Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.4 Text Book Back Questions and Answers, Notes.

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

Question 1.
If sin x = $$\frac{15}{17}$$ and cos y = $$\frac{12}{13}$$, 0 < x < $$\frac{\pi}{2}$$, 0 < y <$$\frac{\pi}{2}$$, find the value of
(i) sin(x + y)
(ii) cos(x – y)
(iii) tan(x + y)
Given sin x = $$\frac{15}{17}$$, 0 < x < $$\frac{\pi}{2}$$
we have cos2x + sin2x = 1
∴ cos2x = 1 – sin2x
= 1 – $$\left(\frac{15}{17}\right)^{2}$$
= 1 – $$\frac{225}{289}$$

Given that 0 < x < $$\frac{\pi}{2}$$, that is x lies in the first quadrant ∴ cos x is positive.
cos x = $$\frac{8}{17}$$
Also given cos y = $$\frac{12}{13}$$, 0 < x < $$\frac{\pi}{2}$$
we have cos2y + sin2y = 1
sin2y = 1 – cos2y

Since 0 < y < $$\frac{\pi}{2}$$, y lies in the first quadrant

(i) sin (x + y)
sin (x + y) = sin x cos y + cos x sin y

(ii) cos (x – y)
cos (x – y) = cos x cos y + sin x sin y

(iii) tan (x + y)

Question 2.
If sin A = $$\frac{3}{5}$$ and cos B = $$\frac{9}{41}$$ 0 < A < $$\frac{\pi}{2}$$, 0 < B < $$\frac{\pi}{2}$$, find the value of
(i) sin (A + B)
(ii) cos (A – B)
Given sin A = $$\frac{3}{5}$$ 0 < A < $$\frac{\pi}{2}$$
we have sin2A + cos2A = 1
cos2A = 1 – sin2A

Since 0 < A < $$\frac{\pi}{2}$$, A lies in the first quadrant cos A is positive. ∴ cos A = $$\frac{4}{5}$$
Also given cos B = $$\frac{9}{41}$$, 0 < B < $$\frac{\pi}{2}$$
We have cos2B + sin2B = 1

(i) sin (A + B)
sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A – B)
cos (A – B) = cos A cos B + sin A sin B

Question 3.
Find cos(x – y),given that cos x = –$$\frac{4}{5}$$ with π < x < $$\frac{3 \pi}{2}$$ and sin y = –$$\frac{24}{25}$$ with π < y < $$\frac{3 \pi}{2}$$
Given cos x = –$$\frac{4}{5}$$, π < x < $$\frac{3 \pi}{2}$$
we have cos2 x + sin2 x = 1

Since π < x < $$\frac{3 \pi}{2}$$, x lies in the third quadrant.
Since x is negative in the third quadrant. ∴ sin x = –$$\frac{3}{5}$$

Question 4.
Find sin(x – y) , given that sin x = $$\frac{8}{17}$$ with 0< x < $$\frac{\pi}{2}$$, and cos y = –$$\frac{24}{25}$$, x < y < $$\frac{3 \pi}{2}$$

Question 5.
Find the value of
(i) cos 105°
cos 105° = cos(90° + 150)
= -sin 15°
= -sin(45°- 30°)
= -[sin 45° cos 30° – cos 45° sin 30°]

(ii) sin 105°
sin 105° = sin (90° + 15°)
= cos 15°
= cos (45° – 30°)
= cos 45° cos 30° + sin 45° sin 30°

(iii) tan $$\frac{7 \pi}{12}$$

Question 6.
prove that
(i) cos (30° + x) = $$\frac{\sqrt{3} \cos x-\sin x}{2}$$
(ii) cos (π + θ) = – cos θ
(iii) sin (π + θ) = – sin θ
(i) cos (30° + x) = $$\frac{\sqrt{3} \cos x-\sin x}{2}$$
cos ( 30° + x) = cos 30°. cos x – sin 30° sin x

