Samacheer Class 11 Maths - Trigonometry

(i) 25°
25° First quadrant
(ii) 825°
825° = 9 × 90° + 15°
825° = 2 × 360° + 105°
∴ 825° lies in the second quadrant.
iii) -55°
-55° lies in the fourth quadrant
iv) 328°
328° = 270° + 58° lies in the fourth quadrant.
v) -230°
– 230° = – 180° + (- 50°) lies in the second quadrant.

(i) 395°
395° = 360° + 35°
395° – 35° = 360°
∴ Coterminal angle for 395° is 35°
(ii) 525°
525° = 360° + 165°
360° – 165° = 360°
∴Coterminal angle for 525° is 165°
(iii) 1150°
1150° = 360° + 360° + 360° + 70°
1150° = 3 × 360° + 70°
1150° – 70° = 3 × 360°
∴ Coterminal angle for 1150° is 70°.
(iv) – 270°
– 270° = 360° + 90°
– 270° – 90° = 360°
∴ Coterminal angle for -270° is 90°
(v) – 450°
– 450° = – 720° + 270°
– 450° – 270° = – 2 × 360°
∴ Coterminal angle for – 450° is 270°

a cos θ – b sin θ = c
(a cos θ – b sin θ) 2 + (a sin θ + b cos θ) 2 = a 2 cos 2 θ – 2 ab sin θ cos θ + b 2 sin 2 θ + a 2 sin 2 θ + b 2 cos 2 θ + 2 ab sin θ cos θ
c 2 + (a sin 0 + b cos θ ) 2 = a 2 cos 2 θ + a 2 sin 2 θ + b 2 sin 2 θ + b 2 cos 2 θ
= a 2 (cos 2 θ + sin 2 θ) + b 2 (sin 2 θ + cos 2 θ)
c 2 + (a sin θ + b cos θ )2 = a 2 + b 2
(a sin θ + b cos θ) 2 = a 2 + b 2 – c 2
a sin θ + b cos θ = ± \(\sqrt{\mathbf{a}^{2}+\mathbf{b}^{2}-\mathbf{c}^{2}}\)

tan 2 θ = 1 – k 2
1 + tan 2 θ = 1 + 1 – k 2
sec 2 θ = (2 – k 2 )
sec 2 θ = (2 – k 2 ) 1/2
tan 2 θ = 1 – k 2
When θ = \(\frac{\pi}{2}\), tan \(\frac{\pi}{2}\) = ∞, not defined 2
When θ = 0, tan 2 0 = 1 – k 2
1 – k 2 = 0 ⇒ k 2 = 1 ⇒ k = ± 1
When θ = 45°, tan 2 45° = 1 – k 2
1 – k 2 = 1 ⇒ – k 2 = 0 ⇒ k = 0
When θ > 45°, say θ = 60°
tan 2 60° = 1 – k 2 = (√3) 2 = 1 – k 2
3 = 1 – k 2 ⇒ k 2 = 1 – 3 = – 2
∴ θ > 45°, k 2 is negative ⇒ k is imaginary
∴ k lies between -1 and 1 ⇒ k ∈ [-1, 1]

Given sec θ + tan θ = p
We have sec 2 θ – tan 2 θ = 1
(sec θ + tan θ) (sec θ – tan θ) = 1
p (sec θ – tan θ) = 1
sec θ – tan θ = \(\frac{1}{p}\)
(sec θ – tan θ) + (sec θ – tan θ) = p + \(\frac{1}{p}\)

Let the radius of the circular path be = r m.
Length of the circular path s = 1 k. m
s = 1000 m.
Athlete runs 5 times around the path to cover 1 k. m distance
∴ θ = 360° × 5
θ = 360° × 5 × \(\frac{\pi}{180}\) radians
θ = 10 π radians
s = r θ
1000 = r 10 π
r = \(\frac{1000}{10 \pi}\)
r = \(\frac{1000 \times 7}{10 \times 22}=\frac{350}{11}\)
r = 31.818 meters
Radius of the circular path = 31.82 meters

Given Diameter AB = 40 cm
∴ Radius r = 20 cm
Chord CD = 20 cm
O – Centre of the circle
OC = OD = radius = 20 cm.
∴ Triangle OCD is an equilateral triangle.
To find the length of the minor arc CD.
Let s = minor arc CD.
The arc CD subtends 60° at the centre.
θ = 60°
θ = 60° × \(\frac{\pi}{180}\) radians.
θ = \(\frac{\pi}{3}\) radians
We have s = rθ

Central angle subtended by the arc θ = 41°
θ = 41 × \(\frac{\pi}{180}\) Radians
The radius of the circle r = 10 feet
Length of the arc = s
s = rθ

Let r 1 and r 2 be the radii of the two circles and l be the length of the arc.
Given central angle θ 1 = 60°

Let OAB be the sector of a circle of radius r.
The angle of the sector is θ.
Perimeter of the sector = OA + arc AB + OB arc AB = rθ
∴ Perimeter of the sector = r + r θ + r
= 2r + rθ
= r(2 + θ) ———- (1)
Length of the arc of the semi – circle of radius
l = nπ ——– (2)
Given that perimeter the circular sector = Length of the arc of the semi circle of radius r
From equations (1) and (2), we have
r(2 + θ) = πr
2 + θ = π
θ = π – 2

Given An airplane, the propeller rotates 1000 times per minute.
∴ A point on the edge of the propeller also rotates 1000 times in 1 minute.
∴ In 1 minute the point describes 1000 × 2π radians angle at the centre.
In 60 seconds the point describes 1000 × 2π radians angle.
∴ In 1 second the angle described = \(\frac{1000 \times 2 \pi}{60}\) radians

Radius of the circular track r = 1500 m.
Speed of the train = 66 km/hr
Let θ be the angle made by the path of train at the centre in 20 seconds.
In 1 hr distance moved by train along the circular path = 66 km
In 60 × 60 seconds distance moved = 66 km
∴ In 20 seconds distance moved s = \(\frac{66}{60 \times 60}\) × 20

