle ABC, if s =

2b, then the value of cot A/2 cot C/2 isOptions:A. 2B. 1C. 3D. Sqrt(2)Answer: A questions and answerstrigonometry Doubtnut facilitates RD Sharma Class 11 Solutions in a video tutorial format. Get here all systematically solved exercises of Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles. RD Sharma Solutions are a helpful resource

for CBSE based exam preparation as well for daily homework routine. Doubtnut gives its users access to a profuse supply of solved answers and is created by experts of the subject, hence, sure to help you score well in the exam. The problems provided in RD Sharma Books are prepared in accordance with CBSE syllabus, thus holding higher odds of

appearing in the examination. Not only do these RD Sharma Solutions for Class 11 math will strengthen your foundation in the subject, but they also give them the aptitude to solve various types of questions effortlessly. Our RD Sharma textbook solutions give students an advantage of real-world problems practical questions. The video tutorials are

easy to understand and are curated in a step-by-step manner to describe every problem meticulously. The exercise-wise solutions are ideal for students to prepare for board examination as well as various engineering competitive examinations. It can be your best resource to quickly practice any topic of Chapter 9 Trigonometric Ratios of Multiple and

Submultiple Angles. Class 11 Maths Solutions are as per the latest edition of the RD Sharma book. These solutions have been prepared by our expert mathematicians and board examination experts. Our professional tutors have followed the simplest approach towards solving the questions and designing video tutorials for Chapter 9, Trigonometric

Ratios of Multiple and Submultiple Angles. By following the given tutorials you will find it very easy to solve the math problems. The video tutorials have been prepared to keep in mind student's needs and ensure comprehensive learning of basic concepts. The video tutorials are filled with amazing math tips and tricks to make it easier for you to meet

the answer. It is to give you a competitive edge to ace the examination. You can also install the DOUBTNUT App to watch and learn in the comfort of your home on your mobile or tablet.Class 11 Maths RD Sharma Solutions Trigonometric Ratios of Multiple and Submultiple AnglesIf A is an angle, then 2A, 3A, 4A etc. are called `multiple' angles of A

and A/2, A/3, etc. are called `sub multiple' angles of A. Trigonometry is full of formulae along with multiple and submultiple of an angle. Learning and understanding of these formulae are tremendously important if you wish to earn some direct formula-based marks in your exams. If you are struggling to understand any topic, then you can rely on

these video tutorials of RD Sharma textbook solutions, these are prepared by doubtnut to guide students in easy to understand language. Catered to Class 11 students, RD Sharma Maths Solutions provides step-by-step solutions to unsolved problems. Every exercise is thoroughly covered and if don't understand any step; you can at once clear your

doubts using the Doubtnut App. Each concept is explained to helps students a great deal in preparing for their Class 11 board exams. The RD Sharma Class 11 Solutions - Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles PDF file includes exercises wise solutions that cover the entire curriculum. Attributes of RD Sharma Class 11

Solutions Chapter 9 Trigonometric Ratios of Multiple and Submultiple AnglesAll video solutions are recorded in Hinglish (Hindi-English) language.Fully solved exercises.Strictly as per the CBSE examination guidelines and CBSE marking scheme.Step-by-step detailed explanation for every question. Video tutorials showcasing every brief summary of

concepts and formulae applied. The tutorials have been answered by the subject matter experts of IIT cadre. Videos are full of tips and tricks to help you gain the competitive edge to ace the board examination. Diagrams are given to help the students in visualizing the solutions.We hope that the given RD Sharma Class 11 Solutions Chapter 9

Trigonometric Ratios of Multiple and Submultiple Angles, help you. If you have any query then feel to reach for expert's advice on the Doubtnut App, it will get resolved at earliest. Install the Doubtnut App Now!For the latest updates about our newest published video tutorial, you can install the Doubtnut app on your smartphone. With the highest 4.4

play store rating, it is the best e-learning app for Math's learning and practice. DOUBTNUT App facilitates easily downloadable and sharable amazing video solutions for RD Sharma Class 11 Solutions - Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles. Solution:Let us solve LHS,= cos2 (/4 ? x) ? sin2(/4 ? x)As we know that,cos2 A ?

