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Trigonometric Ratios(Qns.& Ans) V.S.A.Q.’S

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These solutions designed by the ‘Basics in Maths‘ team.These notes to do help the intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in I.P.E examinations.


Trigonometric Ratios Up to Transformations

 Question 1

Find the value of sin2(π/10) + sin2(4π/10) + sin2(6π/10) + sin2(9π/10)

Sol:

 sin2(π/10) + sin2(4π/10) + sin2(6π/10) + sin2(9π/10)

= sin2(π/10) + sin2(π/2 – π/10) + sin2(π/2+ π/10) + sin2(π – π/10)

= sin2(π/10) + cos2(π/10) + cos2(π/10) + sin2(π/10)

= 1 + 1 = 2

 Question 2

If sin θ = 4/5 and θ not in the first quadrant, find the value of cos θ

Sol:

Given sin θ = 4/5 and θ not in the first quadrant

⇒ θ in the second quadrant

⇒ cos θ < 0

    cos2θ = 1 – sin2 θ

              =1 – (4/5)2

             = 1 – 16/25

∴cos θ   = – 3/5 (∵cos θ < 0)

 Question 3

If 3sin θ + 4 cos θ = 5, then find the value of 4 sin θ – 3cos θ

Sol:

Given, 3sin θ + 4 cos θ = 5

let 4 sin θ – 3cos θ = x

  (3sin θ + 4 cos θ )2 + (4 sin θ – 3cos θ)2 = 52 + x2

 9 sin2 θ + 16 cos2 θ + 12 sin θ cos θ + 16 sin2 θ + 9 cos2 θ – 12sin θ cis θ = 25 + x2

25 sin2 θ + 25 cos2 θ = 25 + x2

25 = 25 + x2

⇒ x2 = 0

 x = 0

∴ 4 sin θ – 3cos θ = 0

 Question 4

If sec θ + tan θ =Trigonometry up to Transformations 1, find the value of sin θ and determine the quadrant in which θ lies

Sol:

Given, sec θ + tan θ =  ———— (1)

 We know that sec2 θ – tan2 θ = 1

⇒ (sec θ + tan θ) (sec θ – tan θ) = 1

 sec θ – tan θ =Trigonometry up to Transformations 2

⇒ sec θ – tan θ = Trigonometry up to Transformations 3———— (2)

 (1) + (2)

⇒ (sec θ + tan θ) + (sec θ – tan θ) = Trigonometry up to Transformations 4

2sec θ =Trigonometry up to Transformations 5

 sec θ =Trigonometry up to Transformations 6

(1) – (2)

⇒ (sec θ + tan θ) – (sec θ – tan θ) = Trigonometry up to Transformations 7

2 tan θ = Trigonometry up to Transformations 8 ⇒ tan θ =Trigonometry up to Transformations 9

Now sin θ = tan θ ÷ sec θ =Trigonometry up to Transformatio

     Sin θ =Trigonometry up to Transformations 11

Since sec θ positive and tan θ is negative θ lies in the 4th quadrant.

 Question 5

Prove that cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16) = 1

Sol:

cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (4π/16). cot (5π/16) cot (6π/16) cot (7π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). cot (π/2 – 3π/16) cot (π/2 – 2π/16) cot (π/2 – π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). tan (3π/16) tan (2π/16) tan (π/16)

= [cot (π/16). tan (π/16)] [cot (2π/16). tan (2π/16)] [cot (3π/16). tan (3π/16]. cot (π/4)

= 1.1.1.1

 =1

 Question 6

If cos θ + sin θ = Trigonometry up to Transformations 12cos θ, then prove that cos θ – sin θ =  sin θ

Sol:

Given, cos θ + sin θ = Trigonometry up to Transformations 12cos θ

Sin θ = Trigonometry up to Transformations 12 cos θ – cos θ

           = ( Trigonometry up to Transformations 12 – 1) cos θ

( Trigonometry up to Transformations 12 + 1) sin θ = ( Trigonometry up to Transformations 12 + 1) ( Trigonometry up to Transformations 12 – 1) cos θ

Trigonometry up to Transformations 12 sin θ + sin θ = cos θ

∴ cos θ – sin θ = Trigonometry up to Transformations 12 sin θ

 Question 7

Find the value of 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ)

Sol:

2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ)

= 2[(sin2 θ)3 + (cos2 θ)3] – 3[(sin2 θ)2 + (cos2)2

= 2[(sin2 θ + cos2 θ)3 – 3 sin2 θ cos2 θ (sin2 θ + cos2 θ)] – 3[(sin2 θ + cos2 θ)2 – 2 sin2 θ cos2 θ]