Radius of the circular metallic plate r = 8 cm
Thickness of the plate h = 6 mm = \(\frac{6}{10}\)
Radius of the Pie l = 16 cm
Thickness of the Pie ( h) = 4mm = \(\frac{4}{10}\) cm
Given volume of the cylinder = Volume of the sector

(i) sin(480°) = sin(360° + 120°) = sin 120°
= sin(90° + 30°) = cos 30° = \(\sqrt{3}\)/2
(ii) sin(-1110°) = -sin(1110°)
= – sin (360° × 3 + 30°)
= -sin 30° = -1/2
(iii) cos(300°) = cos(270° + 30°) = sin 30° = 1/2
(iv) tan (1050°)
tan (1050°) = tan(12 × 90 – 30°)
= – tan30° = – \(\frac{1}{\sqrt{3}}\)
(v) cot 660°
cot 660° = cot (7 × 90 + 30°)
= – tan 30° = – \(\frac{1}{\sqrt{3}}\)
(vi) tan \(\left(\frac{19 \pi}{3}\right)\)
(vii) sin \(\left(\frac{-11 \pi}{3}\right)\)

(i) cos θ = –\(\frac{1}{2}\), θ lies in the III quadrant
(ii) cos θ = \(\frac{2}{3}\), θ lies in the I quadrant
We know that cos 2 θ + sin 2 θ = 1
\(\left(\frac{2}{3}\right)^{2}\) + sin 2 θ = 1
\(\frac{4}{9}\) + sin 2 θ = 1
Since θ lies in the I quadrant all trigonometric functions are positive.
(iii) sin θ = –\(\frac{2}{3}\), θ lies in the IV quadrant
We know that cos 2 θ + sin 2 θ = 1
cos 2 θ + \(\left(-\frac{2}{3}\right)^{2}\) = 1
cos 2 θ + \(\frac{4}{9}\) = 1
Since θ lies in the fourth quadrant cos θ is positive.
(iv) tan θ = – 2, θ lies in the II quadrant
We know that sec 2 θ – tan 2 θ = 1
sec 2 θ – (-2) 2 = 1
sec 2 θ – 4 = 1
sec 2 θ = 1 + 4 = 5
sec θ = ± √5
Since θ lies in the second quadrant sec θ is negative.
(v) sec θ = \(\frac{13}{5}\), θ lies inthe IVquadrant
We know that sec 2 θ – tan 2 θ = 1
\(\left(\frac{13}{5}\right)^{2}\) – tan 2 θ = 1
\(\frac{169}{25}\) – 1 = tan 2 θ

sin 2 θ = \(\frac{3}{4}\) ⇒ sin θ = ± \(\frac{\sqrt{3}}{2}\)
sin 60° = \(\frac{\sqrt{3}}{2}\)
sin 120° = sin (180° – 60°)
= sin 60° = \(\frac{\sqrt{3}}{2}\)
∴ θ = 60° and 120°

= sin 2 10° + sin 2 20° + [cos 20°] 2 + [cos 10°] 2
= sin 2 10° + sin 2 20° + cos 2 20° + cos 2 10°
= sin 2 10° + cos 2 10° + sin 2 20° + cos 2 20°
= 1 + 1 = 2

Given sin x = \(\frac{15}{17}\), 0 < x < \(\frac{\pi}{2}\)
we have cos 2 x + sin 2 x = 1
∴ cos 2 x = 1 – sin 2 x
= 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 cos 2 y + sin 2 y = 1
sin 2 y = 1 – cos 2 y
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)

Given sin A = \(\frac{3}{5}\) 0 < A < \(\frac{\pi}{2}\)
we have sin 2 A + cos 2 A = 1
cos 2 A = 1 – sin 2 A
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 cos 2 B + sin 2 B = 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

Given cos x = –\(\frac{4}{5}\), π < x < \(\frac{3 \pi}{2}\)
we have cos 2 x + sin 2 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 = –\(\frac{3}{5}\)

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 \(\frac{7 \pi}{12}\)

(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

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
x 2 – (cos 15° + sin 15°)x + (cos 15°) (sin 15°) = 0 ——— (3)
Substituting in equation (3) we have

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

(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

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.

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°)

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

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

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

= 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

LHS = sin (A + B) sin (A – B)
= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)
= sin 2 A cos 2 B – cos 2 A sin 2 B
= sin 2 A (1 – sin 2 B) – (1 – sin 2 A) sin 2 B
= sin 2 A – sin 2 A sin 2 B – sin 2 B + sin 2 A sin 2 B
= sin 2 A – sin 2 B = RHS
(ii) cos (A + B). cos (A – B) = cos 2 A – sin 2 B = cos 2 B – 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 = cos 2 A cos 2 B – sin 2 A sin 2 B
= cos 2 A(1 – sin 2 B) – (1 – cos 2 A) sin 2 B
= cos 2 A – cos 2 A sin 2 B – sin 2 B + cos 2 A sin 2 B
cos(A + B). cos(A – B) = cos 2 A – sin 2 B
Also cos(A + B). cos(A – B) = cos 2 A cos 2 B – sin 2 A sin 2 B
= (1 – sin 2 A)cos 2 B – sin 2 A(1 – cos 2 B)
= cos 2 B – sin 2 A cos 2 B – sin 2 A + sin 2 A cos 2 B
cos(A + B). cos(A – B) = cos 2 B – sin 2 A
(iii) sin 2 (A + B) – sin 2 (A – B) = sin 2 A sin 2 B
sin 2 A – sin 2 B = sin (A + B) sin (A – B)
LHS = sin 2 (A + B) – sin 2 (A – B) = sin [(A + B) + (A – B)] [sin (A + B) – (A – B)]
= sin 2A sin 2B = RHS
(iv) cos 8θ. cos 2θ = cos 2 5θ – sin 2 3θ
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
= cos 2 5θ cos 2 3θ – sin 2 5θ sin 2 3θ
= cos 2 5θ (1 – sin 2 3θ) – (1 – cos 2 5θ) sin 2 3θ
= cos 2 5θ – cos 2 5θ sin 2 3θ – sin 2 3θ + cos 2 5θ sin 2 3θ
cos 8θ. cos 2θ = cos 2 5θ – sin 2 3θ

