trigonometric ratios and identities 1
TRANSCRIPT
Mathematics
Trigonometric ratios and Identities
Session 1
Topics
Transformation of AnglesCompound Angles
Definition and Domain and Range of Trigonometric Function
Measurement of Angles
Measurement of Angles
O A
BOA InitialRay
OB Ter minal Ray
Angle is considered as the figure obtained by rotating initial ray about its end point.
J001
Measure and Sign of an Angle
Measure of an Angle :-Amount of rotation from initial side to terminal side.Sign of an Angle :-
O A
B
Rotation anticlockwise – Angle positive
B’
Rotation clockwise – Angle negative
J001
Right Angle
O
Y
X
Revolving ray describes one – quarter of a circle then we say that measure of angle is right angle
J001
Angle < Right angle Acute AngleAngle > Right angle Obtuse Angle
Quadrants
O
Y
Y’
X’ X
II Quadrant( , )
I Quadrant( , )
IV Quadrant( , )
III Quadrant( , )
X’OX – x - axis
Y’OY – y - axis
J001
System of Measurement of Angle
Measurement of Angle
Sexagesimal System or British System
Centesimal System or French System
Circular System or Radian Measure
J001
System of Measurement of Angles
Sexagesimal System (British System)
1 right angle = 90 degrees (=90o)1 degree = 60 minutes (=60’)1 minute = 60 seconds (=60”)
Centesimal System (French System)1 right angle = 100 grades (=100g)1 grade = 100 minutes (=100’)1 minute = 100 Seconds (=100”)
J001
Is 1 minute of sexagesimal1 minute of centesimal ?
=
NO
System of Measurement of Angle
Circular System
J001
O
r
r
r
A
B
1c
If OA = OB = arc ABcThen AOB 1radian( 1 )
System of Measurement of Angle
Circular System
O A
C
B1c
AOC arc ACAOB arcACB
1radian r2right angles r
2right angles radian
J001
180 radian
Relation Between Degree Grade And Radian Measure of An Angle
0 gD G 2C
90 100
OR
0 gD G C
180 200
J002
Illustrative ProblemFind the grade and radian measures of the angle 5o37’30”Solution
o' o30 30 130" 60 60 60 120
o37and37' 60
o oo 37 1 455 37'30" 5 60 120 8
J002
We know thatD G 2C
90 100
Illustrative Problem Find the grade and radian measures of the angle 5o37’30”
g10G D9
g g
g10 45 225 12.5 Ans9 8 18
c
and R D180
c45 radian Ans180 8 32
Solution
J002
Relation Between Angle Subtended by an Arc At The Center of Circle
O A
C
1c
B
Arc AC = r and Arc ACB = AOC arcACAOB arc ACB
J002
1radian r
r
Illustrative Problem A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 88 meters when it has traced out 72o at the center. Find the length of rope. [ Take = 22/7 approx.].Solution
P A
B
72o
Arc AB = 88 m and AP = ?c
o 272 72 rad180 5
J002
arc ABr AP
2 88 22AP 70m [ approx.]5 AP 7
Definition of Trigonometric Ratios
J003
2 2r x y
xO
Y
X
P (x,y)
M
yr
ysin rxcos rytan x
xcot yrsec x
rcosec y
Some Basic Identities
sin cosec 1 ; n ,n I
2 2sin cos 1 2 2sec tan 1
2 2cosec cot 1
sintan ; 2n 1 ,n Icos 2
tan cot 1 ; 2n 1 ; n ,n I2
coscot ; n ,n Isin
cos sec 1 ; 2n 1 ,n I2
Illustrative Problem
Solution3sec tan .cosec
3 cosecsec 1 tan sec
3sec 1 tan cot 2sec 1 tan
2 21 tan 1 tan 3
2 21 tan
3
2 22 e Proved
J0032 2I f tan 1 e ,provethat
3
3 2 2sec tan .cosec 2 e
0 2
Signs of Trigonometric Function In All Quadrants In First Quadrant
xO
Y
X
P (x,y)
M
yr
Here x >0, y>0, 2 2r x y >0
ysin 0r
xcos 0r
ytan 0x
xcot 0y
rsec 0x
rcosec 0y
J004
Signs of Trigonometric Function In All Quadrants In Second Quadrant
Here x <0, y>0, 2 2r x y >0
ysin 0r
rcosec 0y
XX’
Y
Y’
P (x,y)
x
y r
xcos 0r
ytan 0x
xcot 0y
rsec 0x
J004
Signs of Trigonometric Function In All Quadrants In Third Quadrant
Here x <0, y<0, 2 2r x y >0
rcosec 0y
rsec 0x
X’ X
P (x,y)
O
Y’
YM
ysin 0r
xcos 0r
ytan 0x
xcot 0y
J004
Signs of Trigonometric Function In All Quadrants In Fourth Quadrant
Here x >0, y<0, 2 2r x y >0ysin 0r
XO
P (x,y)Y’
M
xcos 0r
ytan 0x
xcot 0y
rsec 0x
rcosec 0y
J004
Signs of Trigonometric Function In All Quadrants
I QuadrantAll Positive
II Quadrantsin & cosec are Positive