sin2 A = cos 2ASo,= cos2 (/4 ? x) sin2 (/4 ? x)= cos 2 (/4 ? x)= cos (/2 ? 2x)= sin 2x

[As we know that, cos (/2 ? A) = sin A]LHS = RHSHence Proved.Question 17. cos 4x = 1 ? 8 cos2x + 8 cos4 xSolution:Let us solve LHS,= cos 4xAs we know that,cos 2x = 2 cos2x ? 1So,cos 4x = 2 cos2 2x ? 1= 2(2 cos2 2x ? 1)2 ? 1 = 2[(2 cos2 2x)2 +

12 ? 2 ? 2 cos2x] ? 1= 2(4 cos4 2x + 1 ? 4 cos2x) ? 1= 8 cos4 2x + 2 ? 8 cos2 x ? 1= 8 cos4 2x + 1 ? 8 cos2xLHS = RHSHence Proved.Question 18. sin 4x = 4 sin x cos3 x ? 4 cos x sin3 xSolution:Let us solve LHS,= sin 4xAs we know that,sin 2x = 2 sin x cos xcos 2x = cos2x ? sin2xSo,sin 4x = 2 sin 2x cos 2x= 2 (2 sin x cos x) (cos2 x ? sin2 x)= 4 sin x

cos x (cos2 x ? sin2 x)= 4 sin x cos3 x ? 4 sin3 x cos xLHS = RHSHence proved.Question 19. 3(sin x ? cos x)4 + 6 (sin x + cos x)2 + 4 (sin6 x + cos6 x) = 13Solution:Let us solve LHS,= 3(sin x ? cos x)4 + 6 (sin x + cos x)2 + 4 (sin6 x + cos6 x) As we know that,(a + b)2 = a2 + b2 + 2ab(a ? b)2 = a2 + b2 ? 2aba3 + b3 = (a+b)(a2 + b2 ? ab)So,= 3{(sinx

? cosx)2}2 + 6 {(sinx)2 + (cosx)2 + 2 sinx cosx} + 4 {(sin2x)3 + (cos2x)3}= 3{(sinx)2 + (cosx)2 ? 2 sinx cosx}2 + 6(sin2x + cos2x + 2 sinx cosx) + 4{(sin2x + cos2x)(sin4x + cos4x ? sin2x cos2x)}= 3(1 ? 2 sinx cosx)2 + 6(1 + 2 sinx cosx) + 4{(1)(sin4x + cos4x ? sini2x cos2x)}Since, sin2x + cos2x = 1So,= 3{12 + (2 sinx cosx)2 ? 4 sinx cosx} + 6(1 +

2 sinx cosx) + 4{(sin2x)2 + (cos2x)2 + 2 sin2x cos2x ? 3 sin2x cos2x}= 3{1 + 4 sin2x cos2x ? 4 sinx cosx} + 6(1 + 2 sinx cosx) + 4{(sin2x + cos2x)2 ? 3 sin2x cos2x}= 3 + 12 sin2x cos2x ? 12 sinx cosx + 6 + 12 sinx cosx + 4{(1)2 ? 3 sin2x cos2x}= 9 + 12 sin2x cos2x + 4(1 ? 3 sin2x cos2x)= 9 + 12 sin2x cos2x + 4 ? 12 sin2x cos2x= 13LHS =

RHSHence proved.Question 20. 2(sin6x + cos6x) ? 3(sin4x + cos4x) + 1 = 0Solution:Let us solve LHS,= 2(sin6 x + cos6 x) ? 3(sin4 x + cos4 x) + 1 As we know that,(a + b)2 = a2 + b2 + 2aba3 + b3 = (a + b) (a2 + b2 ? ab) So, = 2(sin6x + cos6x) ? 3(sin4x + cos4x) + 1 = 2{(sin2x)3 + (cos2x)3} ? 3{(sin2x)2 + (cos2x)2} + 1= 2((sin2x + cos2x)(sin4x +

cos4x ? sin2x cos2x) ? 3{(sin2x)2 + (cos2x)2 + 2 sin2x cos2x ? 2sin2x cos2x} + 1= 2{(1)(sin4x + cos4x + 2 sin2x cos2x ? 3 sin2x cos2x) ? 3((sin2x + cos2x)2 ? 2sin2x cos2x) + 1Sincesin2x + cos2x = 1So, = 2{(sin2x + cos2x)2 ? 3 sin2x cos2x} ? 3{(1)2 ? 2 sin2x cos2x} + 1= 2{(1)2 ? 3 sin2x cos2x} ? 3(1 ? 2 sin2x cos2x) + 1= 2(1 ? 3 sin2x cos2x) ? 3 +