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

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)+
(cos 2 α + sin 2 α) + (cos 2 β + sin 2 β) + (cos 2 γ + sin 2 γ) = 0
(cos 2 α + cos 2 β + cos 2 γ + 2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α) + (sin 2 α + sin 2 β + sin 2 γ + 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

we know sin 2 A + cos 2 A = 1
sin 2 A = 1 – cos 2 A
Since A lies in the first quadrant, sin A is positive
∴ sin A = \(\frac{8}{17}\)
cos 2A = cos 2 A – sin 2 A
(ii) sin A = \(\frac{4}{5}\)
we know sin 2 A + cos 2 A = 1
cos 2 A = 1 – sin 2 A
Since A lies in the first quadrant, cos A is positive
∴ cos A = \(\frac{3}{5}\)
cos 2A = cos 2 A – sin 2 A
(iii) tan A = \(\frac{16}{63}\)

Given A + B = 45°
tan(A + B) = tan 45°
tan A + tan B = 1 – tan A. tan B —— (1)
(1 + tan A)(1 + tan B) = 1 + tan B + tan A + tan A tan B
= 1 + (tan A + tan B) + tan A tan B
= 1 + 1 – tan A tan B + tan A tan B (By equation (1))
= 2

1 + tan 44° = 1 + tan (45° – 1°)
(1 + tan 1°)(1 + tan 44°) = 2
Similarly (1 + tan 2°) (1 + tan 43°) = 2
(1 + tan 3°) (1 + tan 42°) = 2
(1 + tan 22°) (1 + tan 23°) = 2
= (1 + tan 1°) (1 + tan 2°)… (1 + tan 44°) = 2 × 2 × … 22 times
It is a multiple of 4.

(i) sin 35°. cos 28°
We know
sin A cos B = \(\frac { 1 }{ 2 }\) [sin (A + B) + sin (A – B)]
Take A = 35° and B = 28°
sin 35°cos 28° = \(\frac { 1 }{ 2 }\)[sin(35° + 28°) + sin(35° – 28°)]
sin 350 cos 28° = \(\frac { 1 }{ 2 }\)[sin 63° + sin 7°]
(ii) sin 4x cos 2x
We know
sin A cos B = \(\frac { 1 }{ 2 }\) [sin (A + B) + sin (A – B )]
Take A = 4x, B = 2x
sin 4x. cos 2x = \(\frac { 1 }{ 2 }\)[sin(4x + 2x) + sin(4x – 2x)]
sin 4x. cos 2x = \(\frac { 1 }{ 2 }\)[sin 6x + sin 2x]
(iii) 2 sin 10θ. cos 2θ
We know
2 sin A cos B = sin (A + B) + sin (A – B)
Take A = 10θ, B = 2θ
2 sin 10θ. cos 2θ = sin (10θ + 2θ) + sin (10θ – 2θ)
2 sin 10θ. cos 2θ = sin 12 θ + sin 8θ
2 sin 10θ. cos 2θ = \(\frac { 1 }{ 2 }\)[sin 12θ + sin 8θ]
(iv) cos 5θ. cos 2θ
We know.
cosA cosB = \(\frac { 1 }{ 2 }\) [cos (A + B) + cos (A – B)]
Take A = 5θ, B = 2θ
cos 5θ. cos 2θ = \(\frac { 1 }{ 2 }\) [cos (5θ + 2θ) + cos(5θ – 2θ)]
cos 5θ. cos 2θ = \(\frac { 1 }{ 2 }\) [cos 7θ + cos 3θ]
(v) sin 5θ. sin 4θ
we know
sin A sin B = \(\frac { 1 }{ 2 }\) [cos (A – B) – cos (A + B)]
Take A = 5θ, B = 4θ
sin 5θ. sin 4θ = \(\frac { 1 }{ 2 }\) [cos (5θ – 4θ) – cos (5θ + 4θ)]
sin 5θ. sin 4θ = \(\frac { 1 }{ 2 }\) [cos θ – cos 9θ]

sin 12°. sin 48°. sin 54° = sin 48°. sin 12°. sin (90° – 36°)
= \(\frac { 1 }{ 2 }\) [cos (48° – 12°) – cos (48° + 12°)] cos 36°
= \(\frac { 1 }{ 2 }\) [cos 36° – cos 6o°] cos 36°
= \(\frac { 1 }{ 2 }\) [cos 36° – \(\frac { 1 }{ 2 }\)] cos 36°
= \(\frac { 1 }{ 2 }\) [cos 2 36° – \(\frac { 1 }{ 2 }\) cos 36°]

sin x + sin 2x + sin 3x = sin x + 2 sin x cos x + 3 sin x – 4 sin 3 x
= sin x [1 + 2 cos x + 3 – 4 sin 2 x]
= sin x [2 cos x + 4 – 4 sin 2 x ]
= sin x [2 cosx + 4(1 – sin 2 x)]
= sin x [2 cos x + 4 cos 2 x]
= 2 sin x cos x [1 + 2 cos x]
= sin 2x (1 + 2 cosx)

