1 / 10

Using Basic trick for trigonometric functions- Fullmakrseducare

In mathematics, the trigonometric functions are functions of an angle. This chapter explains how the trig functions of sin, cos, tan, cosec, sec and cot can be used to solve problems.

fullmarkseducare
Download Presentation

Using Basic trick for trigonometric functions- Fullmakrseducare

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter-3 Trigonometric Functions Class– XI r Exercise 3.1 Ans: We know that θ = l Q1: Find the radian measures corresponding to the following degree measures: (i) 25° (ii) – 47° 30' Ans: We know that 180° = π rad 5 25º 25 = rad 180 36 (ii) – 47° 30' 95 2 95π 95 = = × 2 180  π Here, r = 100 cm, l = 22 cm. We have 22 θ = 100 180 7 22degree 22 100 12º36' = Thus, the required angle is 12°36′. (iii) 240° (iv) 520° 180 π 22 100 × degree = rad π π = × (i) × × × = ° − Q5: In a circle of diameter 40 cm, the length of a chord is 20 cm. Find out the length of minor arc of the chord. Ans: Diameter of the circle = 40 cm Radius of the circle = 40 2 Let AB be the chord. Given AB = 20 cm = [1° = 60'] – 47° 30' ° − − −      19 72 = π rad rad 2 4 3 26 9 = × π (iii) 240º 240rad = rad 180 π = 20 cm = × π (iv) 520º 520rad= rad 180 Q2: Find the degree measures corresponding to the following radian measures.   (i) 11 16 Ans: We know that π rad = 180° (i) 11rad 16 16 π 3 60' 39º 8 1 39º 22' minutes 2 = 39º 22' 30" 180 4rad= ( 4)º π 1 60 229º 11 5 229º 5' minutes 11 229º5'27" = − 5 rad 3 π 7 180 7 rad 6 π Q3: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? Ans: Number of revolutions made in 1 minute = 360 ∴ Revolutions made per second = 360 22 7    (iii) 5 Useπ= π π (iv) 7 (ii) – 4 3 6 In ∆OAB, OA and OB are radii of the circle. Hence, OA = OB = 20 cm. Also, AB = 20 cm. Thus, ∆OAB is an equilateral triangle. 180 11 × 315º 8 = = × = + [1º=60'] π OAB ∠ = Therefore, = rad 60º 3 + + = As we know, θ = l r [1'=60"] π π AB 20 20 = ⇒ AB = cm × − °= ° 3 3 180 7( 4) 22 1 − − × − (ii) = 22911 Thus, the length of the minor arc of the chord is 20 cm. 3 π × = l = − + minutes Q6: If in two circles, arcs of the same lengths subtend angles of 60° and 75° at the centres, find the ratio of their radii. Ans: Let the radii of the two circles be r1 and r2. Let the arc length be l. It subtends an angle of 60° at the centre of the circle of radius r1, and an angle of 75° at the centre of the circle of radius r2. π and 75º = 5 = − + π π 180 5 = × = (iii) 300º 3 π π = × = (iv) 210º 6 π rad Now, 60º = rad 12 3 l r = θ We know that θ = l r or π π 25 12 r r = 6 ∴ = = 1 3 r and l l 60 Q 4: 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.      7 Angle swept by the wheel in 1 revolution = 2π rad. Angle covered in 6 revolutions = 6 × 2π =12 π rad Thus, the wheel covers 12π rad in one second. π π 25 12 r ⇒ = 1 3 r r 5 4 ⇒ = 1 2 Q7: Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length Thus, the ratio of their radii is 5:4. 22  Useπ= http://www.fullmarkseducare.com/Ph:. 9911968787 NCERT SOLUTIONS