6 sin2x cos2x + 1= 2 ? 6 sin2x cos2x ? 2 + 6 sin2x cos2x= 0 LHS = RHSHence Proved.Question 21. cos6 x ? sin6 x = cos 2x (1 ? 1/4 sin2 2x)Solution:Let us solve LHS,= cos6 x ? sin6 xAs we know that,(a + b)2 = a2 + b2 + 2aba3 ? b3 = (a ? b) (a2 + b2 + ab)So,cos6 x ? sin6 x = (cos2 x)3 ? (sin2 x)3 = (cos2x ? sin2x) (cos4x + sin4x + cos2x sin2x)As we

know that,cos 2x = cos2x ? sin2xSo,= cos 2x [(cos2x)2 + (sin2x)2 + 2 cos2x sin2x ? cos2x sin2x]= cos 2x [(cos2x)2 + (sin2x)2 ? 1/4 ? 4 cos2x sin2x]As we know that,sin2x + cos2x = 1So,= cos2x [(1)2 ? 1/4 ? (2 cosx sinx)2]As we know that,sin2x = 2 sinx cosxSo,= cos 2x [1 ? 1/4 ? (sin 2x)2]= cos 2x [1 ? 1/4 ? sin22x]LHS = RHSHence

proved.Solution:Let us solve LHS,= tan (/4 + x) + tan (/4 ? x)As we know that,tan (A + B) = (tan A + tan B)/(1 ? tan A tan B)tan (A ? B) = (tan A ? tan B)/(1 + tan A tan B) So, = Since, tan /4 = 1So,= = By using formulas, we get(a ? b)(a + b) = a2 ? b2(a + b)2 = a2 + b2 + 2ab & (a ? b)2 = a2+ b2 ? 2abSo,= = = As we know that,tan x = sin x/cos

xSo,= = = By using the formulas, we getcos2x+ sin2x = 1 & cos 2x = cos2x ? sin2xSo,= = 2/cos2x= 2 sec2xLHS = RHSHence Proved.Question 23. cot2x ? tan2x = 4cot2x cosec2xSolution:Let us solve LHS,= cot2x ? tan2x= cos2x/sin2x ? sin2x/cos2x= [(cos2x)2 ? (sin2x)2] / sin2xcos2x= [(cos2x + sin2x)(cos2x ? sin2x)] / sin2xcos2x= (1 ? cos2x) /

sin2xcos2x= 4cos2x / 4sin2xcos2x= 4(cos2x) / (sin2x)2= 4(cos2x) / (sin2x) ? 1 / (sin2x)= 4 cot2x cosex2x LHS = RHSHence ProvedQuestion 24. cos4x ? cos4 = 8(cosx ? cos)(cosx + cos)(cosx ? sin)(cosx + sin)Solution:Let us solve RHS,= 8(cosx ? cos)(cosx + cos)(cosx ? sin)(cosx + sin)= 8(cos2x ? cos2)(cos2x ? sin2)= 8(cos4x ? cos2x ?

sin2 ? cos2 ? cos2x + cos2 ? sin2)= 8{cos4x ? cos2x(sin2 + cos2) + cos2 ? sin2}= 8{cos4x ? cos2x + cos2 ? (1 ? cos2)}= 8{cos4x ? cos2x + cos2 ? cos4)}= 8{cos2x(cos2x ? 1) + cos2 ? (1 ? cos2)}= 8{1/2 cos2x (2cos2x ? 1 ? 1) ? 1/2 cos2 (2cos2 ? 1 -1)}= 8{1/2 cos2x (cos2x ? 1) ? 1/2cos2 (cos2 ? 1)}= 8[1/4 {2cos2x (cos2x ? 1)

