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.
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
(1) + (2)
⇒ (sec θ + tan θ) + (sec θ – tan θ) =
(1) – (2)
⇒ (sec θ + tan θ) – (sec θ – tan 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