  2. Chapter-3 Trigonometric Functions Class– XI Ans: We know that in a circle of radius r unit, θ = l (i) 10 cm (ii) 15 cm (iii) 21 cm ⇒ 1 sin = − 2 2 cos x x 2 r     3 5 ⇒ = −  2 cos 1 x Where, θ is the measure of the angle in radians subtended by an arc of length l at the centre of the circle. Given r = 75 cm (i) Here, l = 10 cm 10 2 θ = rad = rad 75 15 (ii) Here, l = 15 cm 15 1 θ = rad = rad 75 5 (iii) Here, l = 21 cm 21 7 θ = rad = rad 75 25 16 25 4 5 ⇒ = ⇒ = ± 2 cos cos x x As x is in the 2nd quadrant, therefore value of cos x will be negative. 4 cos 5 1 5 sec cos 4 x 3 sin 5 tan cos x  −   1 4 cot tan 3 x ⇒ = − x = = − x       3 4 x = = = − = x    4 5 = = − = x Exercise 3.2 Q3: Find the values of other five trigonometric functions if 3 cot 4 3 cot 4 1 tan cot x 1 tan sec x x + = 2 2 4 1 sec 3   25 sec 9 5 sec 3 Since x lies in the 3rdquadrant, the value of sec x will be negative. 5 sec 3 1 3 cos sec 5 x sin tan cos x ⇒ 4 3 3 5   1 cosec sin x Q4: Find the values of other five trigonometric functions if 13 sec 5 13 sec 5 1 5 cos sec 13 x sin 1 cos = − x x Q1: Find the values of other five trigonometric functions if 1 cos 2 1 cos 2 1 sec cos x Now, sin cos x x + = sin 1 cos x ⇒ = − x = , and x lies in the third quadrant. x = − and x lies in the third quadrant. x = Ans: Given: x = − Ans: Given: 4 3 = = x ∴ = = − 2 x 2 2 2 2 1 x     2 2 + = x ⇒ 2    1 4    1 2 3 4 ⇒ = − − 2 sin 1 x ⇒ = 2 x ⇒ = − = 2 sin 1 x ⇒ = ± x 3 ⇒ = ± sin x 2 As x lies in the 3rd quadrant, the value of sin x will be negative. 3 sin 2 1 2 cosec sin 3 x  −    = =  −   1 1 cot tan 3 x ∴ = − x ∴ = − x = =− x = =− x x = x     3 sin 4 5 x ⇒ = − = sin x 2 1 2 sin cos   x x = − tan 3 x     5 4 = = − x = = x Q2: Find the values of other five trigonometric functions if 3 sin 5 3 sin 5 1 cosec sin x sin cos 1 x x + = x = , and x lies in the fourth quadrant. x = , and x lies in second quadrant. x = Ans: Given: x = Ans: Given: 5 3 = = x = = x 2 2 2 2 http://www.fullmarkseducare.com/Ph:. 9911968787 NCERT SOLUTIONS

  3. Chapter-3 Trigonometric Functions Class– XI π 25 169 12 13 144 169 19 ⇒ = − = 2 Q8: Find the value of the trigonometric function sin 1 tan x 3 Ans: The values of tan x repeat after an interval of π or 180°. 19 tan tan 6 3 3   ⇒ = ± sin x π π π     ∴ = + = = = π tan tan60º 3 Since x lies in the 4th quadrant, the value of sin x will be negative. 12 sin 13 −    = =    1 1 cot 12 tan  −   3 1 13 12 π       11 ∴ = − = = − ⇒ cosec x x − Q9: Find the value of the trigonometric function sin sin x 3    12 13 5 13 π π π 3             11 11 = = ∴ − = − 2 2 + × π sin3 Ans: sin sin sin cos 12 5 x x 2 =− 3 3 tan x    π       15 − Q10: Find the value of trigonometric function cot 5 4 = = = − x    12 x Ans: The values of cot x repeat after an interval of π or 180°. 15 15 cot cot 4    Exercise 3.3 π 5 π π π          ∴ − = − + π = = 4 cot 1 Q5: Find the values of other 5 trigonometric functions if 5 tan 12 5 tan 12 1 cot tan x 1 tan sec x + = 25 1 sec 144 13 sec 12 Since x lies in the 2nd quadrant, the value of sec x is negative. 13 12 1 cos sec x sin tan cos x 5 sin 12 12 13   5 sin 12    1 cosec sin x 4 4 x = − , and x lies in second quadrant. π π 1 2 x = − Ans: Given: + − = − 2 2 2 Q1: Prove that: sin cos tan 6 3 4 12 5 x π π π = = − x + − 2 2 2 Ans: LHS = sin cos tan 6 3 4 2 2 2 2       1 4       1 2 1 2 = + − 2 (1) ⇒ + = 2 x 1 4 1 2 = + − = − 1 = RHS ⇒ = ± x π π π 7 3 2 + = 2 2 2 Q2: Prove that: 2sin cosec cos 6 6 3 7 − ∴ sec x = π π π + 2 2 2 Ans: LHS = 2sin cosec cos 12 13 6 6 3 = = − x 2 2 π       1 4    π       1 2 1 2 = + π + 2 2 cosec x 6 = x 2             3 2 1 4 = × + − 2 cosec x ⇒ − = 6     −       1 2 1 4 = = ( 2) + − 2     × −      12 13 1 5 13 5 ⇒ = − = x = RHS 13 13 5 π π π 5 = = = x + + = 2 2 Q3: Prove that: cot cos 3tan 6 ec       6 6 6 π π π 5 + + 2 2 Ans: LHS = cot cos 3tan ec Q6: Find the value of the trigonometric function sin 765° Ans: Values of sin x repeat after an interval of 2π or 360°. 6 ) 6 6   2    π (       1 2 π = + − +  3 cosec 3 6 3 1 ∴ = sin(2 360º 45º) × + = = sin765º sin45º ecπ 1 3 2 3 cos = + + × 3 Q7: Find value of trigonometric function cosec (–1410°) Ans: The values of cosec x repeat after an interval of 2π or 360°. cosec(–1410°) cosec( 1410º 4 360º) = − cosec( 1410 = − cosec30º = 6 6 3 2 1 = + + = = RHS π π π 3 + + = 2 2 2 Q4: Prove that: 2sin 2cos 2sec 10 + × 1440º) °+ = 4 4 3 π π π 3 + + 2 2 2 Ans : LHS = 2sin 2cos 2sec 2 4 4 3 http://www.fullmarkseducare.com/Ph:. 9911968787 NCERT SOLUTIONS

