Trigonometry Assignment: Detailed Solutions for Tasks 1, 2, and 3

This document contains solutions for a trigonometry assignment, covering a range of topics. The solutions include detailed steps for solving problems related to trigonometric functions, identities, and angle calculations. Task 1 involves calculating trigonometric ratios given specific values, as well as solving problems involving angles of depression. Task 2 focuses on trigonometric identities and their application. Task 3 incorporates trigonometric functions and their graphical representations. The assignment covers various trigonometric concepts, providing a comprehensive guide for students seeking to understand and solve related problems. The document provides a breakdown of the solutions, making it a valuable resource for students studying trigonometry.

Trigonometry

TABLE OF CONTENTS
TASK 1............................................................................................................................................1
AC 1.1..........................................................................................................................................1
AC 1.2..........................................................................................................................................3
TASK 2............................................................................................................................................3
AC 2.1 .........................................................................................................................................3
AC 2.2..........................................................................................................................................4
AC 2.3..........................................................................................................................................6
AC 2.4 .........................................................................................................................................8
AC 3.1, 3.2...................................................................................................................................8
TASK 3............................................................................................................................................9
AC 4.1, 4.2...................................................................................................................................9
AC 5.1 .......................................................................................................................................10
AC 5.2 .......................................................................................................................................11

TASK 1
AC 1.1
1.)
cos x = 9/41
In the triangle xyz:
xz = 41 xy= 9
Solution
cos x = base of the triangle / Hypotenuse of right angle triangle
Thus, from the given value of cos x it can be observed that:
Base (B) = 9
Hypotenuse (H) = 41
Perpendicular (P) = √ (41)² - (9)² = 40
Sin x = P/H = 40/41
tan x = P/B = 40/9
cosec x = 1/sinx = H/P = 41/40
sec x = 1/ cos x = H/B = 41/9
cot x = 1/ tan x = B/P = 9/40
2.)
sin θ = 0.625
cos θ = 0.5
Solution
cosec θ = 1/sin θ = 1/0.625 = 1.6
sec θ = 1/cos θ = 1/0.5 = 2
Using the identity: 1+tan² θ = sec² θ
The value of tan θ = 1.73
cot θ = 1/tanθ = 1/1.73 = 0.57
3.)
Angle of depression = 30°
Height of cliff = 75 m
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After 1 minute (60 seconds):
Angle of depression = 20°
Solution
Let the intitial position of ship is at point C and final position at point D.
Alternate angles between parallel lines are equal. Thus, angle ACB = 30° and angle ADB = 20°
From the figure it can be observed that
AB = 75
Tan 30 = AB/BC = 75/BC
So BC = 129.9 m, which represents the first position of the ship
Similarly,
Tan 20 = AB / (BC +CD)
On substituting values we get:
129.9 + CD = tan 20
CD = 76.16 m
Thus, in one minute (60 seconds) ship sails a distance of 76.16 m
Hence,
Speed = distance /time = 76.16/60 = 1.26 m/sec
Speed = 4.57 km/hour
2