? 2cos2x (cos2 ? 1)}]= 8[1/4 {(1 + cos2x)(cos2x ? 1) ? (1 + cos2)(cos2 ? 1)}]= 8[1/4 { cos22x ? 1 ? cos22 + 1}]= 8[1/8 {2cos22x ? 2cos22}]= [{(1 + cos4x) ? (1 + cos4)}]= [1 + cos4x ? 1 ? cos4]= cos4x ? cos4 LHS = RHSHence provedQuestion 25. sin3x + sin2x ? sinx = 4 sinx cos(x/2) cos(3x/2)Solution:Let us solve LHS,= sin3x + sin2x ?

sinx= sin3x + 2sin(2x ? x)/2 cos(2x + x)/2= sin3x + 2sin(x/2) cos(3x/2)= 2sin(3x/2) cos(3x/2) + 2sin(3x/2) cos(x/2)= 2cos(3x/2)[sin(3x/2) cos(x/2)]= 2cos(3x/2)[2sin(3x/2+x/2)/2 cos(3x/2 ? x/2)/2]= 2cos(3x/2)[2sinx cos(x/2)]= 4 sinx cos(x/2) cos(3x/2)LHS = RHSHence proved.Question 26. = (3 + 2)(2 + 1) = 2 + 3 + 4 + 6Solution:Let us solve

LHS,tan(82.5)? = tan(90 ? 7.5)? = cot(7.5)? = 1/ tan(7.5)?We have,tan(x/2) = sinx/(1 + cosx)Now on putting x = 15?, we gettan(15/2) = sin15?/(1 + cos15?)= sin(45-30)?/{1 + cos(45-30)?} = (sin45?cos30? ? sin30?cos45?) / (1 + cos45? sin30?)= = = Now,tan(82.5)? = 1/tan(7.5)? = (22 + 3 + 1)/(3 ? 1)= (22 + 3 + 1)/(3 ? 1) ? (3 + 1)/(3 + 1)=

[3 + 1(22 + 3 + 1)] / [(3)2 ? 12]= (26 + 3 + 3 + 22 + 3 + 1) / (3 ? 1)= (26 + 23 + 22 + 4) / (2)= 6 + 3 + 2 + 2= 2 + 3 + 4 + 6

.....(i)= 6 + 3 + 2 + 2= 3(2 + 1) + 2(2 + 1)= (3 + 2)(2 + 1) .....(ii)From equ (i) and (ii), we get tan(82.5)? = (3 + 2)(2 + 1) = 2 + 3 + 4 + 6LHS = RHSHence provedQuestion

27. = 2 + 1Solution:As we know that, /8 = = 45?Let A = By using the identity cot2A = (cot2A ? 1)/2cotA, we getcot45? = {cot2()? ? 1} / 2cot()? 1 = {cot2()? ? 1} / 2cot()? 2cot()? ? cot2()? + 1 = 0 cot2()? ? 2cot()? ? 1 = 0 { cot2() ? 2cot()? + 1} ? 2 = 0 { cot()? ? 1}2 = 2 cot()? ? 1 = 2 cot()? = 2 + 1LHS = RHSHence provedQuestion 28

(i). If cosx = (-3/5) and x lies in the 3rd quadrant, find the values of cos(x/2), sin(x/2), sin2x.Solution:Given that, cosx = (-3/5) cosx = cos2(x/2) ? sin2(x/2) -3/5 = 2cos2(x/2) ? 1 1 ? 3/5 = 2cos2(x/2) 2/5 = 2cos2(x/2) 1/5 = cos2(x/2) cos(x/2) = ? (1/5)Also, given that x lies in 3rd quardant, so x/2 lies in 2nd quadrant.cos(x/2) = ?

(1/5) Again,cosx = cos2(x/2) ? sin2(x/2) -3/5 = (- (1/5))2 ? sin2(x/2) ? 3/5 = 1/5 ? sin2(x/2) -1/5 -3/5 = -sin2(x/2) 4/5 = sin2(x/2) sin(x/2) = ? 2/5It is given x lies in 3rd quardant, so x/2 lies in 2nd quadrant.sin(x/2) = 2/5Now,sinx = (1 ? cos2x)= (1 ? (-3/5))2= (1 ? 9/25)= ? 4/5It is given x llies in 3rd quardant, so sinx is negative.sinx = ?