  4. Chapter-3 Trigonometric Functions Class– XI π 2 2                           π           1 + tan tan x = π − + + 2 2 sin    = 2 2(2) 4 4 2 π             π + tan x 1 tan − tan x 2 π    1 2 4 π 4 + × + 2 sin      1 1 8 = + + = 2 8 = Now LHS= 4 π − − tan tan tan x x 2 4 4    1 = 1 8 + + 2 π 1 tan + tan x 2 4 = Q5: Find the value of (i)sin 75° (ii) tan 15° Ans: (i) sin 75° = sin (45° + 30°) We know that sin (x + y) = sin x cos y + cos x sin y sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° 1 2    10 R.H.S             + − − 1 tan 1 tan 1 tan 1 tan + x x x x 2 +       1 tan 1 tan − x x = = = RHS                      3 1 1 2 = + + − cos(π )cos( x x ) = 2 Q8: Prove that: 2 cot x 2       ) π 2 − + sin(π )cos x x + 3 1 3 1 2 2 = + = [ ( ][ − ] ) − π + − cos sin cos sin x x x cos( )cos( x x 2 2 2 2 Ans: LHS = = )( π (ii) tan 15° = tan (45° – 30°) tan45º tan30º 1 tan45ºtan30º +       x π − + sin( )cos x x 2       − − tan 1 tan tan + tan x x y − 2 cos sin x x = − = tan( ) x y = = 2 cot RHS x =− y 2 1 Q9: Prove that: 3 cos 2  Ans: LHS = ( ) 3 1 − − 2 1 3 1 − − 3 1 3 1 + 3 1 = = = ( )( )      3 1 + 1 1 + π  π               3 3 + π + − + π + = cos(2 ) cot   cot(2 ) 1 x x x x 2 = − Q6: Prove that: π  −   2 3 π  π          x       3 3 + π + − + π + cos cos(2 ) cot   ] cot x cos sin + cot(2 ) x x x x sin       =   π π π                       2 2 − − − − = + cos cos sin sin( ) x y x y x y [    4 4    4 π 4    = + sin cos x tan x π π π                sin cos    x x x x − − − − − Ans: LHS = cos cos sin sin x y x y = + sin cos x x 4    4 4 4 π π       − = −    Let Aand B x y 2 2 sin cos x x = = = 1 RHS (sin cos ) x x 4 4 sin cos x x Therefore, LHS = cosA cosB − sinA sinB = cos (A + B) Q10: Prove that: sin(n + 1)x sin (n + 2)x + cos(n + 1)x cos (n + 2)x = cos x Ans: LHS = sin (n + 1)x.sin(n + 2)x + cos (n + 1)x.cos (n + 2)x Applyingcos(C−D) = cosC.cosD + sinC.sinD LHS = { cos ( 1) ( 2) n x n + − + cos( ) cos RHS x x = − = = 3 cos 4   Ans: We know: ( ) cos A+ B − Therefore, 3 cos 4   3 2sin sin 4   π π π       − + x − = cos y 4 4 π π          =cos + − + ( ) x y } 4 π 4 x    =cos − + ( ) x y 2 = sin (x + y ) π π           3 + − − = − Q11: Prove that: cos 2 sin x x x 4 − = RHS ( ) 2sinA.sinB = − cos A B π         =    + tan x 2 π π           3 +      1 tan 1 tan − x x 4 π + − − LHS = cos x x Q7: Prove that: 4 − tan x π π           4 = − = − π − 2sin sin x x Ans: We know that 4 − tanA+ tanB 1 tanAtanB − tan A tanB 1 tanA tanB + 1 + = = − andtan(A B) tan(A B) = − = − × × 2sin sin x 2 sin x 4 2 = − = 2 sin RHS x http://www.fullmarkseducare.com/Ph:. 9911968787 NCERT SOLUTIONS