(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°.
sin 15° = sin (45° – 30°)
= sin 45°. cos 30° – cos 45°. sin 30°

cos 15° = cos(45° – 30°)
= cos 45° . cos 30° + sin 45° . sin 30°

The quadratic whose roots cos 15° and sin 15° is
x2 – (cos 15° + sin 15°)x + (cos 15°) (sin 15°) = 0 ——— (3)

Substituting in equation (3) we have

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 = $$\frac{\pi}{2}$$
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) = √2 sin θ
(ii) sin (30° + θ) + cos (60° + θ) = cos θ
(i) sin (45° + θ) – sin (45° – θ) = √2 sin θ
sin(45° + θ) – sin(45° – θ) = (sin 45° cos θ + cos 45° sin θ) – (sin 45° cos θ + cos 45° sin θ)
= sin 45° cos θ + cos 45° sin θ – sin 45° cos θ + cos 45° sin θ
= 2 cos 45° sin θ
= 2 × $$\frac{1}{\sqrt{2}}$$ sin θ
= $$\frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$$ × sin θ
sin (45° + θ) – sin (45° – θ) = $$\frac{2 \sqrt{2}}{2}$$ sin θ
= √2 sin θ

(ii) sin (30° + θ) + cos (60° + θ) = cos θ
sin (30° + θ) + cos (60° + θ)
= sin 30° cos θ + cos 30° sin θ + cos 60° cos θ – sin 60° sin θ
= $$\frac{1}{2}$$ cos θ + $$\frac{\sqrt{3}}{2}$$ sin θ + $$\frac{1}{2}$$ cos θ – $$\frac{\sqrt{3}}{2}$$ sin θ
= cos θ

Question 10.
If a cos (x + y) = b cos(x – y), show that (a + b) tan x = (a – b) cot y.
a cos (x + y) = b cos (x – y)
a [cos x cos y – sin x sin y] = b [cos x cos y + sin x sin y]
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 cos x cos y = a sin x sin y + b sin x sin y
(a – b) cos x cos y = (a + b) sin x sin y
(a – b) $$\frac{\cos y}{\sin y}$$ = (a + b) $$\frac{\sin x}{\cos x}$$
(a – b) cot y = (a + b) tan x
(a + b) tan x = (a – b) cot y .

Question 11.
Prove that sin 105° + cos 105° = cos 45°
sin 105° + cos 105° = sin (90° + 15°) + cos ( 90° + 15°)
= cos 15° – sin 15°
= cos (45° – 30°) sin (45° – 30°)
= (cos 45° . cos30° + sin 45° sin 30°) – (sin 45° cos 30° – cos 45° sin 30°)

Question 12.
Prove that sin 75° – sin 15° = cos 105° + cos 15°
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
tan 75° = tan (45° + 30°)

Question 14.
Prove that cos(A + B). cos C – cos(B + C) cos A = sin B sin (C – A)
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
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.
If x cos θ = y cos$$\left(\theta+\frac{2 \pi}{3}\right)$$ = z cos $$\left(\theta+\frac{4 \pi}{3}\right)$$ find the value of xy + yz + zx.

= cos θ + cos (θ + 120°) + cos (θ + 240°)
= cos θ + cos θ cos 120° – sin θ sin 120° + cos θ . cos 240° – sin θ sin 240°
= cos θ + cos θ cos (180° – 60°) – sin θ sin( 180°- 60°) + cos θ cos (180°+ 60°) – sin θ sin (180° + 60°)
= cos θ + cos θ × – cos 60° – sin θ × sin 60° + cos θ × – cos 60°- sin θ (- sin 60°)
= cos θ – cos θ cos 60° – sin θ sin 60° – cos θ cos60° + sin θ sin 60°
= cos θ – 2 cos θ cos 60°
= cos θ – 2 cos θ × $$\frac { 1 }{ 2 }$$ = cos θ – cos θ = 0