4/5sin2x = 2 sinx cosx= 2 (-4/5) (-3/5)= 24/25Hence, the value of cos(x/2) = ? (1/5), sin(x/2) = 2/5, and sin2x = 24/25.Question 28 (ii). If cosx = (-3/5) and x lies in the 3rd quadrant, find the values of sin2x and sin(x/2).Solution:Given that, cosx = (-3/5) sinx = sinx = ? 4/5Here, x lies in the second quadantSo, sinx = 4/5As we know that, sin2x = 2

sinx cosxsin2x = 2 ? 4/5 ? (-3/5) = (-24/25)Now,cosx = 1 ? 2 sin2(x/2) 2sin2(x/2) = 1 ? (-3/5) = 8/5 sinx2(x/2) = 4/5 sin(x/2) = ? 2/5Since x lies in the second quadrant,x/2 lies in the first quadrantSo, sin(x/2) = 2/5Hence, the value of sin2x = (-24/25) and sin(x/2) = 2/5Question 29. If sinx = 5/3 and x lies in 2rd quadrant, find the values of

cos(x/2), sin(x/2) and tan(x/2).Solution:Given that, sinx = 5/3As we know that sinx = P/H So, P = 5, H = 3 and B = 2Now, cosx = B/H = -2/3So, cos(x/2) = {(1 + cosx)/2} = {(1 ? 2/3)/2} = 1/6sin(x/2) = {(1 ? cosx)/2} = {(1 + 2/3)/2} = (5/6)tan(x/2) = sin(x/2)/cos(x/2) = {(5/6)} / (1/6) = 5Attention reader! Don't stop learning now. Join the

First-Step-to-DSA Course for Class 9 to 12 students , specifically designed to introduce data structures anf algorithms to the class 9 to 12 studentsPage 2Solution:Given that, sinx = 1/4As we know that, sinx = (1 ? cos2x)So, (1/4)2 = (1 ? cos2x) (1/16) ? 1 = ? cos2xcosx = ? 15/4It is given that x is in 2nd quadrant, so cosx is negative.cosx = ?

15/4Now,As we know that, cosx = 2 cos2(x/2) ? 1So, ? 15/4 = 2cos2(x/2) ? 1 cos2(x/2) = ? 15/8 + 1/2cos(x/2) = ? (4-15)/8It is given that, x is in 2nd quadrant, so cos(x/2) is positive.cos(x/2) = (4 ? 15)/8Again,cosx = cos2(x/2) ? sin2(x/2) ? 15/4 = {(4 ? 15)/8}2 ? sin2(x/2) sin2(x/2) = (4 + 15)/8 sin(x/2) = ? {(4 + 15)/8} = {(4 +

15)/8}Now,tan(x/2) = sin(x/2) / cos(x/2)= = = = = = 4 + 15Hence, the value of cos(x/2) = (4 ? 15)/8, sin(x/2) = {(4 + 15)/8}, and tan(x/2) = 4 + 15 .Question 30(ii). If cosx = 4/5 and x is acute, find tan2x.Solution:Given that, cosx = 4/5As we know that, sinx = (1 ? cos2x)So, = (1 ? (4/5)2)= (1 ? 16/25)= {(25 ? 16)/25}= (9/25)= 3/5Since,

tanx = sinx/cosx, so= (3/5) / (4/5)= 3/4As we know that,tan2x = 2tanx / (1 ? tan2x)= 2(3/4) / {1 ? (3/4)2}= 2(3/4) / (1 ? 9/16)= (3/2) / (7/16)= 24/7Hence, the value of tan2x is 24/7Solution:Given that, sinx = 4/5As we know that, sinx = (1 ? cos2x)So, (4/5)2 = 1 ? cos2x 16/25 ? 1 = -cos2x 9/25 = cos2x cosx = ?3/5It is given that, x is ln the 1st