AC 1.2
sin² x + cos² x = 1
1.)
a. ) 1 + tan² θ = sec² θ
Solution
sin² θ + cos² θ = 1 (i)
Dividing both the sides of equation (i) by cos² θ we get:
tan² θ + 1 = sec² θ
(as sin² θ / cos² θ = tan² θ and 1/ cos² θ = sec² θ)
b.) cot² θ +1 = cosec² θ
Solution
sin² θ + cos² θ = 1 (ii)
Dividing both the sides of equation (ii) by sin² θ we get:
cot² θ +1 = cosec² θ
(as cos² θ / sin² θ / = cot² θ and 1/ sin² θ = cosec² θ)
TASK 2
AC 2.1
1A.) sin (π + α)
Solution
sin (A+B) = sin A cos B + cos A sin B
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sin (π + α) = sin π cos α + cos π sin α
sin π = 0 and cos π = -1
so
sin (π + α) = -sin α
1B.) -cos (90 +β)
Solution
cos (A+B) = cos A cos B – sin A sin B
-cos (90 +β) = - [cos 90 cos β – sin 90 sin β]
sin 90 = 1 and cos 90= 0
-cos (90 +β) = - [– sin β]
-cos (90 +β) = sin β
1C.) tan (x +π /4) tan (x -π /4)
Solution
tan (A+B) = [tan A +tan B] / [1-tan A tan B]
tan (A-B) = [tan A -tan B] / [1+tan A tan B]
tan (x +π /4) = [tan x +tan π /4] / [1-tan x tan π /4]
tan π /4 = 1
tan (x +π /4) = [tan x +1] / [1-tan x] = 1
tan (x -π /4) =[tan x -tan π /4] / [1+tan x tan π /4]
tan (x -π /4) = [tan x -1] / [1+tan x]
Thus,
tan (x +π /4) tan (x -π /4) = [tan x -1] / [1+tan x]
AC 2.2
1A.) sin 2A
Solution
sin 2A = sin (A+A) = sin A cos A + cos A sin A
sin 2A = 2 sin A cos A
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1B.) sin 8A
Solution
sin 8A = sin (4A +4A)
= sin 4A cos 4A + cos 4A sin 4A
sin 8A = 2 sin 4A cos 4A
1C.) cos 2A
Solution
cos 2A = cos (A+A) = cos A.cos A – sin A.sin A
cos 2A = cos² A - sin² A
= 2 cos² A – 1 or (Using identity cos² θ + sin² θ = 1)
= 1- 2sin² A
1D.) cos 4A
cos 4A = cos (2A+2A) = cos 2A.cos 2A – sin 2A.sin 2A
cos 4A = cos² 2A - sin² 2A
1E.) tan 2A
Solution
tan 2A = tan (A+A) = [tan A +tan A] / [1-tan A tan A]
tan 2A = 2 tan A / [1-tan² A]
1F.) tan 5A
Solution
tan (5A) = tan (2A +3A)
= [tan 2A +tan 3A] / [1-tan 2A tan 3A] (i)
[tan 2A +tan 3A] = 2tanA / (1-tan²2A) + (3tanA - tan²3A) / (1-3tan²2A)
= (tan⁵A – 10 tanA + 5tan A) / [(1-tan²2A) (1-3tan²2A)] (ii)
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Similarly,
[1-tan 2A tan 3A]= 1 - [2tanA (3tanA - tan A)] / [(1 - tan²2A)(1 - 3tan²A]
= (1 - 10tan²A + 5tanA) / [(1 - tan²A)(1 - 3tan²A)] (iii)
On substituting equations (ii) and (iii) in (i) we get:
tan 5A= (tan⁵A – 10 tanA + 5tan A) / (1 - 10tan²A + 5tanA)
2.) Isin 3θ when I = 1
sin 3θ = ??
sin 3θ = sin (θ +2θ ) = sin θ cos 2θ + cos θ sin 2θ
sin 2θ = 2 sinθ cos θ
and cos 2θ = 1- 2sin² θ
so
sin 3θ = sin (θ +2θ ) = sinθ [1- 2sin² θ] + cosθ [2sinθ cosθ]
= sinθ - 2 sinθ + 2 sinθ cos²θ
putting cos²θ = 1- sin²θ
sin 3θ = sin 3θ - 2 sinθ + 2 sinθ [1- sin²θ ]
sin 3θ =3sin 3θ - 4 sinθ
AC 2.3
1A.) [1- cos 2θ] / sin 2θ = tan θ
Solution
sin 2θ = 2sinθ cosθ
cos 2θ = 1- 2sin²θ
On substituting these values in LHS of the given equation:
[1- (1- 2sin²θ)] / 2sinθ cosθ = tan θ
2sin²θ / 2sinθ cosθ = tan θ
tan θ =tan θ
Hence, proved
[1- cos 2θ] / sin 2θ = tan θ
6

1B.) cot 2θ + cosec 2θ = cot θ
Solution
cot 2θ + cosec 2θ
cot 2θ = cos 2θ / sin 2θ
cosec 2θ = 1/sin 2θ
cot 2θ + cosec 2θ = (cos 2θ / sin 2θ) + (1/sin 2θ )
= (cos 2θ +1) / sin 2θ
cos 2θ = 2 cos²θ – 1 and sin 2θ = 2 sinθ cosθ
cot 2θ + cosec 2θ = (2 cos²θ – 1 +1) / 2 sinθ cosθ
cot 2θ + cosec 2θ = cot θ Hence proved.
2)
sin p = 0.8142 cos q = 0.4432
a.) sin (p-q)
Solution
sin p = 0.8142 cos q = 0.4432
Using the identity cos² θ + sin² θ = 1
cos p = 0.5805 sin q = 0.7461
sin (p-q) = sin p cos q - cos p sin q
= (0.8142) (0.4432) - (0.5805)(0.7461)
= 0.360 -0.433
sin (p-q) = -0.073
2b.) cos (p +q)
Solution
cos (p +q) = cos p.cos q – sin p.sin q
= (0.5805) (0.4432) - (0.8142) (0.7461)
= 0.2572 – 0.6074
cos (p +q) = -0.350
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2c.) tan (p +q)
Solution
tan (p +q) = [tan p +tan q] / [1-tan p tan q]
tan θ = sin θ / cosθ
tan p = 0.8142 /0.5805 = 1.4025
tan q= 0.7461/0.4432= 1.6834
tan (p +q) = [1.4025 +1.6834] / [1- (1.4025) (1.6834)]
= 3.085 / (-1.360)
tan (p +q) = - 2.268
AC 2.4
sinθ +2cosθ = 1
Solution
if t = tan (θ /2)
sin x = 2t / (1+ t²)
cos x = (1- t²) /(1+ t²)
On putting values of sin and cos in terms of t in the given identity we get:
sinθ +2cosθ = 1
[2t / (1+ t²)] + 2 [(1- t²) /(1+ t²)] = 1
2t +2 - 2 t² = 1 + t
On solving we get: t = 1 and t = -1/3
t = tan θ/2
When t = 1 θ = 22.5 degree
as θ is in the range 0 to 360 only the value of t= -1/3 is discarded as it gives negative value of
θ. Thus value of θ= 22.5
AC 3.1, 3.2
1A.) 3 sin wt + 4 cos wt
Using the identity:
a sin θ + b cosθ = R sin (θ + α) where
8