4 cos x cos 2x. cos 3x = 4 cos x. cos 3x. cos 2x = 4 cos x. [cos (3x + 2x) + cos (3x – 2x)]
2 cos x. [cos 5x + cos x] = 2 cos 5x. cos x + 2 cos 2 x
= 2 × \(\frac { 1 }{ 2 }\) [cos (5x + x) + cos (5x – x)] + 1 + cos 2x
= cos 6x + cos 4x + 1 + cos 2x
= 1 + cos 2x + cos 4x + cos 6x

cos(30° – A) cos(30° + A) + cos(45° – A). cos(45° + A)
= cos (30° + A) cos (30°- A) + cos (45° + A) cos (45° – A)
= \(\frac { 1 }{ 2 }\) [cos (30° + A + 30° – A) + cos ( 30° + A – (30° + A ))] + \(\frac { 1 }{ 2 }\) [cos (45° + A + 45° – A) + cos (45° + A – (450 + A))
= \(\frac { 1 }{ 2 }\) [cos 60° + cos (30° + A – 30° + A)] + \(\frac { 1 }{ 2 }\)[cos 90° + cos(45° + A – 45° + A)]
= \(\frac { 1 }{ 2 }\)[cos 60° + cos 2A] + \(\frac { 1 }{ 2 }\)[cos 90° + 2A]
= \(\frac { 1 }{ 2 }\) cos 60° + \(\frac { 1 }{ 2 }\) cos 2A + \(\frac { 1 }{ 2 }\) cos 90° + \(\frac { 1 }{ 2 }\) cos 2A
= \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) + cos 2A + \(\frac { 1 }{ 2 }\) × o
= \(\frac { 1 }{ 4 }\) + cos 2A

sin 2A + sin 2 B + sin 2 C = 2 sin \(\left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right)\). cos \(\left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)\) + 2 sin C cos C
= 2 sin (A + B). cos (A – B) + 2 sin C cos C
= 2 sin( 180° – C) cos (A – B) + 2 sin C cos C (∴ A + B + C = 180°)
= 2 sin C cos (A – B) + 2 sin C cos C
= 2 sin C (cos(A – B) + cos C)
= 2 sin C [(cos (A – B) + cos (180° – (A + B))]
= 2 sin C [cos (A – B) – cos (A + B)]
= 2 sin C. 2 sin A. sin B
= 4 sin A sin B sin C
(ii) cos A + cos B – cos C = -1 + 4 cos \(\frac{\mathbf{A}}{2}\) cos \(\frac{B}{2}\) sin \(\frac{\mathbf{c}}{2}\)
(iii) sin 2A + sin 2B + sin 2C = 2 + 2 cos A cos B cos C
(iv) sin 2A + sin 2B – sin 2C = 2 sin A sin B cos C
Given A + B + C = 180° ⇒ c = 180° – (A + B)
(v)
Given A + B + C = 180° ⇒ A + B = 180° – C
(vi) sin A + sin B + sin C = 4 cos \(\frac{\mathbf{A}}{2}\) cos \(\frac{\mathbf{B}}{2}\) cos \(\frac{\mathbf{c}}{2}\)
(vii) sin(B + C – A) + sin(C + A – B) + sin(A + B – C) = 4 sin A sin B sin C
Given A + B + C = 180°
sin ( B + C – A) = sin (180° – A – A) = sin (180° – 2A) = sin 2A
sin ( C + A – B) = sin (180° – B – B) = sin ( 180° – 2B) = sin 2B
sin (A + B – C) = sin (180° – C – C) = sin ( 180° – 2C) = sin 2C
∴ sin ( B + C – A) + sin ( C + A – B) + sin (A + B – C) = sin 2A + sin 2B + sin 2C
= 2 sin \(\left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right)\). cos \(\left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)\) + 2 sin C. cos C
= 2 sin(A + B). cos(A – B) + 2 sin C cos C
= 2 sin(180° – C) cos(A – B) + 2 sin C cos C
= 2 sin C cos(A – B) + 2 sin C cos C
= 2sin C(cos(A – B) + cos C)
= 2 sin C [cos(A – B) + cos (180° -(A + B))]
= 2 sin C [cos(A – B) – cos(A + B)]

Given A + B + C = 2s, we have sin A sin B = \(\frac { 1 }{ 2 }\) [cos ( A – B) – cos ( A + B)]
sin(s – A) sin(s – B) + sins. sin(s – C) = \(\frac { 1 }{ 2 }\) [cos((s – A) – (s – B)) – cos(s – A + s – B)] + \(\frac { 1 }{ 2 }\) [cos (s – (s – C)) – cos (s + s – C)]
= \(\frac { 1 }{ 2 }\) [cos (s – A – s + B) – cos(2s – A – B)] + \(\frac { 1 }{ 2 }\) [cos(s – s + C) – cos(2s – C)]
= \(\frac { 1 }{ 2 }\) [cos (B – A) – cos(2s – A – B)] + \(\frac { 1 }{ 2 }\) [cos C – cos(2s – C)]
= \(\frac { 1 }{ 2 }\) [cos(A – B) – cos(A + B + C – A – B)] + \(\frac { 1 }{ 2 }\) [cos C – cos (A + B + C – C)]
= \(\frac { 1 }{ 2 }\) [cos (A – B) – cos C] + \(\frac { 1 }{ 2 }\) [cos C – cos(A + B)]
= \(\frac { 1 }{ 2 }\) [cos (A – B) – cos C + cos C – cos (A + B)] = \(\frac { 1 }{ 2 }\) [cos(A – B) – cos(A + B)]

Given x + y + z = x y z,
Let x = tan A,
y = tan B,
z = tan C
x + y + z = xyz ⇒ tan A + tan B + tan C = tan A tan B tan C
tan A + tan B + tan C – tan A tan B tan C = 0 ——– (1)
tan A + tan B + tan C – tan A tan B tan C = 0 ——– (2)
A + B + C = 180° ⇒ tan A + tan B + tan C – tan A tan B tan C = 0
⇒ tan (A + B + C) = 0
⇒ tan A + tan B + tan C – tan A tan B tan C = 0
⇒ tan (A + B + C) = 0
∴ tan 2(A + B + C) = o
⇒ tan (2A + 2B + 2C) = 0
⇒ tan 2A + tan 2B + tan 2C – tan 2A tan 2B tan 2C = 0 By eqn (1)
tan 2 A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

∆ ABC is a right triangle. Given ∠A = 90°
we know A + B + C = 180°
∴ B + C = 180° – A
B + C = 180° – 90° = 90°
(ii) sin 2 B + sin 2 C = 1
(iii) cos B – cos C = -1 + 2√2 cos \(\frac{\mathbf{B}}{2}\). sin \(\frac{\mathbf{C}}{2}\)