quadrant So, cosx = 3/5Now,sin4x = 2 sin2x cos2x= 2 (2 sinx cosx)(1 ? 2sin2x)= 2(2 ? 4/5 ? 3/5)(1 ? 2(4/5)2)= 2(24/25)(1-32/25)= 2(24/25)((25-32)/25)= 2(24/25)(-7/25)= -336/625Hence, the value of sin4x is (- 336/625)Question 31. If tanx = b/a, then find the value of Solution:We have to find the value of So, = It is given that tanx = b/a,

so= = = = = = Hence, the value of is Solution:Given that, tanA = 1/7 and tanB = 1/3Show: cos2A = sin4BAs we know that, tan2B = 2tanB / (1 ? tan2B)= (2 ? 1/3)(1 ? 1/9) = 3/4So, cos2A = (1 ? tan2A)/(1 + tan2A) = {1-(1/7)2}/{1+(1/7)2}

= 48/50

= 24/25And sin4B = 2tan2B / (1 + tan22B)= {2 ? 3/4}{1 +

(3/4)2}= 24/25Hence, cos2A = sin4BQuestion 33. cos7? cos14? cos28? cos56? = sin68?/16cos83?Solution:Lets solve LHS = cos7? cos14? cos28? cos56?On dividing and multiplying by 2sin7?, we get= ? 2sin7? ? cos7? ? cos14? ? cos28? ? cos56?= ? cos28? ? cos56?= ? cos56?= = = LHS = RHSHence proved.Question 34. Proved that,

cos(2/15)cos(4/15)cos(8/15)cos(16/15) = 1/16Solution:Let's solve LHS = cos(2/15)cos(4/15)cos(8/15)cos(16/15)On dividing and multiplying by 2sin(2/15), we get= = = = = = = 1/16 LHS = RHSHence proved.Question 35. Proved that, cos(/5)cos(2/5)cos(4/5)cos(8/5) = -1/16Solution:Lets solve LHS =

cos(/5)cos(2/5)cos(4/5)cos(8/5)On dividing and multiplying by 2sin(2/5), we get= ? 2sin(/5)cos(/5)cos(2/5)cos(4/5)cos(8/5)= (sin(2/5)cos(2/5)cos(4/5)cos(8/5))= [2sin(2/5)cos(2/5)cos(4/5)cos(8/5)]= [sin(4/5)cos(4/5)cos(8/5)]= [2sin(4/5)cos(4/5)cos(8/5)]= [sin(8/5)cos(8/5)] =[2sin(8/5)cos(8/5)]= == = -1/16 LHS =

RHSHence proved.Solution:Lets solve LHS = cos(/65)cos(2/65)cos(4/65)cos(8/65)cos(16/65)cos(32/65)Now on dividing and multiplying by 2sin(/65), we get= ? 2sin(/65)cos(/65)cos(2/65)cos(4/65)cos(8/65)cos(16/65)cos(32/65)= ? [cos(2/65) ? cos(4/65) ? cos(8/65) ? cos(16/65) ? cos(32/65)]= ? cos(4/65) ? cos(8/65) ?

cos(16/65) ? cos(32/65)= ? cos(8/65) ? cos(16/65) ? cos(32/65)= ? cos(16/65) ? cos(32/65)= ? cos(32/65)= = = = 1/64 LHS = RHSHence provedQuestion 37. If 2tan = 3tan, prove that tan( ? ) = sin2 / (5 ? cos2)Solution:Given that,2tan = 3tanProve: tan( ? ) = sin2 / (5 ? cos2)Proof:Lets solve

LHS = = = = = = = = = = = = = = LHS = RHSHence proved.Question 38(i). If sin + sin = a and cos + cos = b, prove that sin( + ) = 2ab/(a2 + b2)Solution:Given that, sin + sin = a and cos + cos = bProve: sin( + ) = 2ab/(a2 + b2)Proof:As we know that, So ......(i)Now, using the identity .....(ii)Now on dividing eq(i) and (ii), we

gettan( + )/2 = a/bAs we know that,sin2x = 2tanx/(1 + tan2x)= = 2ab/(a2 + b2)LHS = RHSHence provedQuestion 38(ii). If sin + sin = a and cos + cos = b, prove that cos( ? ) = (a2 + b2 ? 2)/2Solution:Given that, sin + sin = a ......(i) cos + cos = b .......(ii)Now on squaring eq(i) and (ii) and then adding them, we getsin2 + sin2 +

