trigonometry book notes, Study Guides, Projects, Research of Mathematics

this is a detailed document it has all trigonometry nits and questions it also has topics which come under trigonometry

Typology: Study Guides, Projects, Research

2020/2021

Uploaded on 10/22/2021

strawbwrry
strawbwrry 🇮🇳

4 documents

1 / 14

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Unit 2
Trigonometry
Course Outcome: Utilize basic concepts of trigonometry to solve elementary engineering
problems.
Unit outcome:
a. Apply the concept of compound angle, allied angle, and multiple angles to solve the given
simple engineering problem(s).
b. Apply the concept of Sub- multiple angle to solve the given simple engineering related
problem(s).
c. Employ concept of factorization and de-factorization formulae to solve the given simple
engineering problem(s).
d. Investigate given simple problems utilizing inverse trigonometric ratios.
Introduction: Trigonometry is a study of relationships in Mathematics involving lengths, heights
and angles of different triangles. Presently Trigonometry finds wide applications in engineering
faculties like Applied Mechanics, Electrical Technology, Basic Electronics, and Computer
Engineering etc.
Subtopic: Compound angles, Allied angles, Multiple and Submultiple angles.
Significance : This sub-topic is used to find trigonometric ratios of angles other than standard
angles.
Some important formulae:
I. Fundamental Identities:
𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠2𝜃 = 1
1 + 𝑡𝑎𝑛2𝜃 = 𝑠𝑒𝑐2𝜃
1 + 𝑐𝑜𝑡2𝜃 = 𝑐𝑜𝑠𝑒𝑐2𝜃
II. For a rightangled triangle
sin θ = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos θ =𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
tan θ = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
sec θ = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
cosec θ = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
cot θ = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
Also we have,
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe

Partial preview of the text

Download trigonometry book notes and more Study Guides, Projects, Research Mathematics in PDF only on Docsity!

Unit 2

Trigonometry

Course Outcome : Utilize basic concepts of trigonometry to solve elementary engineering problems. Unit outcome:

a. Apply the concept of compound angle, allied angle, and multiple angles to solve the given

simple engineering problem(s). b. Apply the concept of Sub- multiple angle to solve the given simple engineering related

problem(s).

c. Employ concept of factorization and de-factorization formulae to solve the given simple

engineering problem(s).

d. Investigate given simple problems utilizing inverse trigonometric ratios.

Introduction: Trigonometry is a study of relationships in Mathematics involving lengths, heights and angles of different triangles. Presently Trigonometry finds wide applications in engineering faculties like Applied Mechanics, Electrical Technology, Basic Electronics, and Computer Engineering etc. Subtopic : Compound angles, Allied angles, Multiple and Submultiple angles. Significance : This sub-topic is used to find trigonometric ratios of angles other than standard angles. Some important formulae:

I. Fundamental Identities:

 𝑠𝑖𝑛^2 𝜃 + 𝑐𝑜𝑠^2 𝜃 = 1

 1 + 𝑡𝑎𝑛^2 𝜃 = 𝑠𝑒𝑐^2 𝜃

 1 + 𝑐𝑜𝑡^2 𝜃 = 𝑐𝑜𝑠𝑒𝑐^2 𝜃

II. For a right–angled triangle

 sin θ =

𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒

 cos θ =

𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒

 tan θ =

𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒

 sec θ =

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒

 cosec θ =

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒

 cot θ =

𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒

Also we have,

 cosec θ = 1 sin θ  sec θ = 1 cos θ

 cot θ = 1 tan θ

 tan θ = 𝑠𝑖𝑛 θ cos θ

 cot θ =

𝑐𝑜𝑠 θ sin θ  cosec θ × sin θ = 1  sec θ × cos θ = 1  cot θ × tan θ = 1 III. Measures of an angle : In practice we use two systems to measure the angle. a) Sexagesimal system (Degree) : In this system , the unit of measurement is degree. b) Circular systems (Radian) : In this system , the unit of measurement is radian. Relation between degrees and radians:

Notation : Any angle can be measured in degrees = θο^ and in radians = θ𝒸

  1. Conversion of angle in degrees to angle in radians 𝛳𝒸^ = θο^ × 𝜋 180
  2. Conversion of radians to degree 𝛳𝜊^ = θ𝒸^ × 180 𝜋 The following table shows the conversion of degree measure to radian measure of standard

angles

𝛳𝜊^ 𝛳𝒸

Compound angle:

Definition: The angle obtained by addition or subtraction of given angles is called compound

angle.

For e.g. A+B, A-B are called compound angles

e. g. A = 45, B = 30 then

A + B = 45 + 30 = 75 and A  B = 45  30  = 15 are compound angles.