(i) sin θ = – \(\frac{1}{\sqrt{2}}\)
We know that principal of sin θ lies in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
sin θ = – \(\frac{1}{\sqrt{2}}\) < 0
∴ The principal value of sin θ lies in the IV quadrant.
sin θ = – \(\frac{1}{\sqrt{2}}\)
= – sin \(\left(\frac{\pi}{4}\right)\)
sin o = sin \(\left(-\frac{\pi}{4}\right)\)
Hence θ = \(-\frac{\pi}{4}\) is the principal solution.
The general solution is
θ = nπ + (- 1) n. \(\left(-\frac{\pi}{4}\right)\), n ∈ Z
θ = nπ + (- 1) n + 1. \(\frac{\pi}{4}\), n ∈ Z
(ii) cot θ = √3
The principal value of tan θ lies in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
Since tan θ = \(\frac{1}{\sqrt{3}}\) > 0
The principal value of tan θ lies in the I quadrant.
The general solution of tan θ is
θ = nπ + \(\frac{\pi}{6}\), n ∈ Z
(iii) tan θ = –\(\frac{1}{\sqrt{3}}\)
The principal value of tan θ lies in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
Since tan θ = – \(\frac{1}{\sqrt{3}}\) > 0
The principal value of tan θ lies in the IV quadrant.
The general solution of tan θ is
θ = nπ – \(\frac{\pi}{6}\), n ∈ Z

sin 4 x – sin 2 x = 0
sin 2 x (sin 2 x – 1) = 0
sin 2 x [ – (1 – sin 2 x)] = 0
sin 2 x × – cos 2 x = 0
– sin 2 x cos 2 x = 0
(sin x cos x) 2 = 0
(\(\frac { 1 }{ 2 }\) × 2 sin cos x) 2 = 0
\(\frac { 1 }{ 4 }\) sin 2x = 0
sin 2x = 0
The general solution is
2x = nπ, n ∈ Z
x = \(\frac{\mathrm{n} \pi}{2}\), n ∈ Z
(ii) 2 cos 2 x + 1 = – 3 cos x
2 cos 2 x + 1 = – 3 cos x
2 cos 2 x + 3 cos x + 1 = 0
2 cos 2 x + 2 cos x + cos x + 1 = 0
2 cos x (cos x + 1) + 1 (cos x + 1) = 0
(2 cos x + 1) (cos x + 1) = 0
2 cos x + 1 = 0 or cos x + 1 = 0
cos x = \(-\frac{1}{2}\) or cos x = – 1
To find the solution of cos x = \(-\frac{1}{2}\)
cos x = \(-\frac{1}{2}\)
To find the solution of cos x = – 1
cos x = – 1
cos x = cos π
The general solution is
x = 2nπ ± π, n ∈ Z
x = 2nπ + π or x = 2nπ – π, n ∈ Z
Consider x = 2nπ + π
when n = 0, x = 0 + π = π ∈ (0°, 360°)
when n = 1, x = 2π + π = 3π ∉ (0°, 360°)
Consider x = 2nπ – π
when n = 0, x = 0 – π ∉ (0°, 360°)
when n = 1, x = 2π – π = π ∈ (0°, 360°)
when n = 2, x = 4π – π = 3π ∉ (0°, 360°)
∴ The required solution are x = \(\frac{2 \pi}{3}\), \(\frac{4 \pi}{3}\), π
(iii) 2 sin 2 x + 1 = 3 sin x
2 sin 2 x – 3 sin x + 1 = 0
2 sin 2 x – 2 sin x – sin x + 1 = 0
2 sin x (sin x – 1) – 1 (sin x – 1) = 0
(2 sin x – 1)(sin x – 1) = 0
2 sin x – 1 = 0 or sin x – 1 = 0
sin x = \(\frac { 1 }{ 2 }\) or sin x = 1
To find the solution of sin x = \(\frac { 1 }{ 2 }\)
sin x = \(\frac { 1 }{ 2 }\)
sin x = sin \(\left(\frac{\pi}{6}\right)\)
The general solution is x = nπ + (-1) n \(\frac{\pi}{6}\), n ∈ z
(iv) cos 2 x = 1 – 3 sin x
1 – 2 sin 2 x = 1 – 3 sinx
2 sin 2 x – 3 sin x = 0
sin x(2 sin x – 3) = 0smx =
sin x = 0 or 2 sin x – 3 = 0
sin x = 0 or sin x = \(\frac{3}{2}\)
sin x = \(\frac{3}{2}\) is not possible since sin x ≤ 1
∴ sin x = 0 = sin 0
The general solution is x = nit,
When n = 0, x = 0 ∉ (0°, 360°)
When n = 1, x = π ∈ (0°, 360°)
When n = 2, x = 2π ∉ (0°, 360°)
∴ The required solutions is x = π

