# Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.4

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.4

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

since x is in III quadrant
Both sin x and cos x are negative

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

(ii) cos (π + θ) = – cos θ
cos (π + θ) = cos π cos θ – sin π sin θ
= (- 1) cos θ – (0) sin θ
cos (π + θ) = – cos θ

(iii) sin (π + θ) = – sin θ
sin (π + θ) = sin π cos θ + cos π sin θ
= (0) cos θ + (- 1) sin θ
sin (π + θ) = 0 – sin θ = – sin θ

Question 7.
Find a quadratic equation whose roots are sin 15° and cos 15°
Solution:

Question 8.
Expand cos(A + B + C). Hence prove that cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B, if A + B + C = $$\frac{\pi}{2}$$
Solution:
cos(A + B + C) = cos (A + (B + C))
= cos A cos (B + C) – sin A sin (B + C)
= cos A [ cos B cos C – sin B sin C] – sin A [sin B cos C + cos B sin C]
= cos A cos B cos C – cos A sin B sin C – sin A sin B cos C – sin A cos B sin C
Given A + B + C = $$\frac{\pi}{2}$$
∴ cos$$\left(\frac{\pi}{2}\right)$$ = cos A cos B cos C – cos A sin B sin C – sin A sin B cos C – sin A cos B sin C
0 = cos A cos B cos C – cos A sin B sin C – sin A sin B cos C – sin A cos B sin C
cos A cos B cos C = cos A sin B sin C + sin A sin B cos C + sin A cos B sin C

Question 9.
Prove that
(i) sin(45° + θ) – sin(45° – θ) = $$\sqrt{2}$$sin θ.
(ii) sin(30° + θ) + cos(60° + θ) = cos θ.
Solution:

Question 10.
If a cos(x + y) = b cos(x – y), show that (a + b) tan x = (a – b) cot y.
Solution:
a cos (x + y) = b cos (x – y)
a[cos x cos y – sin x sin y] = 6[cos x cos y + sin x sin y]
(i.e) a cos x cos y – a sin x sin y = b cos x cos y + b sin x sin y
a cos x cos y – b sin x sin y = a sin x sin y + b cos x cos y

⇒ a cot y – b tan x = a tan x + b cot y
a cot y – b cot y = a tan x + b tan x
⇒ (a + b) tan x = (a – b) cot y.

Question 11.
Prove that sin 105° + cos 105° = cos 45°.
Solution:
sin 105° = sin (60°+ 45°)
= sin 60° cos 45° + cos 60° cos 45°

= cos 45° = RHS

Question 12.
Prove that sin 75° – sin 15° = cos 105° + cos 15°.
Solution:
sin 75° – sin 15° = sin (45° + 30°) – sin (45° – 30°)
= (sin 45° cos 30° + cos 45° sin 30°) – (sin 45° cos 30° – cos 45° sin 30°)
= sin 45° cos30° + cos 45° sin 30° – sin 45° cos 30° + cos 45° sin 30°
= 2 cos 45° sin 30°
sin 75° – sin 15° = 2 × $$\frac{1}{\sqrt{2}}$$ × $$\frac{1}{2}$$ = $$\frac{1}{\sqrt{2}}$$ ——— (1)
cos 105° + cos 15° = cos (90° + 15°) + cos 15°
= – sin 15° + cos 15°
= cos 15° – sin 15°
= cos(45° – 30°) – sin(45° – 30°)
= (cos 45° cos 30°+ sin 45° sin 30°) – (sin 45° cos 30° – cos 45° sin 30°)

cos 105° + cos 15° = $$\frac{2}{2 \sqrt{2}}$$ = $$\frac{1}{\sqrt{2}}$$ ——— (2)
From equations (1) and (2)
sin 75° – sin 15° = cos 105° + cos 15°

Question 13.
Show that tan 75° + cot 75° = 4
Solution:

Question 14.
Prove that cos(A + B) cos C – cos(B + C) cos A = sin B sin(C – A).
Solution:
cos (A + B) cos C = (cos A cos B – sin A sin B) cos C
cos (A + B) cos C = cos A cos B cos C – sin A sin B cos C ——— (1)
cos (B + C) cos A = (cos B cos C – sin B sin C) cos A
cos (B + C) cos A = cos A cos B cos C – cos A sin B sin C ——— (2)
Equation (1) – (2) ⇒
Cos (A + B) cos C – cos (B + C) cos A = cos A cos B cos C – sin A sin B cos C – cos A cos B cos C + cos A sin B sin C
= sin A sin B cos C + cos A sin B sin C
= sin B (cos C cos A + sin C sin A)
= sin B cos (C – A)

Question 15.
Prove that sin(n + 1) θ sin(n – 1) θ + cos(n + 1) θ cos(n – 1)θ = cos 2θ, n ∈ Z.
Solution:
cos (n + 1)θ cos(n – 1)θ + sin(n + 1)θ sin(n – 1)θ
= cos [(n + 1)θ – (n – 1)θ]
= cos [nθ + θ – nθ + θ]
= cos 2θ, n ∈ Z

Question 16.

