Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.12 Text Book Back Questions and Answers, Notes.
Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.12
Choose the correct or the most suitable answer:
Question 1.
(1) √2
(2) √3
(3) 2
(4) 4
Answer:
(4) 4
Explaination:
Question 2.
If cos 28° + sin 28°= k3, then cos 17° is equal to
(1) \(\frac{\mathbf{k}^{3}}{\sqrt{2}}\)
(2) \(-\frac{\mathbf{k}^{3}}{\sqrt{2}}\)
(3) \(\pm \frac{\mathbf{k}^{3}}{\sqrt{2}}\)
(4) \(-\frac{\mathbf{k}^{3}}{\sqrt{3}}\)
Answer:
(1) \(\frac{\mathbf{k}^{3}}{\sqrt{2}}\)
Explaination:
cos 28° + sin 28° = k3
cos 28° + sin (90° – 62°) = k3
cos 28° + cos 62° = k3
2 cos 45° . cos 17° = k3
2 × \(\frac{1}{\sqrt{2}}\) cos 17° = k3
√2 cos 17° = k3
cos 17° = \(\frac{\mathrm{k}^{3}}{\sqrt{2}}\)
Question 3.
The maximum value of
4 sin2x + 3 cos2x + sin \(\) + cos \(\) is
(1) 4 + √2
(2) 3 + √2
(3) 9
(4) 4
Answer:
(1) 4 + √2
Explaination:
4 sin2x + 3 cos2x + sin \(\frac{x}{2}\) + cos \(\frac{x}{2}\)
= sin2x + 3 sin2x + 3 cos2x + sin \(\frac{x}{2}\) + cos \(\frac{x}{2}\)
= sin2x + 3(sin2x + cos2x) + sin \(\frac{x}{2}\) + cos \(\frac{x}{2}\)
= 3 + sin2x + sin \(\frac{x}{2}\) + cos \(\frac{x}{2}\) —– (1)
Maximum value of sin x = 1
sin x = 1 when x = \(\frac{\pi}{2}\)
Maximum value of sin2x = 1
Maximum value is obtained when x = \(\frac{\pi}{2}\)
∴ (1) ⇒ 4 sin2 x + 3 cos2 x + sin \(\frac{x}{2}\) + cos \(\frac{x}{2}\)
= 3 + 1 + sin \(\left(\frac{90^{\circ}}{2}\right)\) + cos \(\left(\frac{90^{\circ}}{2}\right)\)
= 4 + sin 5° + cos 45°
= 4 + \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) = 4 + \(\frac{2}{\sqrt{2}}\)
= 4 + √2
Question 4.
(1) \(\frac{1}{8}\)
(2) \(\frac{1}{2}\)
(3) \(\frac{1}{\sqrt{3}}\)
(4) \(\frac{1}{\sqrt{2}}\)
Answer:
(1) \(\frac{1}{8}\)
Explaination:
Question 5.
If π < 2θ < \(\frac{3 \pi}{2}\), \(\sqrt{2+\sqrt{2+2 \cos 4 \theta}}\) equals to
(1) – 2 cos θ
(2) – 2 sin θ
(3) 2 cos θ
(4) 2 sin θ
Answer:
(1) – 2 cos θ
Explaination:
∴ θ lies in the second quadrant, cos θ is negative in the IInd quadrant.
∴ \(\sqrt{2+\sqrt{2+2 \cos 4 \theta}}\) = – 2 cos θ
Question 6.
If tan 40° = λ, then
(1) \(\frac{1-\lambda^{2}}{\lambda}\)
(2) \(\frac{1+\lambda^{2}}{\lambda}\)
(3) \(\frac{1+\lambda^{2}}{2 \lambda}\)
(4) \(\frac{1-\lambda^{2}}{2 \lambda}\)
Answer:
(4) \(\frac{1-\lambda^{2}}{2 \lambda}\)
Explaination:
Question 7.
cos 1° + cos 2° + cos 3° + ….. + cos 179° =
(1) 0
(2) 1
(3) – 1
(4) 89
Answer:
(1) 0
Explaination:
cos 1° + cos 2° + cos 3° + ……………… + cos 179°
= (cos 1° + cos 179°) + (cos 2° + cos 178°) + (cos 3° + cos 177°) + …………..
