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

Trigonometric Ratios(Qns.& Ans) V.S.A.Q.’S designed by the ‘Basics in Maths‘ team.These notes to do help 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 =, 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 =

⇒ sec θ – tan θ = ———— (2)

(1) + (2)

⇒ (sec θ + tan θ) + (sec θ – tan θ) =

2sec θ =

sec θ =

(1) – (2)

⇒ (sec θ + tan θ) – (sec θ – tan θ) =

2 tan θ =  ⇒ tan θ =

Now sin θ = tan θ ÷ sec θ =

Sin θ =

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 = cos θ, then prove that cos θ – sin θ =  sin θ

Sol:

Given, cos θ + sin θ = cos θ

Sin θ =  cos θ – cos θ

= (  – 1) cos θ

(  + 1) sin θ = (  + 1) (  – 1) cos θ

sin θ + sin θ = cos θ

∴ cos θ – sin θ =  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 that

Sol:

Given tan 200 = λ

=

=

=

=

###### 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

⇒  = 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 + cos2   – tan2

Sol:

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 = (1 – sin4 α)

Sol:

cos4 α + 2 cos2 α

= 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 =

⇒ period of tan 5x =

(ii) f(x) =

we know that period of =

period of =

=

(iii) f(x) = 2 sin + 2 cos

period of sin  =  = 8

period of cos  =  = 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

= tanx

we know that period of tan kx =

Period of tan  =

=

Question 15

Prove that sin2(52 ½)0 – sin2 (22 ½)0 =

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

=

∴ sin2(52 ½)0 – sin2 (22 ½)0 =

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) =

⇒ Tan 500 =

⇒ 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 = , sin β =  and α, β are acute, show that α + β =

Sol:

Given sin α =                                               sin β =

tan α = 1/3                                                          tan β = ½

tan (α + β) =

tan (α + β) = 1

∴ α + β =

#### Question 18

Find tan in terms of tan A

Sol:

tan  =

=

Question 19

Prove that = cot 360

Sol:

=

(on dividing numerator and denominator by cos 90)

=

= tan (450 + 90)

= tan 540

= tan (900 – 360)

= cot 360

∴  = 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 sin θ + cos θ as a single of an angle

Sol:

sin θ + cos θ = 2(  sin θ + cos 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

= 5

∴ minimum value = –5 and maximum value = 5

(ii) cos (x + ) + 2  sin (x + ) – 3

a= 1, b = 2  and c = – 3

∴ 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

∴ 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

Sol:

=

=

= tan θ

Question 26

For what values of x in the first quadrant is positive?

Sol:

> 0 ⟹ tan 2x > 0

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

⟹ 0 < x < π/4

Question 27

If cos θ = and π < θ < 3π/2, find the value of tan θ/2.

Sol:

cos θ =

π < θ < 3π/2 ⟹ π/2 < θ/2 < 3π/4

tan θ/2 < 0

tan θ/2 =

=–  (tan θ/2 < 0)

=–

= – 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 =

=

=

=

=

= 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

= 2

=

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  = 2 cos (θ/2)

Sol:

=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 = 4

Sol:

= 4

Question 34

Prove that sin 780 + cos 1320 =

Sol:

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

= sin 780 – sin 420

= 2 cos  sin

= 2 cos 600 sin 180

= 2

=

Question 35

Find the value of sin 340 + cos 640 – cos40

Sol:

sin 340 + cos 640 – cos40

= sin 340 –2 sin  sin

= 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) =

Sol:

4(cos 660 + sin 840)

=4(cos 660 + sin (900 – 60)

=4(cos 660 + cos (60)

= 4[ 2 cos  cos ]

=4[ 2 cos  cos ]

=8 cos 360 cos 300

= 8

=

Question 35

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

Sol:

tan θ + cot θ =

=

=

= sec θ. cosec θ

(tan θ + cot θ)2   = sec2 θ. cosec2 θ

sec2 θ + cosec2 θ =

= sec2 θ. cosec2 θ