  5. Chapter-3 Trigonometric Functions Class– XI Q12: Prove that: sin2 6x – sin24x = sin 2x sin10x Ans: LHS. = sin2 6x – sin2 4x = (sin 6x + sin 4x)(sin 6x – sin 4x) 6 2sin =     6 2cos     = (2 sin 5x cos x)(2 cos 5x sin x) = (2 sin 5x cos 5x)(2 sin x cos x) = sin 10x sin 2x = RHS Q13: Prove that: cos2 2x – cos2 6x = sin 4x sin 8x Ans. LHS. = cos2 2x – cos2 6x = (cos 2x + cos 6x)(cos 2x – cos 6x) + −             9 5 9 5 x x x x 2sin − .sin 2 + 2 − =             17 3 17 3 x x x x  + −       2cos .sin                    4 6 4 x x x x × cos 2 2 2 + 2 − − 2sin7 .sin2 2cos10 .sin7 sin2 cos10 x x x x x x =   4 6 4 x x x x sin 2 2 = − = RHS + + + + 5 sin5 cos5 sin3 cos3 sin3 cos3 x + x x x x       x x x x x = Q17: Prove that: tan 4 x =sin5 cos5 Ans: LHS + − A B A B −                   3 5 3 x x + = cosA cos B 2cos cos , Applying: 2sin .cos 2 + 2 − 2 + 2 − = A B A B cosA cosB − = − 2sin sin 5 3 5 3 x x x x 2cos .cos 2 2 2 2 2 ( ) −      2 6 x x    + −    x +                2sin 4 sin( 2 ) x    2 6 2 6 6 x x x x x x = − sin 4 cos4 x x 2cos cos 2 sin 2sin4 .cos 2cos4 .cos tan4 x x x = x x = = 2 2 2 2 = [ ][ ] 2cos 4 cos( 2 ) x = [2cos 4x cos 2x] [–2sin 4x (–sin 2x)] = (2sin 4x cos 4x) (2sin 2x cos 2x) = sin 8x.sin 4x = RHS – – – x = Q18: Prove that: sin RHS Q14: Prove that: sin 2x + 2sin 4x + sin 6x = 4cos2x sin 4x Ans: LHS = sin 2x + 2 sin 4x + sin 6x = [sin 2x + sin 6x] + 2 sin 4x − + − sin cos sin cos x + x x − + y y x y = tan cos 2 =sin cos x x y y y Ans:LHS −       x                   x y 2cos .sin − A+B 2 x A B 2 6 x sin A sinB + = Applying: 2 + 2 − 2sin cos = x y x y + − 2cos .cos 2 6 2 x x = + 2sin cos 2sin4 x 2 y 2 2 2 −             = 2 sin 4x cos (–2x) + 2 sin 4x = 2 sin 4x (cos2x + 1) = 2 sin 4x (2cos2x) = 4cos2x sin 4x = RHS sin − Q15: Prove that: Ans: LHS       2 − x y = = = tan RHS 2 x y cos 2 Q19: Prove that sin − + sin3 cos3 sin3 cos3 + x +    +    x x x x x x cot 4x (sin 5x + sin 3x) = cot x (sin5x – sin 3x) = cot 4x (sin 5x + sin 3x) cos4 5 3 2sin sin4 2   x cos 4 2sin4 cos sin4 x = 2 cos 4x cos x = cot x (sin 5x – sin 3x) cot 5 3 2cos sin 2    x [ ] 2cos4 sin sin x = 2 cos 4x.cos x LHS = RHS = tan 2 x cos   + −               5 3 x x x x x − = sin x x x x       = cos Ans: LHS 2 cos x −             3 3 x x x = x x 2sin cos 2 2 − = RHS 3 3 x x x x 2cos cos 2 2   + −              5 3 x x x x x sin 2 cos2 = sin = = tan 2 = RHS x 2 cos x = x x − sin sin sin sin sin3 cos − sin3 cos − x +    x x x x x = Q20: Prove that: 2sin x 2 2 x Q16 : Prove that: cos9 − x Ans: LHS = x    2 2 − cos5 sin3 − cos5 sin3 − x sin2 cos10 x x x x = − x −       3 3 x x x 2cos sin sin17 x 2 − 2 = − Ans: LHS =cos9 x x x cos2 − x x x sin17 2cos2 sin( cos2 − ) x = http://www.fullmarkseducare.com/Ph:. 9911968787 NCERT SOLUTIONS