= 2 cos 90° cos 89° + 2 cos 90° . cos 88° + …………….
= 2 × 0 × cos 89°+ 2 × 0 × cos 88° + …………..
= 0
Question 8.
Let fk(x) = \(\frac{1}{k}\)[sinkx + coskx] where x ∈ R and k ≥ 1. Then f4(x) – f6(x) =
(1) \(\frac{1}{4}\)
(2) \(\frac{1}{12}\)
(3) \(\frac{1}{6}\)
(4) \(\frac{1}{3}\)
Answer:
(2) \(\frac{1}{12}\)
Explaination:
Question 9.
Which of the following is not true?
(1) sin θ = – \(\frac{3}{4}\)
(2) cos θ = – 1
(3) tan θ = 25
(4) sec θ = \(\frac{1}{4}\)
Answer:
(4) sec θ = \(\frac{1}{4}\)
Explaination:
We know |cos θ| < 1
sec θ = \(\frac{1}{4}\)
⇒ \(\frac{1}{\cos \theta}\) = \(\frac{1}{4}\)
⇒ cos θ = 4
which is not possible.
Question 10.
cos 2θ cos 2Φ + sin2(θ – Φ) – sin2(θ + Φ) is equal to
(1) sin 2 (θ + Φ)
(2) cos 2 (8 + Φ)
(3) sin 2 (θ – Φ)
(4) cos 2(θ – Φ)
Answer:
(2) cos 2 (8 + Φ)
Explaination:
cos 2θ cos 2Φ + sin2(θ – Φ) – sin2(θ + Φ)
= cos 2θ cos 2Φ – sin 2θ sin 2Φ
= cos(2θ + 2Φ)
= cos 2(θ + Φ)
Question 11.
(1) sin A + sin B + sin C
(2) 1
(3) 0
(4) cos A + cos B + cos C
Answer:
(3) 0
Explaination:
Question 12.
If cos pθ + cos qθ = o and if p ≠ q then θ is equal to(n is any integer)
(1) 
(2) 
(3) 
(4) 
Answer:
Given cos pθ + cos qθ = o
Question 13.
If tan α and tan β are the roots of x2 + ax + b = 0 then \(\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}\) is equal to
(1) \(\frac{\mathbf{b}}{\mathbf{a}}\)
(2) \(\frac{\mathbf{a}}{\mathbf{b}}\)
(3) –\(\frac{\mathbf{a}}{\mathbf{b}}\)
(4) –\(\frac{\mathbf{b}}{\mathbf{a}}\)
Answer:
(3) –\(\frac{\mathbf{a}}{\mathbf{b}}\)
Explaination:
x2 + ax + b = 0
Given tan α and tan β are the roots of the above equation. Then
Question 14.
In a triangle ABC, sin2 A + sin2 B + sin2 C = 2 then the triangle is .
(1) equilateral triangle
(2) isosceles triangle
(3) right triangle
(4) scalene triangle
Answer:
(3) right triangle
Explaination:
Given sin2 A + sin2 B + sin2 C = 2
Suppose the given triangle is a right angle triangle with ∠C = 90°, then
sin2 C = sin 2 90° = 1 …….. (1)
∴ sin2 A + sin2 B + 1 = 2
sin2 A + sin2B = 1
Also A + B = 90° ⇒ A = 90° – B
sin A = sin (90° – B) = cos B ——- (2)
sin2 A + sin2 B + sin2 C = 2
Using equations (1) and (2)
⇒ cos2 B + sin2 B + sin2 90° = 2
1 + 1 = 2
2 = 2
∴ sin2 A + sin2 B + sin2 C = 2 is true.
