Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.4

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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) L.HS = cos(π + θ) = cos(180° + θ)
= cos 180° cos θ – sin 180° sin θ
= (-1) cos θ – (0) sin θ
= – cos θ = RHS

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

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:
Taking A + B = X and C = Y
We get cos (X + Y) = cos X cos Y – sin X sin Y
(i.e) cos (A + B + C) = cos (A + B) cos C – sin (A + B) sin C
= (cos A cos B – sin A sin B) cos C – [sin A cos B + cos A sin B] sin C
cos (A + B + C) = cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C If (A + B + C) = π/2 then cos (A + B + C) = 0
⇒ cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C = 0
⇒ cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B

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:
RHS = cos 105° + cos 15° = cos (90° + 15°) + cos (90° – 75°)
= – sin 15° + sin 75°
= sin 75° – sin 15° = LHS

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:
LHS = (cos A cos B – sin A sin B) cos C – (cos B cos C – sin B sin 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
= cos A sin B sin C – sin A sin B cos C
= sin B [sin C cos A – cos C sin A]
= sin B [sin (C – A)] = RHS

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:
Taking (n + 1) θ = A and (n – 1) θ = B
LHS = sin A sin B + cos A cos B
= cos (A – B)
= cos[(n + 1) – (n – 1)]θ
= cos (n + 1 – n + 1)θ = cos 2θ = RHS

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 3θ
Solution:
(i) LHS = sin (A + B) sin (A – B)
= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)
= sin²A cos² B – cos² A sin² B
= sin² A (1 – sin² B) – (1 – sin² A) sin² B
= sin² A – sin² A sin² B – sin² B + sin² A sin² B
= sin² A – sin² B = RHS

(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 – sin² B = sin (A + B) sin (A – B)
LHS = sin² (A + B) – sin² (A – B) = sin [(A + B) + (A – B)] [sin (A + B) – (A – B)]
= sin 2A sin 2B = RHS

(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:
LHS = cos² A + cos² B – 2 cos A cos B [cos A cos B – sin A sin B]
= cos² A + cos² B – 2 cos² A cos² B + 2 sin A cos A sin B cos B
= (cos² A – cos² A cos² B) + (cos² B – cos² A cos² B) + 2 sin A cos A sin B cos B
= cos² A (1 – cos² B) + cos² B (1 – cos² A) + 2 sin A cos A sin B cos B
= cos² A sin² B + cos² B sin² A + 2 sin A cos B sin B cos A
= (sin A cos B + cos A sin B)²
= sin² (A + B) = RHS

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