  6. Chapter-3 Trigonometric Functions Class– XI 2 ( sin ) = − × − = = 2sin x x RHS − 2 4tan (1 tan (1 tan − ) x x = − 2 2 2 ) 4tan − x x + + + + + + + x cos4 sin4 cos3 sin3 cos3 sin3 + cos2 ) cos3 sin2 ) sin3 + x x cos2 sin 2 cos2 sin2 + + x x x x x x x x x x x x = Q21: Prove that: 2 cot3 x 4tan (1 tan 1 tan + − x 4tan (1 tan 1 6tan − x ) x x = x − 4 2 2 2tan 4tan x ) x = cos4 sin4 = (cos4 (sin4 − 2 x x Ans: LHS = + 2 4 tan + x x x Q24: Prove that cos 4x = 1 – 8sin2x cos2x Ans: LHS = cos 4x = cos 2(2x) = 1 – 2 sin22x = 1 – 2(2 sin x cos x)2[sin2A = 2sin A cos A] = 1 – 8sin2x cos2x = RHS Q25: Prove that cos6x = 32cos6x – 48cos4x + 18cos2x – 1 Ans: LHS. = cos 6x = cos 3(2x) = 4 cos3 2x – 3 cos 2x [cos 3A = 4cos3A – 3cosA] = 4[(2 cos2x – 1)3 – 3 (2 cos2 x – 1) = 4[(2 cos2x)3– 13– 3(2 cos2x)2 + 3(2 cos2x)] – = 4[8cos6x –1 – 12cos4x + 6 cos2x] – 6cos2x + 3 = 32cos6x – 4 – 48cos4x + 24cos2x – 6cos2x + 3 = 32cos6x – 48cos4x + 18cos2x – 1 = RHS. Exercise 3.4 Q1: Find the principal and the general solutions of the equation tan 3 x = Ans: Given: tan 3 x = We know that π tan 3  Applying: + −       x             x x             A B A B [cos 2A = 1 – 2 sin2 A] cosA cosB + = 2cos cos , 2 + 2 − A B A B sinA sinB + = 2sin cos 2 2 +   +   + −             cos3 sin3 1) 1) + 4 2 4 2 x x x 2cos cos cos3 x 2 + 2 − = LHS 4 2 4 2 x x x x 2sin cos sin3 x 2 2 + + 2cos3 cos 2sin3 cos =cos3 (2cos sin3 (2cos x x x x = [cos 2x = 2 cos2x – 1] + x x x x = = cot3 RHS x 6cos2x + 3 Q22: Prove that: cot x.cot 2x – cot2x cot 3x – cot 3x cot x = 1 Ans: LHS = cot x cot 2x – cot 2x cot 3x – cot 3x cot x = cot x cot 2x – cot 3x (cot 2x + cot x) = cot x cot 2x – cot (2x + x) (cot 2x + cot x)  = −      = cot x cot 2x – (cot 2x cot x – 1) = 1 = RHS 4tan (1 tan tan4 1 6tan − 2tanA tan2A 1 tan A − LHS. = tan 4x = tan 2(2x) 2tan 2 1 tan 2 − 2tan 2 1 tan 2tan 1 1 tan −  x 4tan 1 tan 4tan 1 (1 tan x −  4tan 1 tan (1 tan ) (1 tan −  −    cot2 cot cot x 1 x x + cot cot2 x (cot2 cot ) x x x + cot2 x cotA cotB 1 cotA +    − + = cot(A B) π π      4 cotB + = = π 3= tan 3 tan 3 and 3 πand4 π Therefore, the principal solutions are x =3 π 3 − x 2 ) x x = Q23: Prove that x x Now, tan x = tan 3 Therefore, the general solution is + 2 4 tan = Ans: We know: 2 π π = + ∈ where x n n Z 3 x Q2: Find the principal and general solutions of the equation sec 2 x = Ans: Given: sec 2 x = We know that 5 sec 2 and sec 3 3 = x 2              x − 2 x = 2 x π π π π −     = = = = π se c 2   – sec 2 2 3 3    x πand5 π Therefore, the principal solutions are x = 3 − 2 x = 3      2 x π − Now, sec x = sec3 2 2 ) x       π π ⇒ = ± ∈ 2 where Z x n n − 2 x 3 = Therefore, the general solution is π π , where n∈ Z      − − 2 2 2 4tan ) x x x 2 2 = ± 2 x n 3 http://www.fullmarkseducare.com/Ph:. 9911968787 NCERT SOLUTIONS