2 cos 3 x. sin 2x = cos 3 x
2 cos 3x. sin 2x – cos3x = 0
cos 3x (2 sin 2x – 1) = 0
cos 3x = 0 or 2 sin 2x – 1 = 0
cos 3x = 0 or sin 2x = \(\frac { 1 }{ 2 }\)
To find the general solution of cos 3x = 0
The general solution of cos 3x = 0 is
3x = (2n + 1)\(\frac{\pi}{2}\), n ∈ Z
x = (2n + 1)\(\frac{\pi}{6}\), n ∈ Z
To find the general solution of sin 2x = \(\frac{1}{2}\)
sin 2x = \(\frac{1}{2}\)
sin 2x = sin \(\left(\frac{\pi}{6}\right)\)
The general solution is
∴ The required solutions are
(ii) 2 cos 2 θ + 3 sin θ – 3 = θ
2 cos 2 θ + 3 sin θ – 3 = θ
2(1 – sin 2 θ)+ 3 sin θ – 3 = θ
2 – 2 sin 2 θ + 3 sin θ – 3 = θ
– 2 sin 2 θ + 3 sin θ – 1 = θ
2 sin 2 θ – 3 sin θ + 1 = θ
2 sin 2 θ – 2 sin θ – sin θ + 1 = θ
2 sin θ (sin θ – 1) – (sin θ – 1) = θ
(2 sin θ – 1) (sin θ – 1) = 0
2 sin θ – 1 = 0 or sin θ – 1 = θ
sin θ = \(\frac { 1 }{ 2 }\) or sin θ = 1
To find the general solution of’ sin θ = \(\frac { 1 }{ 2 }\)
sin θ = \(\frac { 1 }{ 2 }\)
sin θ = sin \(\frac{\pi}{6}\)
The general solution is θ = nπ + (-1) n \(\frac{\pi}{6}\), n ∈ Z
To find the general solution of sin θ = 1
sin θ = 1
sin θ = \(\frac{\pi}{2}\)
The general solution is θ = nπ + (-1) n \(\frac{\pi}{2}\), n ∈ Z
∴ The required solutions are
θ = nπ + (-1) n \(\frac{\pi}{6}\), n ∈ Z (or)
θ = nπ + (-1) n \(\frac{\pi}{6}\), n ∈ Z
(iii) cos θ + cos 3θ = 2 cos 2θ
cos 3θ + cos θ = 2 cos 2θ
2 cos 2θ. cos θ = 2 cos 2θ
cos 2θ. cos θ – cos 2θ = θ
cos 2θ (cos θ – 1) = θ
cos 2θ = θ or cos θ – 1 = θ
cos 2θ = θ or cos θ = 1
To find the general solution of cos 2θ = θ
The general solution is
2θ = (2n + 1)\(\frac{\pi}{2}\), n ∈ Z
θ = (2n + 1)\(\frac{\pi}{4}\), n ∈ Z
To find the general solution of cos θ = 1
cos θ = 1
cos θ = cos 0
The general solution is θ = 2nπ, n ∈ Z
∴ The required solutions are
θ = (2n + 1)\(\frac{\pi}{4}\), n ∈ Z (or)
θ = 2nπ, n ∈ Z
(iv) sin θ + sin 3θ + sin 5θ = 0
sin 5θ + sin 3θ + sin θ = 0
2 sin 3θ. cos 2θ + sin 3θ = 0
sin 3θ (2 cos 2θ + 1) = θ
sin 3θ = 0 or 2 cos 2θ + 1 = θ
sin 3θ = 0 or cos 2θ = –\(\frac { 1 }{ 2 }\)
To find the general solution of sin 3θ = 0
The general solution is
3θ = nπ, n ∈ Z
θ = \(\frac{\mathbf{n} \pi}{3}\), n ∈ Z
To find the general solution of cos 2θ = –\(\frac { 1 }{ 2 }\)
The general solution is
∴ The required solutions are
(v) sin 2θ – cos 2θ – sin θ + cos θ = θ
(vi) sin θ + cos θ = √2
The general solution is
(vii) sin θ + √3 cos θ = 1
Divide each term by 2
(viii) cot θ + cosec θ = √3
(ix) tan θ + tan \(\left(\theta+\frac{\pi}{3}\right)\) + tan \(\left(\theta+\frac{2 \pi}{3}\right)\) = √3
(x) cos 2θ = \(\frac{\sqrt{5}+1}{4}\)
we know cos 36° = \(\frac{\sqrt{5}+1}{4}\), 36° = \(\frac{\pi}{5}\)
cos 2θ = cos 36° = cos \(\left(\frac{\pi}{5}\right)\)
The general solution is
2θ = 2nπ ± \(\frac{\pi}{5}\), n ∈ Z
θ = nπ ± \(\frac{\pi}{10}\), n ∈ Z
(xi) 2cos 2 x – 7 cos x + 3 = 0
2 cos 2 x – 7 cos x + 3 = 0
2 cos 2 x – 6 cos x – cos x + 3 = 0
2 cos x (cos x – 3) – 1 (cos x – 3) = 0
(2 cos x – 1) (cos x – 3) = 0
2 cos x – 1 = 0 or cos x – 3 = 0
cos x = \(\frac { 1 }{ 2 }\) or cos x = 3
Since – 1 ≤ cos x ≤ 1, we have
cos x = 3 is not possible.
∴ cos x = \(\frac { 1 }{ 2 }\)
cos x = cos \(\frac{\pi}{3}\)
The general solution is x = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z

\(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\)
sin A. sin (B – C) = sin C. sin (A – B)
sin (180° – (B + C)). sin (B – C) = sin (180° – (A + B)). sin (A – B)
sin (B + C) sin (B – C) = sin (A + B) sin (A – B) ——— (1)
sin(B + C). sin(B – C) = (sin B cos C + cos B sin C) × (sin B cos C – cos B sin C)
= (sin B cos C) 2 – (cos B sin C) 2
= sin 2 B cos 2 C – cos 2 B sin 2 C
= sin 2 B (1 – sin 2 C) – (1 – sin 2 B) sin 2 C
= sin 2 B – sin 2 B sin 2 C – sin 2 C + sin 2 B sin 2 C
sin ( B + C). sin ( B – C) = sin 2 B – sin 2 C
Similarly,
sin (A + B ). sin (A – B) = sin 2 A – sin 2 B
(1) ⇒ sin 2 B – sin 2 C = sin 2 A – sin 2 B
sin 2 B + sin 2 B = sin 2 A + sin 2 C
2 sin 2 B = sin 2 A + sin 2 C ——— (2)

Given that the angles A, B, C are in A. P.
∴ 2B = A + C
Also A + B + C = 180°
B + (A + C) = 180°
B + 2B = 180°
3B = 180° ⇒ B = 60°
A + C = 2B = 2 × 60° = 120°
A + 45° = 120°
A = 120° – 45° = 75°
A = 75°

a 2 + b 2 – c 2 = a 2
b 2 – c 2 = 0
b 2 = c 2 ⇒ b = c
Two sides of is ∆ ABC are equal.
∴ ∆ ABC is an isosceles triangle.