= 2[1 – 3 sin2 θ cos2 θ] – 3 [1 – 2 sin2 θ cos2 θ]

= 2 – 6 sin2 θ cos2 θ – 3 + 6 sin2 θ cos2 θ

= – 1

 Question 8

If tan 200 = λ, then show thatTrigonometry up to Transformations 13    

Sol:

Given tan 200 = λ

Trigonometry up to Transformations 19 =Trigonometry up to Transformations 14

                               =Trigonometry up to Transformations 15

                              =Trigonometry up to Transformations 16

                              =Trigonometry up to Transformations 18

 Question 9

If sin α + cosec α = 2, find the value of sinn α + cosecn α, n∈ Z

Sol:

Given sin α + cosec α = 2

 ⇒ sin α + 1/ sin α = 2

 ⇒Trigonometry up to Transformations 17  = 2

      sin2 α + 1= 2 sin α

       sin2 α – 2 sin α + 1= 0

  (sin α – 1 )2 = 0

⇒ sin α – 1 = 0

sin α = 1 ⇒ cosec α = 1

 sinn α + cosecn α = (1)n + (1)n =1 + 1 =2

∴ sinn α + cosecn α = 2

 Question 10

Evaluate sin2 Trigonometry up to Transformations 20+ cos2 Trigonometry up to Transformations 21  – tan2   Trigonometry up to Transformations 22

Sol:

 Trigonometry up to Transformations 23

Trigonometry up to Transformations 24

 Question 11

Find the value of sin 3300. cos 1200 + cos 2100. Sin 3000

Sol:

 sin 3300. cos 1200 + cos 2100. Sin 3000

=sin (3600 – 300). cos (1800 – 600) + cos (1800 + 300). sin (3600 – 600)

= (– sin 300). (– cos 600) + (– cos300). (– sin600)

= sin 300.  cos 600 + cos300.  Sin600

= sin (600 + 300) = sin 900

=1

 Question 12

Prove that cos4 α + 2 cos2 α Trigonometry up to Transformations 25= (1 – sin4 α)

Sol:

cos4 α + 2 cos2 α Trigonometry up to Transformations 25

= cos4 α + 2 cos2 α (1 – cos2 α)

= (cos2 α)2 + 2 (1 – sin2 α) (sin2 α)

= (1 – sin2 α)2 + 2 sin2 α – 2sin4 α

= 1 + sin4 α – 2 sin2 α + 2 sin2 α – 2sin4 α

= 1 – sin4 α

 Question 13

Eliminate θ from x = a cos3 θ and y = b sin3 θ

Sol:

Given x = a cos3 θ and y = b sin3 θ

 cos3 θ = x/a and sin3 θ = y/b

 cos θ = (x/a)1/3 and sin θ = (y/b)1/3

we know that sin2 θ + cos2 θ = 1

 ⇒ [(y/b)1/3]2 + [(x/a)1/3]2 = 1

   (x/a)2/3 + (y/b)2/3 = 1

 Question 14

Find the period of the following functions

Sol:

(i) f(x) = tan 5x

we know that period of tan kx =Trigonometry up to Transformations 26

⇒ period of tan 5x =Trigonometry up to Transformations 27

(ii) f(x) =Trigonometry up to Transformations 28

we know that period of Trigonometry up to Transformations 29= Trigonometry up to Transformations 31

period of Trigonometry up to Transformations 28 = Trigonometry up to Transformations 32

                                      =Trigonometry up to Transformations 33

(iii) f(x) = 2 sinTrigonometry up to Transformations 34 + 2 cosTrigonometry up to Transformations 35

period of sin Trigonometry up to Transformations 34 = Trigonometry up to Transformations 36 = 8

period of cos Trigonometry up to Transformations 35 = Trigonometry up to Transformations 37 = 6

period of given function is = LCM (8, 6) = 24

 

(iv) f(x) = tan (x + 4x + 9x +…. + n2x)

f(x) = tan (x + 4x + 9x +…. + n2x)

       = tan (1 + 4 + 9 + … + n2) x

= tanTrigonometry up to Transformations 38x

we know that period of tan kx =Trigonometry up to Transformations 26

Period of tan  =Trigonometry up to Transformations 39

                        = Trigonometry up to Transformations 40

 Question 15

Prove that sin2(52 ½)0 – sin2 (22 ½)0 =Trigonometry up to Transformations 41

Sol:

We know that sin2 A – sin2B = sin (A +B) sin (A – B)

 ⇒ sin2(52 ½)0 – sin2 (22 ½)0

= sin (52 ½+ 22 ½) sin (52 ½ – 22 ½)