  7. Chapter-3 Trigonometric Functions Class– XI Q3: Find the principal and general solutions of the equation cot 3 x = − Ans: Given: cot 3 x = − π + −    x    cos2 − 1) − or       3 3 x x x x ⇒ − − = 2cos cos cos2 0 x 2 2 0 ⇒ ⇒ ⇒ = 2cos2 cos cos2 (2cos x cos2 x x x x 0 = 6= π We know thatcot 3 = − = 1 2 0 2cos 1 0 x − = − cot6 3 ⇒ = = cos2 0 cos x or x π       π π π − = − cot 3 ∴ = + = ∈ 2 (2 1) cos cos ,where Z x n or x n 6 2 3 π π π       11 π = − = − − ⇒ = + Also, cot 2 cot 3 (2 1)4 π x n 6 6 π π 5 11 π = ± ∈ Or 2 ,where Z x n n = − = − Hence, cot 3 and cot 3 3 6 6 Q7: Find the general solution of the equation sin 2x + cos x = 0 Ans: Given: sin 2x + cos x = 0 2sin cos x ⇒ cos (2sin x ⇒ cos x ⇒ = Now, cos = x πand11 π Therefore, the principal solutions are x =5 6 6 π Now, cot x = cot5 6 π + = cos 1) + 0 x x x       5 1 = or 0 ⇒ = = tan tan cot x x 6 5 tan x + = 0 0 2sin 1 0 x π π ⇒ = ∈ + where Z x n n 6 π ⇒ = + ∈ cos (2 1) ,where Z x n n Therefore, the general solution is 2 π 5 x+ = 2sin 1 0 1 2 π = ∈ + where Z x n n 6 − π π       ⇒ = = − = π + sin sin sin x Q4: Find the general solution of cosec x = – 2 Ans: Given: cosec x = –2 π= – 2 6 6 π π       7 = π + = sin sin 6 6 We know: cosec 6 π 27 π ⇒ = ( 1) + − ∈ ,where Z x n n π π          7 6 ∴ π + = − = − cosec cosec 2 Therefore, the general solution is 6 π 6 11 π π 7 π    π + ( 1) + − ∈ n (2 1) n or n n Z π − = − = − Also, cosec 2 cosec 2 2 6 6 6 Q8: Find the general solution of the equation sec2 2x = 1− tan 2x Ans: sec2 2x = 1− tan 2x 1 tan 2 ⇒ + tan 2 ⇒ tan 2 (tan 2 x ⇒ tan 2 x ⇒ Now, tan2 = x 2 ⇒ = x n n x πand11 π. Therefore, the principal solutions are x =7 6 6 π Now, cosec x = cosec7 6 7 6 1 tan 2 = − tan 2 x 1) x + 0 0 0, where + π π 2 x x π + = 2 0 x ⇒ = sin sin x = or 0 Hence, the general solution is = + = tan 2 1 0 x π 7 = ( 1) + − π ∈ n ,where Z x n n 6 ∈ . n Z Q5: Find the general solution of the equation cos 4x = cos 2x Ans: Given: cos 4x = cos2x cos4 cos2 x x ⇒ − 4 2 2sin 2  sin3 sin 0 x x ⇒ = sin3 0 x ⇒ = n x or ⇒ = ∈ where n Z 2 x =    0 ⇒ + = tan 2 1 0 π + −         4 2 x x x x π π       3 ⇒ − = sin 0 ⇒ = − =− = − = π tan2 1 tan tan tan x 2 4 4 4 π 3 π ⇒ = + ∈ 2 , where Z x n n = sin 0 or x 4 π π 3 n π ⇒ = + ∈ x where n Z , where π ∴ = = ∈ x n n Z 2 8 3 Therefore, the general solution is n n or Q6: Find the general solution of the equation cos 3x + cos x − cos 2x = 0 Ans: Given: cos 3x +cos x − cos 2x = 0 π π π 3 + ∈ , n Z 2 2 8 . http://www.fullmarkseducare.com/Ph:. 9911968787 NCERT SOLUTIONS