Question 17.
Prove that
(i) sin(A + B) . sin(A – B) = sin2A – sin2B
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) cos (A + B) . cos (A – B) = cos2A – sin2B = cos2B – sinA
cos(A + B) . cos(A – B) = (cos A cos B – sin A sin B) (cos A cos B + sin A sin B)
= (cos A cos B )2 – (sin A sin B )2 = cos2 A cos2 B – sin2 A sin2 B
= cos2A(1 – sin2B) – (1 – cos2A) sin2B
= cos2A – cos2A sin2B – sin2B + cos2A sin2B
cos(A + B) . cos(A – B) = cos2A – sin2B
Also cos(A + B) . cos(A – B) = cos2A cos2B – sin2A sin2B
= (1 – sin2A)cos2B – sin2A(1 – cos2B)
= cos2B – sin2A cos2B – sin2A + sin2A cos2B
cos(A + B) . cos(A – B) = cos2B – sin2A

(iii) sin2 (A + B) – sin2(A – B) = sin 2 A sin 2 B
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) cos 8θ . cos 2θ = cos2 5θ – sin2
cos8θ . cos2θ = cos(5θ + 3θ) cos(5θ – 3θ)
= (cos 5θ . cos 3θ – sin 5θ sin 3θ) (cos 5θ . cos3θ + sin 5θ sin 3θ)
= (cos 5θ . cos 3θ)2 – (sin 5θ sin 3θ)2
= cos2 5θ cos23θ – sin2 5θ sin2
= cos2 5θ (1 – sin2 3θ) – (1 – cos2 5θ) sin2
= cos2 5θ – cos2 5θ sin2 3θ – sin2 3θ + cos2 5θ sin2
cos 8θ . cos 2θ = cos2 5θ – sin2

Question 18.
Show that cos2A + cos2B – 2 cos A cos B cos (A + B) = sin2(A + B)
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) = [lαtex]-\frac{3}{2}[/lαtex],then prove thαt
cos α + cos β + cos γ = sin α + sin β + sin γ = 0
Given cos( α – β) + cos (β – γ) + cos (γ – α) = [lαtex]-\frac{3}{2}[/lαtex]
cos α cos β + sin α sin β + cos β cos γ + sin β sin γ + cos γ cos α + sin γ sin α = [lαtex]-\frac{3}{2}[/lαtex]
2 [cos α cos β + sin α sin β + cos β cos γ + sin β sin γ + cos γ cos α + sin γ sin α] = – 3
(2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α) + (2 sin α sin β + 2sin β sin γ + 2 sin γ sin α) + 3 = 0
(2 cos α cos β + 2 cos β cos γ + 2cos γ cos α) + (2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α)+
(cos2 α + sin2 α) + (cos2β + sin2 β) + (cos2 γ + sin2 γ) = 0
(cos2 α + cos2β + cos2γ + 2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α) + (sin2α + sin2β + sin2γ + 2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α) = 0
(cos α + cos + cos γ)2 + (sin α + sin β + sin )2 = 0
cos α + cos β + cos γ = 0 and sin α + sin β + sin γ = 0
cos α + cos β + cos γ = sin α + sin + sin γ = 0

Question 20.
Show that
(i)

(ii)

Question 21.
Prove that cot (A + B) = $$\frac{\cot A \cot B-1}{\cot A+\cot B}$$

Question 22.
If tan x = $$\frac{n}{n+1}$$ and tan y = $$\frac{1}{2 n+1}$$ find tan (x + y).
Prove that tan $$\left(\frac{\pi}{4}+\theta\right)$$ tan $$\left(\frac{3 \pi}{4}+\theta\right)$$ = -1
Find the value of tan (α + β), given that cot α = $$\frac{1}{2}$$, α ∈ $$\left(\pi, \frac{3 \pi}{2}\right)$$ and sec β = –$$\frac{5}{3}$$ β ∈ $$\left(\frac{\pi}{2}, \pi\right)$$
If θ + Φ = α and tan θ = k tan Φ, then prove that sin (θ – Φ ) = $$\frac{k-1}{k+1}$$ sin α