III Quadranttan & cot are Positive
IV Quadrantcos & sec are Positive
X
Y’
X’
Y
O
J004
ASTC :- All Sin Tan Cos
Illustrative Problem
lies in secondIf cot = 12 ,5
quadrant, find the values of other five trigonometric functionSolution
12 5cot tan5 12 2 2 2 169sec 1 tan sec 144
13 13sec sec liesinsecondquadrant12 12
12Whichgivescos 13
13cosec 5
J004
Method : 1
5 12 5Thensin tan cos 12 13 13
Illustrative Problem
lies in secondIf cot = 12 ,5
quadrant, find the values of other five trigonometric function
Solution
J004
Method : 2
Y
XX’Y’
P (-12,5)
-12
5 r
Here x = -12, y = 5 and r = 13
y 5sin r 13
x 12cos r 13
y 5tan x 12
r 13sec x 12
r 13cosec y 5
Functions
Domain Range
sin R [-1,1]cos R [-1,1]
sec R : (2n 1) 2
R-(-1,1)
cosec R : n R-(-1,1)
tan R : (2n 1) 2
R
cot R : n R
Domain and Range of Trigonometric Function
J005
Illustrative problemProve that
22 (x y)sin 4xy
is possible for real values of x and y only when x=ySolution
2 2(x y) 1 x y 4xy4xy
2sin 1
2 2x y 4xy 0 x y 0
But for real values of x and y 2x y is not less than zero 2x y 0 x y Pr oved
J005
Trigonometric Function For Allied Angles
Trig. ratio - 90o- 90o+ 180o- 180o+ 360o- 360o+
cos cos sin - sin - cos - cos cos cos
tan - tan cot - cot -tan tan - tan tan
sin - sin cos cos sin - sin - sin sin
If angle is multiple of 900 then sin cos;tan cot; sec cosecIf angle is multiple of 1800 then sin sin;cos cos; tan tan etc.
Trigonometric Function For Allied Angles
Trig. ratio - 90o- 90o+ 180o- 180o+ 360o- 360o+
sec sec cosec - cosec - sec - sec sec sec
cosec - cosec sec sec cosec -cosec - cosec cosec
cot - cot tan -tan -cot cot - cot cot
Periodicity of Trigonometric Function
Periodicity : After certain value of x the functional values repeats itself
Period of basic trigonometric functions
sin (360o+) = sin period of sin is 360o or 2cos (360o+) = cos period of cos is 360o or 2tan (180o+) = tan period of tan is 180o or
J005If f(x+T) = f(x) x,then T is called period of f(x) if T is the smallest possible positive number
Trigonometric Ratio of Compound AngleAngles of the form of A+B, A-B, A+B+C, A-B+C etc. are called compound angles(I) The Addition Formula
sin (A+B) = sinAcosB + cosAsinB cos (A+B) = cosAcosB - sinAsinB
tanA tanBtan A B 1 tanA tanB
J006
sin(A B)Pr oof: tan A B cos(A B)
sinA cosB cos A sinBcos A cosB sinA sinB
Trigonometric Ratio of Compound Angle
J006sinA cosB cos A sinBcos A cosB sinA sinB
r rDividing N and D by cos A cosB
We get tanA tanB
1 tanA tanB
Proved
cotBcot A 1cot A B cotB cot A
Illustrative problemFind the value of(i) sin 75o
(ii) tan 105o
Solution(i) Sin 75o = sin (45o + 30o) = sin 45o cos 30o + cos 45o sin 30o
1 3 1 1 3 12 22 2 2 2
(ii) Ans: 2 3
Trigonometric Ratio of Compound Angle
(I) The Difference Formula
sin (A - B) = sinAcosB - cosAsinB cos (A - B) = cosAcosB + sinAsinB
tanA tanBtan A B 1 tanA tanB
Note :- by replacing B to -B in addition formula we get difference formula
cotBcot A 1cot A B cot A cotB
Illustrative problem
If tan (+) = a and tan ( - ) = b
Prove that a btan2 1 ab
Solution
tan2 tan
tan tan1 tan tan
a b1 ab
Some Important Deductions
sin (A+B) sin (A-B) = sin2A - sin2B = cos2B - cos2A
cos (A+B) cos (A-B) = cos2A - sin2B = cos2B - sin2A
tanA tanB tanC tanA tanB tanCtan A B C 1 tanA tanB tanB tanC tanC tanA
To Express acos + bsin in the form kcos or sinacos +bsin
2 22 2 2 2a ba b cos sin
a b a b
2 2 2 2a bLet cos , thensin
a b a b
2 2acos bsin a b cos cos sin sin
2 2a b cos
Similarly we get acos + bsin = sin
2 2k cos , where k a b ,
Illustrative problem
7cos +24sin
Find the maximum and minimum values of 7cos + 24sinSolution
2 22 2 2 2
7 247 24 cos sin7 24 7 24
7 2425 cos sin25 25
7 24Let cos sin25 25
7cos 24sin 25 cos cos sin sin
Illustrative problem Find the maximum and minimum value of 7cos + 24sin
Solution 25cos 25cos where
1 cos 1 25 25cos 25
Max. value =25, Min. value = -25 Ans.