Trigonometric ratios of compound angles (Without proof) :

  1. sin (A + B) = sin A  cos B + cos A  sin B

  2. sin (A  B) = sin A  cos B  cos A sin B

  3. cos (A + B) = cos A  cos B  sin A  sin B

  4. cos (A  B) = cos A cos B + sin A sin B

  5. tan (A + B) =

tan A + tan B 1  tan A  tan B

  1. tan (A  B) =

tan A  tan B 1 + tan A  tan B

SOLVED EXAMPLES:

Without using calculator, find the value of

  1. sin 15 2) cos 75

Solution : 1) 15 = 45  30 

 sin 15 = Sin (45  30 )

= sin 45  cos 30  cos 45  sin 30

2 ^ 

sin 15=

√𝟑−𝟏 𝟐√𝟐

  1. 75= 45 + 30

cos 75 = cos (45+30)

= cos 45  cos 30  sin 45  sin 30

2 ^ 

 cos 75 =

√3− 2√

  1. If tan A =

2 , tan B =

3 find tan (A + B)

Solution : tan (A + B) =

tan A + tan B 1  tan A  tan B

2 ^

  1. If tan (x + y) =

4 and tan (x^ ^ y) =

15 then show that tan 2x =

Solution : As 2x = (x + y) + (x  y)  L.H.S.= tan 2x = tan[(x + y) + (x  y)]

=

tan (x + y) + tan (x  y) 1  tan (x + y)  tan (x  y)

4 ^

cos A =

13 sin B =^

∵ A and B are obtuse (More than 90º and less than 180º)  A is the second quadrant, cos A = – 5/ B is the second quadrant, sin B = 3/

cos A =

13 sin B =^

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

= 

13 ×^ 

5 +^ 

65 +^

65 =^

− 65

EXERCISE:

  1. Without using calculator, find the value
    1. cos 105 2) tan 75 3) tan 15
      1. sin 75 5) tan 105 6) sin 105
  2. Evaluate without using calculator

tan 85  tan 40 1 + tan 85  tan 40

  1. Prove that

1  tan 2  tan  1 + tan 2  tan 

cos 3 cos 

  1. If tan A =

3 , tan B =

4 , where 0 < A <^

2 ,^ ^ < B <

2 , find sin (A + B)

  1. If sin =

13 , cos^ ^ =^

25 and^ ,^ ^ lies in the third quadrant, find sin (^ -^ )

  1. If A and B both obtuse angles and sin A =

13 and cos B =^

5 , then find sin (A + B) and the quadrant of angle A + B.

  1. If A and B are obtuse angles such that sin A =

13 and cos B =^

5 , find tan (A + B).

Allied Angles

If the sum or difference of the measures of two angles is either zero or is an integral multiple of

90 , that is  n 

2 then these angles are called Allied angles.

For any angle , let  be It’s allied angle then

 +  = n 

2 ^ ^ = n^ ^

2 –^ ^ allied angle of^ 

and  –  = n 

2 ^ ^ = n^ ^

2 +^ ^ allied angle of^ .

For any angle of  ; n 

2 ^ ^ are allied angle of^ .

In general the above results can be written as

If n is an even integer

sin 

n

2 ^ ^ ^ sin^ ^ cos^ 

n

2 ^ ^ =^ ^ cos^ 

If n is an odd integer

sin 

n 

2 ^ ^ =^ ^ cos^ ^ cos^ 

n 

2 ^ ^ =^ ^ sin^ 

The algebraic sign is settled down by knowing the quadrant in which the angle (^) ^ n (^)   2 ^ ^ lies.

Note that sin n = 0

cos n =  1 If n is odd integer cos n = 1 If n is even integer

For the angles 

2 ^ ^ i.e. (90^ +^ ) and^ 

2 ^ ^ i.e. (270^ ^ )

Make the changes as follows :

sin  cos, tan  cot, sec  cosec and for the angles (  )

i.e. (180  ) and (2  ) i.e. (360  ) there is no change in trigonometric ratios.

i.e. sin  sin, cos  cos etc.

 The algebraic sign is settled down by knowing the quadrant in which the angle lies.

e.g. sin (^) 

2 ^ ^ =?

For angle 

2 ^ ^ sin changes to cos and angle^ 

2 ^ ^ lies in III

rd (^) - quadrant and in IIIrd

quadrant sin ratio is negative.