 = sin 750 sin 300

 = Trigonometry up to Transformations 42      

∴ sin2(52 ½)0 – sin2 (22 ½)0 =Trigonometry up to Transformations 41

 Question 16

Prove that tan 700 – tan200 = 2 tan 500

Sol:

500 = 700 – 200

Tan 500 = tan (700 – 200)

We know that tan (A –B) =Trigonometry up to Transformations 43

  ⇒ Tan 500 =Trigonometry up to Transformations 44

 ⇒ tan 700 – tan 200 = tan 500 (1 + tan700 tan 200)

     tan 700 – tan 200 = tan 500 [1 + tan700 cot (900 – 200)]

     tan 700 – tan 200 = tan 500 [1 + tan700 cot 700]

     tan 700 – tan 200 = tan 500 [1 + 1]

∴ tan 700 – tan200 = 2 tan 500

 Question 17 

If sin α = Trigonometry up to Transformations 45, sin β =Trigonometry up to Transformations 46  and α, β are acute, show that α + β =Trigonometry up to Transformations 47

Sol:

Given sin α =    Trigonometry up to Transformations 45                                           sin β =Trigonometry up to Transformations 46

Trigonometry up to Transformations 48

 tan α = 1/3                                                          tan β = ½

tan (α + β) =Trigonometry up to Transformations 49

      Trigonometry up to Transformations 50 

  tan (α + β) = 1

∴ α + β =Trigonometry up to Transformations 47

 Question 18

Find tanTrigonometry up to Transformations 51 in terms of tan A

Sol:

 tan Trigonometry up to Transformations 51 =Trigonometry up to Transformations 52

                        =Trigonometry up to Transformations 53

 Question 19

Prove thatTrigonometry up to Transformations 54 = cot 360

Sol:

Trigonometry up to Transformations 54  = Trigonometry up to Transformations 55

(on dividing numerator and denominator by cos 90)

      = Trigonometry up to Transformations 56

   = tan (450 + 90)

    = tan 540

 = tan (900 – 360)

 = cot 360

 ∴Trigonometry up to Transformations 54  = cot 360      

 Question 20

Show that cos 420 + cos 780 + cos 1620 = 0

Sol:

cos 420 + cos 780 + cos 1620

= cos (600 – 180) + cos (600 + 180) + cos (1800 – 180)

=cos 600 cos180 + sin 600 sin 180 + cos 600 cos 180 – sin 600 sin 180 – cos 180

= 2 cos 600 cos 180 – cos 180

= 2 (1/2) cos 180 – cos 180

 = cos 180 – cos 180

= 0

 Question 21

Express Trigonometry up to Transformations 57sin θ + cos θ as a single of an angle

Sol:

Trigonometry up to Transformations 57sin θ + cos θ = 2(Trigonometry up to Transformations 58  sin θ + Trigonometry up to Transformations 59cos θ)

                                = 2(cos 300 sin θ + sin 300 cos θ)

                                = 2 sin (θ + 300)

 Question 22

Find the maximum and minimum value of the following functions

(i) 3 sin x –4 cos x

a= 3, b = –4 and c = 0

Trigonometry up to Transformations 60  

                                     = 5

∴ minimum value = –5 and maximum value = 5

(ii) cos (x + ) + 2  sin (x + ) – 3

a= 1, b = 2  and c = – 3

Trigonometry up to Transformations 61

 ∴ minimum value = –6 and maximum value = 0

 Question 23

Find the range of the function f(x) = 7 cos x – 24sin x + 5

Sol:

Given f(x) = 7 cos x – 24sin x + 5

a= 7, b = –24 and c = 5

Trigonometry up to Transformations 62

∴ Range = [–20, 30]    

 Question 24

Prove that sin2α + cos2 (α + β) + 2 sin α sin β cos (α + β) is independent of α

Sol:

sin2α + cos2 (α + β) + 2 sin α sin β cos (α + β)

= sin2α + cos (α + β) [ cos (α + β) +2 sin α sin β]

= sin2α + cos (α + β) [ cos α cos β – sin α sin β +2 sin α sin β]

=sin2α + cos (α + β) [ cos α cos β + sin α sin β]

=sin2α + cos (α + β) cos (α –β)

= sin2 α + cos2 α – sin2 β

=1 – sin2 β

= cos2 β

 Question 25

Simplify Trigonometry up to Transformations 63

Sol:

Trigonometry up to Transformations 63   =Trigonometry up to Transformations 64

                 =Trigonometry up to Transformations 65

                 = tan θ

Question 26

For what values of x in the first quadrantTrigonometry up to Transformations 66 is positive?