Transformation Formulae
Transformation of product into sum and difference 2 sinAcosB = sin(A+B) + sin(A - B) 2 cosAsinB = sin(A+B) - sin(A - B) 2 cosAcosB = cos(A+B) + cos(A - B)
Proof :- R.H.S = cos(A+B) + cos(A - B)= cosAcosB - sinAsinB+cosAcosB+sinAsinB= 2cosAcosB =L.H.S
2 sinAsinB = cos(A - B) - cos(A+B) [Note]
Transformation Formulae Transformation of sums or difference into products
C D C DsinC sinD 2sin cos2 2
C D C DsinC sinD 2cos sin2 2
C D C DcosC cosD 2cos cos2 2
C D C DcosC cosD 2sin sin2 2
C D D CcosC cosD 2sin sin2 2
or Note
By putting A+B = C and A-B = D in the previous formula we get this result
Illustrative problemProve thatcos8x cos6x cos 4x cot 6xsin8x sin6x sin4x
Solution(cos8x cos 4x) cos6xL.H.S (sin8x sin4x) sin6x
8x 4x 8x 4x2cos cos cos6x2 28x 4x 8x 4x2sin cos sin6x2 2
2cos6x cos2x cos6x2sin6x sin2x sin6x
2cos6x(cos2x 1)2sin6x(cos2x 1)
cot 6x Proved
Class Exercise - 1If the angular diameter of the moon be 30´, how far from the eye can a coin of diameter 2.2 cm be kept to hide the moon? (Take p = approximately)
227
A
B
E (Eye)
r
Moon
Class Exercise - 1If the angular diameter of the moon be 30´, how far from the eye can a coin of diameter 2.2 cm be kept to hide the moon? (Take p = approximately)
227
A
B
E (Eye)
r
Moon
Solution :-Let the coin is kept at a distance r from the eye to hide the moon completely. Let AB = Diameter of the coin. Then arc AB = Diameter AB = 2.2 cm
c c30 130́ 60 2 180 360
arc 2.2radius 360 r
360 2.2 360 2.2 7r 252 cm22
Class Exercise - 2
Solution :-
Prove that tan3A tan2A tanA = tan3A – tan2A – tanA.
We have 3A = 2A + Atan3A = tan(2A + A) tan3A = tan2A tanA
1– tan2A tanA
tan3A – tan3A tan2A tanA = tan2A + tanA
tan3A – tan2A – tanA = tan3A tan2A tanA (Proved)
Class Exercise - 3If sin = sin and cos = cos, then
–sin 02
(c) –cos 02
(d)
sin 02
(a) cos 02
(b)
Solution :- sin sin cos cosand
sin – sin 0 and cos – cos 0
– –2sin cos 0 and – 2sin sin 02 2 2 2
–sin 02
Class Exercise - 4Prove that
3sin20 sin40 sin60 sin80
16
LHS = sin20° sin40° sin60° sin80°Solution:-
1sin60 [2sin20 sin40 ] sin802
3 1[cos(40 – 20 ) – cos(40 20 )] sin802 2
3 [cos20 – cos60 ]sin804
3 1sin80 cos20 – sin804 2
3 2sin80 cos20 – sin808
Class Exercise - 4Prove that
3sin20 sin40 sin60 sin80
16
Solution:-
3 sin(80 20 ) sin(80 – 20 ) – sin808
3 sin100 sin60 – sin808
3 sin(180 – 80 ) sin60 – sin808
3 3sin80 – sin808 2
316
Proved.
