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 the intermediate First-year Maths students.

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

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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 4^th 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