Sol:

Trigonometry up to Transformations 66 > 0 ⟹ tan 2x > 0

⟹ 0 < 2x < π/2 (∵ x is in first quadrant)

⟹ 0 < x < π/4

Question 27

If cos θ = Trigonometry up to Transformations 67 and π < θ < 3π/2, find the value of tan θ/2.

Sol:

cos θ = Trigonometry up to Transformations 67

π < θ < 3π/2 ⟹ π/2 < θ/2 < 3π/4

tan θ/2 < 0

tan θ/2 =Trigonometry up to Transformations 68

               =– Trigonometry up to Transformations 69 (tan θ/2 < 0)

              =–Trigonometry up to Transformations 70

           = – 2

Question 28

If A is not an integral multiple of π/2, prove that cot A – tan A = 2 cot 2A.

Sol:

cot A – tan A = Trigonometry up to Transformations 71

                         =Trigonometry up to Transformations 72

                          =Trigonometry up to Transformations 73

                          =Trigonometry up to Transformations 74

                           =Trigonometry up to Transformations 75

                           = 2 cot 2A

Question 29

Evaluate 6 sin 200 – 8sin3 200

Sol:

6 sin 200 – 8sin3 200 = 2 (3 sin 200 – 4sin3 200)

                                       = 2 sin 3(200)

                              = 2 sin 600

                              = 2Trigonometry up to Transformations 76 

                              =Trigonometry up to Transformations 77

Question 30

Express cos6 A + sin6 A in terms of sin 2A.

Sol:

cos6 A + sin6 A

= (sin2 A)3 + (cos2 A)3

= (sin2 A + cos2 A)3 – 3 sin2 A cos2 A (sin2 A + cos2 A)

= 1 – 3 sin2 A cos2 A

=1 – ¾ (4 sin2 A cos2 A)

 = 1 – ¾ sin22 A

Question 31

If 0 < θ < π/8, show that Trigonometry up to Transformations 79  = 2 cos (θ/2)

Sol:

Trigonometry up to Transformations 79

Trigonometry up to Transformations 80

Trigonometry up to Transformations 81

 =2 cos (θ/2)

Question 32

Find the extreme values of cos 2x + cos2x

Sol:

cos 2x + cos2x = 2cos2 x– 1 + cos2 x

                              =3cos2 x – 1

We know that – 1 ≤ cos x ≤ 1

 ⟹ 0 ≤ cos2 x ≤ 1

      3×0 ≤ 3×cos2 x ≤ 3×1

      0– 1 ≤3 cos2 x – 1≤ 3– 1

   – 1≤3 cos2 x – 1≤ 2

     Minimum value = – 1

     Maximum value = 2

Question 33

Prove that Trigonometry up to Transformations 82 = 4

Sol:

Trigonometry up to Transformations 82

Trigonometry up to Transformations 83

= 4

Question 34

Prove that sin 780 + cos 1320 =Trigonometry up to Transformations 84

Sol:

 sin 780 + cos 1320 = sin 780 + cos (900 + 420)

                                      = sin 780 – sin 420

                                      = 2 cosTrigonometry up to Transformations 85  sinTrigonometry up to Transformations 86

                                     = 2 cos 600 sin 180

                                      = 2Trigonometry up to Transformations 87

                                      =Trigonometry up to Transformations 84

Question 35

Find the value of sin 340 + cos 640 – cos40

Sol:

sin 340 + cos 640 – cos40

= sin 340 –2 sinTrigonometry up to Transformations 88  sinTrigonometry up to Transformations 89

= sin 340 – 2sin 340 sin 300

= sin 340 – 2 sin 340 (1/2)

=sin 340 – sin 340

=0

Question 36

Prove that 4(cos 660 + sin 840) =Trigonometry up to Transformations 90

Sol:

4(cos 660 + sin 840)  

=4(cos 660 + sin (900 – 60)  

=4(cos 660 + cos (60)  

= 4[ 2 cosTrigonometry up to Transformations 91  cos Trigonometry up to Transformations 92]

=4[ 2 cosTrigonometry up to Transformations 93  cosTrigonometry up to Transformations 94 ]

=8 cos 360 cos 300

= 8Trigonometry up to Transformations 95

=Trigonometry up to Transformations 90

Question 35

Prove that (tan θ + cot θ)2 = sec2 θ + cosec2 θ = sec2 θ. cosec2 θ

Sol:

tan θ + cot θ =    Trigonometry up to Transformations 96

                         =Trigonometry up to Transformations 97

                         =Trigonometry up to Transformations 98

                         = sec θ. cosec θ

(tan θ + cot θ)2   = sec2 θ. cosec2 θ

sec2 θ + cosec2 θ =Trigonometry up to Transformations 99

                               Trigonometry up to Transformations 100

                              = sec2 θ. cosec2 θ


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