Solution:

Question 17.
Prove that
(i) sin(A + B) sin(A – B) = sin2 A – sin2 B
(ii) cos(A + B) cos(A – B) = cos2 A – sin2 B = cos2 B – sin2 A
(iii) sin2(A + B) – sin2(A – B) = sin2A sin2B
(iv) cos 8θ cos 2θ = cos2 5θ – sin2
Solution:
(i) sin(A + B) sin(A – B)= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)
= sin2A cos2B – cos2A sin2B = sin2A (1 – sin2B) – (1 – sin2A) sin2B
= sin2A – sin2A sin2B – sin2B + sin2A sin2 B
= sin2A – sin2 B

(ii) LHS = cos (A + B) cos (A – B) = (cos A cos B – sin A sin B) (cos A cos B + sin (A sin B)
= cos2 A cos2 B – sin2 A sin2 B
= cos2 A (1 – sin2 B) – (1 – cos2 A) sin2 B
= cos2 A – cos2 A sin2 B – sin2 B + cos2 A sin2 B
= cos2 A – sin2 B = RHS
Now cos2 A – sin2 B = (1 – sin2 A) – (1 – cos2 B)
= 1 – sin2 A – 1 + cos2 B
= cos2 B – sin2 A

(iii) sin2(A + B) – sin2(A – B) = (sin(A + B) + sin(A – B)) (sin(A + B) – sin(A – B))
= [sin A cos B + cos A sin B + sin A cos B – cos A sin B] × [(sin A cos B + cos A sin B) – (sin A cos B – cos A sin B)]
= (2 sin A cos B) × [sin A cos B + cos A sin B – sin A cos B + cos A sin B]
= (2 sin A cos B) (2 cos A sin B)
= (2 sin A cos A)(2 sin B cos B)
= sin2A . sin2B

(iv) LHS = cos 8θ cos 2θ
= cos (5θ + 3θ) cos (5θ – 3θ) .
We know cos (A + B) cos (A – B) = cos2 A – sin2 B
∴ cos (5θ + 3θ) cos (5θ – 3θ) = cos2 5θ – sin2 3θ = RHS

Question 18.
Show that cos2 A + cos2 B – 2 cos A cos B cos(A + B) = sin2(A + B).
Solution:
sin2(A + B) = [sin(A + B)]2
= (sin A cos B + cos A sin B )2
= sin2 A cos2 B + cos2 A sin2 B + 2 sin A cos B cos A sin B
= (1 – cos2A) cos2B + cos2 A ( 1 – cos2B) + 2 sin A sin B cos A cos B
= cos2 B – cos2A cos2B + cos2A – cos2A cos2 B + 2 sin A sin B cos A cos B
= cos2 A + cos2 B – 2cos2 A cos2 B + 2 sin A sin B cos A cos B
= cos2 A + cos2 B – 2 cos A cos B (cos A cos B – sin A sin B)
sin2 (A + B) = cos2 A + cos2 B – 2 cos A cos B cos(A + B)

Question 19.
If cos(α – β) + cos(β – γ) + cos(γ – α) = $$-\frac{3}{2}$$, then prove that cos α + cos β + cos γ = sin α + sin β + sin γ
Solution:

2 cos (α – β) + 2cos (β – γ) + 2cos (γ – α) = -3
2cos(α – β) + 2cos(β – γ) + 2cos (γ – α) + 3 = 0
[2 cos α cos β + 2 sin α sin β] + [2 cos β cos γ + 2 sin β sin γ] + [2 cos γ cos α + sin γ sin α] + 3 = 0
= [2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α] + [2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α] + (sin2 α + cos2 α) + (sin2 β + cos2 β) + (sin2 γ + cos2 γ) = 0
⇒ (cos2 α + cos2 β + cos2 γ + 2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α) + (sin2 α + sin2 β) + (sin2 γ + 2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α) = 0
(cos α + cos β + cos γ)2 + (sin α + sin β + sin γ)2 = 0
=(cos α + cos β + cos γ) = 0 and sin α + sin β + sin γ = 0
Hence proved

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

### Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.4 Additinal Questions

Question 1.
Prove that sin (A + B) sin (A – B) = cos2 B – cos2 A
Solution:
LHS = sin (A + B) sin (A – B)
= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)
= sin2 A cos2 B – cos2 A sin2 B
= (1 – cos2 A) cos2 B – cos2 A (1 – cos2 B)
= cos2 B – cos2 A cos2 B – cos2 A + cos2 A cos2 B
= cos2 B – cos2 A = RHS

Question 2.
Prove that
(i) sin A + sin(120° + A) + sin (240° + A) = 0
(ii) cos A + cos (120° +A) + cos (120° – A) = 0
Solution:
(i) sin A + sin(120° + A) + sin (240° + A)
= sin A+ sin 120° cos A + cos 120° sin A + sin 240° cos A + cos 240° sin A …… (1)

By substituting these values in (1), we get,

(ii) cos 120° = cos (180° – 60°) = – cos 60° = $$\frac{-1}{2}$$
LHS = cos A + cos (120° + A) + cos (120° – A)
= cos A + cos 120° cos A – sin 120° sin A + cos 120° cos A + sin 120° sin A
= cos A + 2 cos 120° cos A

Question 3.

Solution:

tan (A + B) = 1 ⇒ A+ B = 45°

Question 4.
If A + B = 45°, show that (1 + tan A) (1 + tan B) = 2 and hence deduce the value of tan $$22 \frac{1^{0}}{2}$$.
Solution:
Given A + B = 45° ⇒ tan (A + B) = tan 45°

(i.e.) tan A + tan B = 1 – tan A.tan B
(i.e.) 1 + tan A + tan B = 2 – tan A tan B (add 1 on both sides)
1 + tan A + tan B + tan A tan B = 2
(i.e.) (1 + tan A) (1 + tan B) = 2