Class Exercise - 5Prove that
n n
n
cos A cosB sin A sinBsin A – sinB cos A – cosB
A B2 cot , if n is even20, if n is odd
Solution :-n ncos A cosB sinA sinBLHS sinA – sinB cos A – cosB
n nA B A – B A B A – B2cos cos 2sin cos2 2 2 2A B A – B A B A – B2cos sin –2sin sin2 2 2 2
n nA– B A – Bcot – cot2 2
Class Exercise - 5
Solution :-
n n nA – B A – Bcot (–1) cot2 2
nn nA – B2cot , if n is evenA – Bcot 1 (–1) 22 0, if n is odd
Prove thatn n
n
cos A cosB sinA sinBsin A – sinB cos A – cosB
A B2cot , if n is even20, if n is odd
Class Exercise - 6The maximum value of 3 cosx + 4 sinx + 5 is
(d) None of these(a) 5 (b) 9(c) 7
2 23cosx 4sinx 3 4 Solution :-
2 2 2 23 4cosx sinx
3 4 3 4
3 45 cosx sinx5 5
5 cosx cos sinx sin
3 4Let cos sin5 5
Class Exercise - 6The maximum value of 3 cosx + 4 sinx + 5 isSolution :-
3 4Let cos sin5 5
5cos(x – )
–1 cos(x – ) 1
–5 5cos(x – ) 5
–5 5 5cos(x – ) 5 10
0 3cosx 4sinx 5 10
Maximum value of the given expression = 10.
Class Exercise - 7If a and b are the solutions of a cos + b sin = c, then show that
2 22 2
a – bcos( )a b
Solution :-We have … (i)acos bsin c
acos c – bsin 2 2 2a cos (c – bsin )
2 2 2 2 2a 1– sin c – 2bcsin b sin
2 2 2 2 2a b sin – 2bcsin (c – a ) 0
are roots of equatoin (i),
Class Exercise - 7If a and b are the solutions of acos + bsin
= c, then show that
2 2
2 2a – bcos( )a b
Solution :-
2 22 2
c – asin sina b
Hence
Again from (i),bsin c – acos
2 2 2b sin (c – acos )
2 2 2 2 2b (1– cos ) c a cos – 2cacos
2 2 2 2 2(a b ) cos – 2accos c – b 0
sin and sin are roots of equ. (ii).
Class Exercise - 7If a and b are the solutions of acos + bsin
= c, then show that
2 2
2 2a – bcos( )a b
Solution :- (iv) and be the roots of equation (i),
cos and cos are the roots of equation (iv).2 22 2
c – bcos cosa b
2 2 2 2 2 22 2 2 2 2 2
c – b c – a a – b–a b a b a b
cos( ) cos cos – sin sin Now
Class Exercise - 8If a seca – c tana = d and b seca + d tana = c, then
(a) a2 + b2 = c2 + d2 + cd
(c) a2 + b2 = c2 + d2
(d) ab = cd
2 22 21 1a b
c d(b)
Class Exercise - 8If a seca – c tana = d and b seca + d tana = c, then
asec – ctan d a csin– dcos cos
Solution :-
bsec dtan c Againb dsin ccos cos
b ccos – dsin ….. ii
a csin dcos ….. (I)
Squaring and adding (i) and (ii), we get
2 2 2 2 2 2 2 2a b c (sin cos ) d (cos sin )
2cd sin cos 2cd sin cos 2 2c d
Class Exercise -9
2 2A Asin – sin –8 2 8 2
The value of
(a) 2 sinA
(c) 2 cosA
1 sinA2
(b)1 cos A2(d)
Class Exercise -9
2 2A Asin – sin –8 2 8 2
Solution :-
A A A Asin – sin – –8 2 8 2 8 2 8 2
1sin sinA sinA4 2
2 2sin A – sin B sin(A B) sin(A – B)
2 2A Asin – sin –8 2 8 2
The value of
Class Exercise -10If , ,
and lie between0 and , then value of tan2 is
4cos( ) 5 5sin( – ) 13
4
(a) 1
(c) 0
(b)5633
(d)3356
Solution :- and between 0 and ,4
– – and 04 4 2
Consequently, cos and sin are positive.2sin( ) 1– cos ( )
Class Exercise -10
Solution :-16 31– 25 5
2 25 12cos( – ) 1– sin ( – ) 1– 169 13
3 5tan( ) , tan( – )4 12
tan2 tan[( ) ( – )]
3 5tan( ) tan( – ) 564 12
3 51– tan( ) tan( – ) 331– 4 12
If , , and lie between0 and , then value of tan2 is
4cos( ) 5 5sin( – ) 13
4