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
(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) $$-\frac{\mathbf{k}^{3}}{\sqrt{3}}$$
(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
(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) $$\frac{1}{\sqrt{2}}$$
(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
(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) $$\frac{1-\lambda^{2}}{2 \lambda}$$
(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
(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) $$\frac{1}{3}$$
(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 = $$\frac{1}{4}$$
(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)
(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
(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)
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) –$$\frac{\mathbf{b}}{\mathbf{a}}$$
(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
(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]
(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
(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.
(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
(1) 10 π seconds

Explanation:
1 rotation makes 2πc
Distance travelled in 1 second = 2 radians
So time taken to complete 10 rotations = 6 × 2π = 20 πc
$$=\frac{20 \pi}{2}=10 \pi$$ 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
(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