LHS = a cos A+ 6 cos B + c cos C
Using sine formula, we get k sin A cos A + k sin B cos B + k sin C cos C k
= \(\frac{k}{2}\) [2 sin A cos A + 2 sin B cos B + 2 sin C cos C]
= \(\frac{k}{2}\) [sin 2A + sin 2B + sin 2C]
= \(\frac{k}{2}\) [2 sin (A + B). cos (A – B) + 2 sin C. cos C]
= \(\frac{k}{2}\) [2 sin (A – B). cos (A – B) + 2 sin C. cos C]
= \(\frac{k}{2}\) [2 sin C. cos (A – B) + 2 sin C. cos C]
= k sin C [cos(A – B) + cos C]
= k sin C [cos (A – B) – cos (A + B)]
= k sin C. 2 sin A sin B
= 2k sin A. sin B sin C
= 2a sin B sin C
= RHS

Let ∆ A B C be the triangular-shaped park.
a, b, c be the length of the sides.
Given perimeter of the park = 120 m
2s = a + b + c = 120m —– (1)
For a fixed perimeter 2s. the area of a triangle is maximum when a = b = c.
(1) = a + a + a = 120
3a = 120
⇒ a = 40m
Length of the sides 40 m, 40 m, 40 rn.

Let a, b, c be the lengths of the sides of the triangle.
Given the perimeter of the triangle
2s = a + b + c = 42m —-—(1)
For a fixed perimeter 2 s, the area of a triangle is maximum
when a = b = c.
(1) ⇒ a + a + a = 42
3a = 42 ⇒ a = \(\frac{42}{3}\)
a = 14m
∴ a = b = c = 14m
∴ The dimensions of the triangle are 14 m, 14 m, 14 m.
Maximum area = 49√3 sq.m.

To prove (a) a = b cos C + c cos B
(b) b = c cos A + a cos C
(c) c = a cos B + b cos A
(i) Using the Law of sines,
(a) b cos C + c cos B = 2 R sin B cos C + 2 R sin C cos B
= 2 R ( sin B cos C + cos B sin C)
= 2R sin (B + C)
= 2R sin (180° – A)
b cos C + c cos B = 2R sin A = a
a = b cos C + c cos B
(b) c cos A + a cos C = 2R sin C cos A + 2R sin A cos C
= 2R (sin C cos A + cos C sin A)
= 2R sin(C + A)
= 2R sin(180° – B)
= 2R sin B = b
∴ b = c cos A + a cos C
(c) a cos B + b cosA = 2R sin A cos B + 2R sin B cos A
= 2R (sin A cos B + cos A sin B)
= 2R sin (A + B)
= 2R sin (180° – C)
= 2R sin C = c
∴ c = a cos B + b cos A
(ii) Using Law of cosines.

In a ∆ ABC, Given
a = √3 – 1, b = √3 + 1
C = 60°
Using cosine formula a
C 2 = a 2 + b 2 – 2 ab cos C
= (√3 – 1) 2 + (√3 + 1) 2 – 2(√3 – 1) × (√3 + 1) cos 6o°
= 3 – 2√3 + 1 + 3 + 2√3 + 1 – 2 (3 – 1) × \(\frac { 1 }{ 2 }\)
c 2 = 8 – 2 = 6 ⇒ c = √6
Using sine formula
sin (45° – 30°) = sin 45°. cos 30° – cos 45° sin 30°
From equations (1) and (2), we have
sin A = sin 15° ⇒ A = 15°
In ∆ ABC, we have A + B + C = 180°
15° + B + 60° = 180°
B = 180°- 75°
B = 105°
∴ The required sides and angles are
c = √6, A = 15°, B = 105°

Let A and B be the two positions of the soldiers.
AC – direction of the intruder seen from A.
BC – the direction of the intruder seen from B.
∠ BAC = 30° angle of elevation of the intruder from A.
∠ PBC = 45° angle of elevation of the intruder from B.
Distance between A and B = 5k.m.
In ∆ ABC, ∠ ABC = 180° – 45° = 135°
∠ BCA = 180° – ( 135° + 30°)
= 180° – 165° = 15°
Using sine formula

A – be the easternmost point on the pond and
B – be the westernmost point on the pond.
AB – Width of the pond
P – Point of observation.
The distance of A from P = 8 km
Distance of B from P = 6km
Angle between the directions PA and PB
∠APB = 60°
In ∆ PAB, using cosine formula
AB 2 = PA 2 + PB 2 – 2PA. PB. cos ∠APB
AB 2 = 8 2 + 6 2 – 2 × 8 × 6. cos 60°
= 64 + 36 – 96 × \(\frac { 1 }{ 2 }\)
= 100 – 48 = 52
AB = \(\sqrt{52}\) = \(\sqrt{4 \times 13}\)
AB = 2\(\sqrt{13}\) k.m.
Width of the pond = 2\(\sqrt{13}\) k.m

A, B are the positions of the helicopter above the sea level.
Distance between A and B = 10 km
C – Position of the boat on the surface of sea.
AC, BC are the directions of the boat as seen from A and B respectively.
Distance of the boat C from A = 6 k.m
∠ ACB = 60°
Using cosine formula
AB 2 = BC 2 + AC 2 – 2 BC. AC cos ∠ACB
c 2 = a 2 + b 2 – 2 ab cos C
10 2 = a 2 + 6 2 – 2a × 6 cos 60°
100 = a 2 + 36 – 12a\(\left(\frac{1}{2}\right)\)
0 = a 2 + 36 – 6a – 100
a 2 – 6a – 64 = 0

p 2 = a 2 + b 2 – 2ab cos P
p 2 = 9 + 25 – 30 Cos 120°
p 2 = 9 + 25 – 30 (-1/2) = 34 + 15 = 49
⇒ p = \(\sqrt{49}\) = 7 km

