Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.12

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 II^nd 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 f_k(x) = \(\frac{1}{k}\)[sin^kx + cos^kx] 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:
On simplifying we get
sin² A + sin² B + sin² C = 2 + 2 cos A cos B cos C
= 2 (given)
⇒ cos A cos B cos C = 0
cos A (or) cos B (or) cos C = 0
⇒ A (or) B (or) C = π/2
⇒ ABC (is a right angled 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 the maximum area with a constant perimeter of 12m
(1) is an equilateral triangle with a side of 4m
(2) is an isosceles triangle with sides 2m, 5m, 5m
(3) is a triangle with sides 3m, 4m, 5m
(4) does not exist.
Answer:
(1) is an equilateral triangle with a side of 4m

Explanation:
A triangle will have a max area (with a given perimeter) when it is an equilateral triangle.

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

Explanation:
1 rotation makes 2π^c
Distance travelled in 1 second = 2 radians
So time taken to complete 10 rotations = 6 × 2π = 20 π^c
\(=\frac{20 \pi}{2}=10 \pi\) 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 an ∆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:
When A + B + C = 180°

When A + B + C = 180° each angle will be lesser than 180°
So sin A, sin B, sin C > 0
⇒ sin A sin B sin C > 0
So both (i) and (ii) are true