Solution : Given sin 420 cos 390 + cos ( 300 ) sin ( 330 )

sin (420) = sin (4  90  + 60) = sin 60 =

cos (390) = cos (360 + 30)

= cos (4  90  + 30) = cos 30 =

cos ( 300 ) = cos (300)

= cos (3  90  + 30) = sin 30 =

sin ( 330 ) =  sin (330) =  sin (360  30 ) =  sin (90  4  30 ) =  ( sin 30)

= sin 30 =

∴ sin 420 cos 390 + cos ( 300 ) sin ( 330 )

=

2 ^

EXERCISE:

  1. Without using calculator, find the value of

  2. sin 210 2) cos 330 3) sec^2 ( 765 )

  3. cot ( 710 ) 5) cosec ( 960 )

  4. Without using calculator, find the value of

1)

sec^2 (135) cos ( 240 )  2 sin (930) 2) Prove that : cos (570)  sin (510) + sin ( 330 )  cos ( 390 ) = 0 3) tan (585)  cot ( 495 )  cot (405)  tan ( 495 ) 4) sin (150)  tan (315) + cos (300) + sec^2 (3660) 5) cos ( 1125 ) + tan ( 945 ) + cos (495) 6) cos (225) cos (675) + sin (585) sin (315) 7) sin 660 sin 480  cos 120 sin 330

Multiple Angles:

Trigonometric ratios of multiple angle: The integral multiple of A i.e. 2A, 3A, 4A, ........... are called multiple angles of A.

e.g. If A = 60 then 2A = 120, 3A = 180, 4A = 240 etc.

Trigonometric ratios of 2:

  1. sin 2 = 2 sin   cos 

=

2 tan  1 + tan^2 

  1. cos 2 = cos^2   sin^2 

= 1  2 sin^2  = 2 cos^2   1 = 1−𝑡𝑎𝑛^2 𝜃 1+𝑡𝑎𝑛^2 𝜃 From the above we can deduce

  1. tan 2 = 2 𝑡𝑎𝑛𝜃 1 −𝑡𝑎𝑛^2 𝜃 Trigonometric ratios of 3:

  2. sin 3 = 3 sin   4 sin^3 

  3. cos 3 = 4 cos^3   3 cos 

  4. tan 3 =

3 tan   tan^3  1  3 tan^2 

SOLVED EXAMPLES:

  1. If sin A = 0.4, find sin 3A Solution: Given sin A= 0.

sin 3A =3 sin A  4 sin^3 A =3(0.4)  4(0.4)^3 = 1.2  0. = 0.

  1. If cos A =

2 , find the value of cos 3A

Solution: Given cos A =

cos 3A=4 cos^3 A  3 cos A

=4 (^) 

3  (^3) 

8 ^

1  cos 2 = 2 sin^2 

1 + cos 2 = 2 cos^2 

L.H.S. =

2 sin 2  cos 2 + sin 2 2 cos^2 2  + cos 2

=

sin 2 (2 cos 2 + 1) cos 2 (2 cos 2 + 1)

=

sin 2 cos 2 = tan 2 = R.H.S. EXERCISE:

  1. If cos  = 0.4, find cos 3

  2. If sin A = 0.4, find cos 2A using multiple angle formula

  3. If sin A =

2 , find sin 3A.

  1. If sin A =

5 , find the value of sin 2A.

  1. If A = 60 verify the result sin 3A = 3 sin A  4 sin^3 A

  2. Prove that

sin 2 + cos  1  cos 2 + sin  = cot^ 

  1. Prove :

1 + sin 2A + cos 2A 1 + sin 2A – cos 2A = cot A

Submultiple Angles:

Let A be the given angle then

A

2 ,^

A

3 ,^

A

4 ,^ .........^ are called submultiples angles of A.

e.g. If A = 60 then

A

A

A

4 = 15^ ...... etc.

Trigonometric ratios of any angle in terms of its sub-multiple angle:

  1. cos  = cos^2 ( 𝜃 2 )^ ^ sin

=2 cos^2 ( 𝜃 2 )^ ^1 = 1  2 sin^2 ( 𝜃 2 )

  1. sin  = 2 sin ( 𝜃 2 )^ cos^ (

𝜃 2 )

  1. tan  =

2 𝑡𝑎𝑛(𝜃 2 ) 1 −𝑡𝑎𝑛^2 (𝜃 2 )

Solved Examples:

  1. If tan 

A

2 =^

, find the value of cos A.

Solution : We know that

cos A =

1 – tan^2 A/ 1 + tan^2 A/

=

1 – (1/ 3)^2

2

  1. Prove that

sin  1 + cos  = tan

Solution : L.H.S.=

sin  1 + cos 

2 sin /2  cos / 2 cos^2 /

sin  = 2 sin /2  cos / 1 + cos  = 2 cos^2 /

2 sin /2  cos / 2 cos /2  cos /

=

sin / cos / = tan ( 𝜃 2 ) EXERCISE:

  1. If  = 45, find the value of cos ( 𝜃 2 )

  2. If tan

2 =^

find the value of sin 

  1. Prove that

1 + sin   cos  1 + sin  + cos  = tan /

Prove that

cos A 1  sin A

1 + tan

A

1  tan

A