Let ∆ ABC be the shape of the land.
Given AB = 120 ft, AC = 60ft
∠ BAC = 60°
Using cosine formula in ∆ ABC
BC 2 = AB 2 + AC 2 – 2AB. AC cos ¿BAC
BC 2 = 120 2 + 60 2 – 2 × 120 × 60 cos (60°)
= 14400 + 3600 – 14400 × \(\frac { 1 }{ 2 }\)
= 18000 – 7200
BC 2 = 10800 = 100 × 2 × 2 × 3 × 3 × 3
BC 2 = 10 2 × 2 2 × 3 2 × 3
BC = \(\sqrt{10^{2} \times 2^{2} \times 3^{2} \times 3}\)
BC = 10 × 2 × 3√3
BC = 60√3 k.m.
Perimeter of the Land = AB + BC + AC
= 120 + 60√3 + 60
= 180 + 60√3
= 60 (3 + √3) feet.
Area of ∆ ABC = \(\frac { 1 }{ 2 }\) × AB × AC × sin ∠ BAC
= \(\frac { 1 }{ 2 }\) × 60 × 120 sin 60°
= 30 × 120 × \(\frac{\sqrt{3}}{2}\)
= 30 × 60 × √3
= 1800 √3 sq. feet.
Cost of 1 sq. feet Rs. 500
∴ Cost of 800 √3 sq. feet = 800 √3 × 500 = 900000√3
Total amount needed = Rs. 900000√3
Perimeter of the land = 60(3 + √3)feet.

Let A be the position of the jet fighter observing the target at an angle of depression 30°.
Also, Let B be the position of the jet 100 k.m away horizontally from A observing the target at an angle of depression 45°.
In ∆ TAB, AB = 100 k.m
∠TAB = 30°
∠ABT = 180°- 45° = 135°
∠ATB = 180° – (135°+ 300) = 180° – 165° = 15°

A, B are the two landmarks,
C – Position of the plane.
The distance of the plane from the landmark A = 1 k.m
The distance of the plane from the landmark B = 2 k.m
∠ACB = 45°
From the ∆ ABC, using cosine formula
AB 2 = AC 2 + BC 2 – 2AC. BC. cos45°
= 1 2 + 2 2 – 2 × 1 × 2
AB 2 = 1 + 4 – 2 × √2 = 5 – 2√2
AB = \(\sqrt{5-2 \sqrt{2}}\)
Distance between the landmarks AB = \(\sqrt{5-2 \sqrt{2}}\) km.

Given In ∆ABC
AC = 4 k.m
∠A = 60°,
∠B = 45°
∠C = 180° – (60° + 45°)
∴ ∠C = 180° – 105° = 75°
Using sine formula

P – Initial point.
PA – The direction of the first vehicle travels with speed km/hr.
PB – The direction of the second vehicle travels with a speed of 80km/hr.
Given in half an hour first vehicle reaches destination A.
∴ PA = \(\frac{60}{2}\) = 30 km.
Also in half an hour the second vehicle reaches the destination B.
∴ PA = \(\frac{80}{2}\) = km.
In ∆ PAB, PA = 30, PB = 40, ∠APB = 60°
Using cosine formula
AB 2 = PA 2 + PB 2 – 2PA PB cos ∠APB
AB = 30 2 + 40 2 – 2 × 30 × 40 cos 60°
= 900 + 1600 – 2400 × \(\frac { 1 }{ 2 }\)
= 2500 – 1200
AB 2 = 1300
AB = \(\sqrt{1300}\) = \(\sqrt{13 \times 100}\)
AB = 10√13 k.m.

O – Centre of Earth,
A – Position of Earth station.
S – Position of the satellite.
Given the radius of Earth
OA = r
The angle of elevation of the satellite from the Earth station = 30°
The distance of the satellite from the Earth station AS = d
The distance of the satellite from the centre of the Earth OS = R.
Angle subtended by the line segment AS at the centre of earth ∠AOS = α
In △ AOS, OA = r, AS = d, OS = R, ∠AOS = α
Using cosine formula
AS 2 = OA 2 + OS 2 – 2 OA. OS cos ∠AOS
d 2 = r 2 + R 2 – 2(r) (R) cos α

Given Width of the Road = x meter
Diameter of the signal AB = a meter
Height of the signal from the eye level = b meter
In ∆ ADC, DC = x, AC = AB + BC = a + b

(1) \(\frac{\mathbf{k}^{3}}{\sqrt{2}}\)
Explaination:
cos 28° + sin 28° = k 3
cos 28° + sin (90° – 62°) = k 3
cos 28° + cos 62° = k 3
2 cos 45°. cos 17° = k 3
2 × \(\frac{1}{\sqrt{2}}\) cos 17° = k 3
√2 cos 17° = k 3
cos 17° = \(\frac{\mathrm{k}^{3}}{\sqrt{2}}\)

(1) 4 + √2
Explaination:
4 sin 2 x + 3 cos 2 x + sin \(\frac{x}{2}\) + cos \(\frac{x}{2}\)
= sin 2 x + 3 sin 2 x + 3 cos 2 x + sin \(\frac{x}{2}\) + cos \(\frac{x}{2}\)
= sin 2 x + 3(sin 2 x + cos 2 x) + sin \(\frac{x}{2}\) + cos \(\frac{x}{2}\)
= 3 + sin 2 x + 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 sin 2 x = 1
Maximum value is obtained when x = \(\frac{\pi}{2}\)
∴ (1) ⇒ 4 sin 2 x + 3 cos 2 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

(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

(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

(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.

(2) cos 2 (8 + Φ)
Explaination:
cos 2θ cos 2Φ + sin 2 (θ – Φ) – sin 2 (θ + Φ)
= cos 2θ cos 2Φ – sin 2θ sin 2Φ
= cos(2θ + 2Φ)
= cos 2(θ + Φ)

(3) right triangle
Explaination:
On simplifying we get
sin 2 A + sin 2 B + sin 2 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).

(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.

(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

(1) b 2 – 1, if b ≤ √2
Explaination:
sin α + cos α = b
(sin α + cos α) 2 = b 2
sinv α + cos 2 α + 2 sin α cos α = b 2
1 + sin 2α = b 2
sin 2α = b 2 – 1
But – 1 ≤ sin 2α ≤ I
– 1 ≤ b 2 – 1 ≤ 1
b 2 – 1 ≤ 1 ⇒ b 2 ≤ 2
⇒ b ≤ √2
∴ sin 2α = b 2 – 1 if b ≤ √2

(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