∴ ∆ ABC is a right angle triangle.
Question 15.
If f(θ) = |sin θ| + |cos θ|, θ ∈ R then f(θ) is in the interval
(1) [0, 2]
(2) [1, √2]
(3) [1, 2]
(4) [0, 1]
Answer:
(2) [1, √2]
Explaination:
f(θ) = |sin θ| + |cos θ|
To find the point of intersection of the sine curve and cosine curve solving
Question 16.
(1) cos 2x
(2) cos x
(3) cos 3x
(4) 2 cos x
Answer:
(4) 2 cos x
Explaination:
Consider the numerator cos 6x + 6 cos 4x + 15 cos 2x + 10
cos 6x + 6 cos 4x + 15 cos 2x + 10 = cos 6x + cos 4x + 5 cos 4x + 5 cos 2x + 10 cos 2x + 10
= (cos 6x + cos 4x) + 5 (cos 4x + cos 2x) + 10(cos 2x + 1)
= 2 cos 5x cos x + 10 cos 3x . cos x + 20 cos2x
= 2 cos x (cos 5x + 5 cos 3x + 10 cos x)
= 2 cos x
Question 17.
The triangle of maximum area with constant perimeter 12m
(1) is an equilateral triangle with side 4m
(2) is an isosceles triangle with sides 2m, 5m ,5m
(3) is a triangle with sides 3m, 4m, 5m
(4) does not exists.
Answer:
(1) is an equilateral triangle with side 4m
Explaination:
Given the perimeter of the triangle is 12m
2s = 12 ⇒ s = 6
Maximum area is obtained when it is an equilateral triangle with side 4m each.
Question 18.
A wheel is spinning at 2 radians / second. How many seconds will it take to make 10 complete rotations.
(1) 10 π seconds
(2) 20 π seconds
(3) 5 π seconds
(4) 15 π seconds
Answer:
(1) 10 π seconds
Explaination:
In 1 second, it rotates = 2 radians
For 2 radians rotation time taken = 1 second
∴ For 1 complete rotation (2 π radians) time taken
= \(\frac { 1 }{ 2 }\) × 2π = π seconds.
∴ For 10 revolution time taken = π × 10
= 10 π seconds.
Question 19.
If sin α + cos α = b, then sin 2α is equal to
(1) b2 – 1, if b ≤ √2
(2) b2 – 1, if b > √2
(3) b2 – 1, if b ≥ √2
(4) b2 – 1, if b < √2
Answer:
(1) b2 – 1, if b ≤ √2
Explaination:
sin α + cos α = b
(sin α + cos α)2 = b2
sinv α + cos2 α + 2 sin α cos α = b2
1 + sin 2α = b2
sin 2α = b2 – 1
But – 1 ≤ sin 2α ≤ I
– 1 ≤ b2 – 1 ≤ 1
b2 – 1 ≤ 1 ⇒ b2 ≤ 2
⇒ b ≤ √2
∴ sin 2α = b2 – 1 if b ≤ √2
Question 20.
In a ∆ ABC
(i) sin \(\frac{\mathbf{A}}{2}\) sin \(\frac{\mathbf{B}}{2}\) sin \(\frac{\mathbf{C}}{2}\) > 0
(ii) sin A sin B sin C > 0,then
(1) Both (i) and (ii) are true
(2) only (1) is true
(3) only (ii) Is true
(4) neither (i) nor (ii) is true
Answer:
(1) Both (i) and (ii) are true
Explaination:
We know in a ∆ ABC
sin \(\frac{\mathbf{A}}{2}\) sin \(\frac{\mathbf{B}}{2}\) sin \(\frac{\mathbf{C}}{2}\) > 0
Also sin A sin B sin C > 0
∴ Statements (i) and (ii) are true.