R = √(a² + b²) and tan α = b/a
For the expression: 3 sin wt + 4 cos wt
a = 3 b= 4 so R = √(3² + 4²) =5
tan α = 4/3 so α = 53.13°
3 sin wt + 4 cos wt = 5 sin (wt + 53.13)
1B.)
Graphical representation of equations:
3 sin wt (violet)
4 cost wt (pink)
5 sin (wt +53.13) (green)
1C.)
Maximum and minimum value of 3 sin wt is +3 and -3 respectively.
Maximum and minimum value of 4 cos wt is +4 and -4 respectively.
Maximum and minimum value of 5 sin (wt +53.13) is +5 and -5 respectively
TASK 3
AC 4.1, 4.2
9

1.) cos θ - √3 sin θ
Solution
The equation can be rewritten as:
cos θ = √3 sin θ
sin θ /cos θ =1-√3
sin θ /cos θ = tan θ
tan θ = 1-√3
θ =30°
2.) x = cos 2θ and y= secθ
using the identity:
cos 2θ = 2 cos²θ – 1
cosθ = 1/ secθ
cos 2θ = [2/ sec²θ] – 1
cos 2θ = x and sec²θ =y²
on putting these values in above equation:
x = [2/y²] -1
AC 5.1
1.)
sin x =x
cos x = 1-x²/2
sin 3θ / (1 + cos2θ)
Solution
sin 3θ= 3sinθ - 4 sinθ
cos 2θ = 2cos²θ – 1
On putting these values in the given exprssion:
sin 3θ / (1 + cos2θ) = [3sinθ - 4 sinθ] / [1+2cos²θ – 1]
Putting approximate value of sin θ and cos θ
sin 3θ / (1 + cos2θ) = [3θ - 4(θ)] / [1-θ²+θ/4]
= 4[3θ - 4(θ)] / [θ -4θ +4]
10

2.)
tan 60 = 1.732
1° = 0.017 rad
Approximate value for tan 61 = ?
Solution
Let x = 60 and dx = 1°
Now
f(x) = tan x
x = 60 degree = π/3
h = 1 degree = π / 180
tan (60 +1) = tan 61
f (x+h) = f(x) + h* df(x)/dx
f(x) = tan x = tan 60 = 1.732
df(x)/dx = sec²x = sec²60 = 4
tan 61= 1.732 + [π / 180]* 4
Approximated value of tan 61=1.8019
AC 5.2
1A.) θ sinθ / sin4θ
Solution
sin 4 θ = 2sin2θ cos2θ
= 2(2sinθ cosθ) cos2θ
= 4sinθ cosθ) cos2θ
θ sinθ / sin4θ = θ sinθ / [(4sinθ cosθ) cos2θ]
cos 2θ = 1-2sin² θ
θ sinθ / sin4θ = θ sinθ / [(4sinθ cosθ) (1-2sin² θ)]
= θ sinθ / [4sinθ cosθ - 8sin³θ cosθ]
Since θ³ is very small 8sin³θ cosθ can be neglected and approximate value of θ sinθ / sin4θ is:
θ sinθ / sin4θ = θ sinθ / 4sinθ cosθ
11

θ sinθ / sin4θ = θ/ 4cosθ
2B.) 2θ/ sin4θ
Solution
Using the value of simplified value of sin 4θ we get:
2θ/ sin4θ = 2θ/ 2sin2θ cos2θ
= θ /[(2sinθ cosθ) (1-2sin² θ)]
= θ /[2sinθ cosθ – 4sin³θ cosθ]
Neglecting 4sin³θ cosθ we get approximate value of 2θ/ sin4θ as:
2θ/ sin4θ = θ /[2sinθ cosθ]
2θ/ sin4θ = θ/ sin θ/2
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