  8. Chapter-3 Trigonometric Functions Class– XI Q9: Find the general solution of the equation sin x + sin 3x + sin 5x = 0 Ans: Given: sin x + sin 3x + sin 5x = 0 Or sin ( x +   ⇒     2sin3 cos2 x ⇒ sin3 (2cos2 x ⇒ sin3 x ⇒ = Now, sin3 0 = x 3 OR ⇒ = x n π + − −          x y x y x y + 2 2 2 = 4cos cos sin 2 + 2 2    x y + = sin5 ) sin x + 3 0 x = = 2 4cos RHS 2 −             5 5 x x x x + = Q4: Prove that 2sin cos sin3 0 x 2 2 = + −       sin3 1) + or 0 x x x 0 x y − + − = 2 2 2 (cos cos ) (sin sin ) 4sin x y x y = 2 − + − 2 2 Ans: LHS = + = (cos cos ) (sin sin ) x y x y 0 2cos2 1 0 x 2 2 + − + −             x y x y x y x y = 2sin − + sin 2cos sin π n 2 y 2 − 2 2 y = ∈ where Z x n 3 + + − x x y x y x + = 2 2 2 2 4sin sin 4cos sin + = Also, 2cos2 1 − 0 x 2 − 2 x 2 + 2 π π       1 +    y    x y y x y ⇒ = = − = π − cos2 cos cos x = + 2 2 2 4sin sin cos 2 3 3 2 x 2 2 π 2 −       ⇒ = cos2 cos x = = 2 4sin RHS 3 2 Q5: Prove that: sin Ans: LHS = sin π 2 π ⇒ = ± ∈ 2 2 , where Z x n n 3 + + + = 4cos cos2 sin 4 x sin7 x sin7 x 5 2 3 7 cos 2  2sin5 cos( 2 ) − x 2sin5 cos2 x x ] sin5 x + 3 5 .cos 2   ] 2sin 4 .cos( ) x x − 4cos2 sin 4 cos x x x = sin3 sin5 sin3 sin5 5 2 sin7 sin5 sin3 π x x + +    x + +    x + + − x x π ⇒ = ± ∈ , x n where n Z x x x x x x 3 = sin nπor n π π ± ∈ Therefore, the general solution is , n Z +       x x x x 3 3 = + 2sin .cos + − MISCELLANEOUS 9 2cos cos 13 13 2cosAcosB = cos A + B + cos (A B)            3 7 x x x x 2sin 2 π π π π 3 13 5 13 + + = Q1: Prove that: cos cos 0 = = = − + x 2sin3 cos( 2 ) x 2sin3 cos2 x [ 2cos2 x x x + x ( ) − Ans: We know, Therefore, sin3 x π π π π + −              9 13    8 13  +   3 13 π 5 13  +   5 13 3 5 x x x x + + LHS = 2cos cos cos cos 2cos2 2sin =   [ 13 π 2 π π π π       9 13 9 13 3 13 5 13 2cos2 x Q6: Prove that (sin7 = = = + + − + cos cos cos cos 13 π 13 3 13 RHS π π π 10 13    = + + + cos cos cos cos + + + + + x sin5 ) (sin9 cos5 ) (cos9 x + sin5 ) cos5 ) + x −   +     −   +     sin3 ) cos3 ) sin3 ) cos3 ) + x + x x x x x x = tan6 x π π π π       3 13 5 13 3 13 3 13 π = 0 = RHS 5 13 (cos7 x = − − + + π π cos cos cos cos + + + (sin7 (cos7 (sin9 (cos9 + x x x x Ans: LHS = π π π 3 13 5 13 5 13 x = − − + + cos cos cos cos       +       −                                                 7 5 7 5 9 3 9 3 x x x x x x x x 2sin .cos 2sin .cos Q2: Prove that (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0 Ans: LHS. = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = sin 3x sin x + sin 2x + cos 3x cos x − cos2x = cos 3x cos x + sin 3x sin x − (cos2x− sin2x) = cos (3x − x) − cos2x = cos 2x − cos 2x = 0 = RH.S. Q3: Prove that: 2 + 2 2 + 2 − = 7 5 7 5 9 3 9 3 x x x x x x x x 2cos .cos 2cos .cos 2 2 2 2 [ [ 2sin6 2cos6 ] [ ] [ + cos3 cos3 + ] ] + 2sin6 .cos 2cos6 .cos 2sin6 .cos3 2cos6 .cos3 x ] ] x x x [ x x x x x = + cos cos x x x x x = = tan 6x = R.H.S. [ + x y + + − = 2 2 2 (cos cos ) (sin sin ) 4cos x y x y 2 3 x x + − = Q7: Prove that sin3 sin2 sin 4sin cos cos x x x x + + − 2 2 Ans: LHS. = (cos cos ) (sin sin ) x y x y + 2 2 2 2 + − − + − 2             Ans: LHS x y x y x y x y = sin3 (sin 2  +  sin ) + x x x x x = + 2cos cos 2cos sin 2 + 2 2 y 2 − −    2 x x sin3 2cos sin x = − + x y x y x x y 2 2 + 2 2 2 2 4cos cos 4cos sin = 2 2 2 2 http://www.fullmarkseducare.com/Ph:. 9911968787 NCERT SOLUTIONS