2sinsin + cos2 + cos2 + 2coscos = a2 + b2 1 + 1 + 2(sinsin + coscos) = a2 + b2 2(sinsin + coscos) = a2 + b2 ? 2 2cos( ? ) = a2 + b2 ? 2 cos( ? ) = (a2 + b2 ? 2)/2 Hence proved.Question 39. If 2tan(/2) = tan(/2), prove that cos = Solution:Given that, 2tan(/2) = tan(/2)Prove: cos = Proof:Let us solve

RHS = = = = = = = = cos RHS = LHSHence proved.Question 40. If cosx = , prove that tan(x/2) = ? tan(/2)tan(/2).Solution:Given that, .....(i) Now, by componendo and dividendo, we get tan2(x/2) = tan2(/2)tan2(/2) tan(x/2) = ?tan(/2)tan(/2)Hence Proved.Solution:Given that, sec(x + ) + sec(x ? ) =

2secxSo, cos2xcos = cos2x(cos2 + sin2) ? sin2 cos2x(1 ? cos) = sin2 = cosx = ? 2 cos(/2)Hence ProvedQuestion 42. If cos + cos = 1/3 and sin + sin = 1/4, prove that cos( ? )/2 = ?5/24.Solution:Given that, cos + cos = 1/3 sin + sin = 1/4, we getProve: cos( ? )/2 = ?5/24Proof:(cos2 + cos2 + coscos) +

(sin2 + sin2 + 2sinsin) = 1/9 + 1/161 + 1 + 2(coscos + sinsin) = 25/1442 + 2cos( ? ) = -263/288 .....(i)Now, = [From (i)]= 25/576= ? 5/24Hence proved.Question 43. If sin = 4/5 and cos = 5/13, prove that cos{( ? )/2} = 8/65.Solution:Given that,sin = 4/5 and cos = 5/13As we know that.cos = (1 ? sin2)So, = {1 ? (4/5)2}=

3/5Also, sin = (1 ? cos2)= {1 ? (5/13)2}= 12/13Now,cos( ? ) = cos cos + sin sin= (3/5)(5/13)(4/5)(12/13)= 63/65Thus,cos{( ? )/2} = = = 8/65Hence Proved.Question 44. If acos2 + bsin2 = c has and as its roots prove that,(i) tan + tan = 2b/(a + c)(ii) tan tan = (c ? a)/(c + a)(iii) tan( + ) = b/aSolution:As we know thatNow

substitute these values in the given equation, we geta(1 ? tan2) + b(2tan) = c(1 + tan2)(c + a)tan2 + 2btan + c ? a = 0 (i) As and are rootsSo, sum of the roots:tan + tan = 2b / (c + a)(ii) As and are rootsSo, product of roots:tan tan = (c ? a) / (c + a) (iii) tan( + )= = = b/aHence proved.Question 45. If cos + cos = 0 = sin + sin,

then prove that cos2 + cos2 = -2cos( + ).Solution:Given that, cos + cos = 0 = sin + sinProve: cos2 + cos2 = -2cos( + )Proof:cos + cos = 0On squaring on both sides, we getcos2 + cos2 + 2 cos cos = 0 ....(i)Similarlysin + sin = 0On squaring on both sides, we getsin2 + sin2 + 2 sin sin = 0 .....(ii)Now, subtract eq (ii)

from (i), we get (cos2 + cos2 + 2 cos cos) ? (sin2 + sin2 + 2 sin sin) = 0 cos2 ? sin2 + cos2 ? sin2 + 2(cos cos ? sin sin) = 0 cos2 + cos2 + 2cos( + ) = 0 cos2 + cos2 = -2cos( + )Hence proved.Attention reader! Don't stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12

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trigonometric ratios of multiple and submultiple angles formula. trigonometric ratios of multiple and submultiple angles pdf

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