  9. Chapter-3 Trigonometric Functions Class– XI 3 x x       1 3 x+ − − sin3 2cos sin = 1 2 3 x 2 x 2 ⇒ = = 2 sin 3 3 3 x x x 2 2 + 2sin .cos 2cos sin = 2 3 2 2 3 2 2         2 2 3 6 x ⇒ = = sin2          x x x 3 = + 2cos sin sin 2      = 1 3 + − 1                      3 3 x x x x 1 cos + x x= 1 3 + − = 2 cos Now, 3 x 2 2 2 2 2 2 2    = 2cos 2sin cos 2 2 2    1 x x ⇒ = − cos cos is negative 2 2 3 3 x x = 2cos .2sin cos x − 3 = 2 2 3 3 x x = = 4sin cos RHS cos x         6 2 2 x sin 3 x 2 x = = = − tan 2         2 3 cos2 − x x xif tan x = 4 3 − Q8: Findsin , cos and tan , and x liesin 3 2 2 2 quadrant II x x xare 6, 3 Thus, the respective values of sin , cos , tan2 π 4 3 x 2 2 π < < − Ans: Given: tan x = and 2 π < x − 3 − and 2 π 3 ⇒ < 4 2 x 2 x x are all positive. 1 4 x x x Therefore, sin ,cos and tan = Q10: Find sin , cos and tan for sin , x in x 2 2 2    3 5 2 2 2 quadrant II. 2 −   4 =25 1 tan = + = + 2 2 sec 1 x x Now, 3 9 1 4 andx lies in the 2nd quadrant. = Ans. Given: sin x 9 25 ∴ = ⇒ = ± 2 cos cos x x π π π x π < < ⇒ < < . ., ie x x Now,cos x = 2cos2 2 4 2 2 2− 1 x x xwill be all positive. Therefore,sin ,cos , tan 3 5 5 5 2 5 x 2 2 2 = − = 2 ⇒ 2cos 1 2 1 1 4 x = sin As x ⇒ = = cos2 5 2       1 4 1 15 16 ∴ 1 sin = − = − = − = 2 2 cos 1 1 x x 2       16 1 x + = Now, 2 sin 1 2 x= − 5 1 5 15 4 ⇒ = − [ cos x is negative in quadrant II] cos x 4 5     = 2 sin 1     2 15 4 − − 1 2 x 1 cos − + 4 15 x x = sin2 = = = 2 sin 5 2 2 2 8          2 x + 4 15       x x sin ⇒ = sin2 sin ispositive x 5  2 x = = = 2 tan2 8 2 1 cos2   + 4 15 2 2 5 = × 8 x x xfor 1 3 + 8 2 15 4 x = − Q9: Findsin , cos and tan , if x liesin the cos = 2 2 2 quadrant III         15 4 + − π π π 1 3 3 x 1, and 3 xare negative, and sin2 1 cos 2 π < < ⇒ < < − Ans: Given: cos x = x 1 cos + − 4 15 x x 2 2 2 4 = = = 2 Now, cos 2 2 2 8 x x is positive. Hencecos ,tan − 4 15       x x 2 2 ⇒ = cos2 cos is positive  − x x 8 2 ⇒ = 2 sin 2 http://www.fullmarkseducare.com/Ph:. 9911968787 NCERT SOLUTIONS

  10. Chapter-3 Trigonometric Functions Class– XI − 4 15 2 2 = × 8 − 8 2 15 4 + =                 8 2 15 4 x sin 8 2 15 + x 2 x = = = tan2 8 2 15 − − 8 2 15 4 cos2 Upon, simplification x 4 = + tan2 15 x x x Thus, the respective values of sin ,cos and tan 2 2 2 + − 8 2 15 4 8 2 15 4 + are , and4 15 http://www.fullmarkseducare.com/Ph:. 9911968787 NCERT SOLUTIONS

More Related