trigonometric ratios and identities 1
TRANSCRIPT
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Mathematics
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Trigonometric ratios and Identities
Session 1
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Topics
Transformation of AnglesCompound Angles
Definition and Domain and Range of Trigonometric Function
Measurement of Angles
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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
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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
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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
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Quadrants
O
Y
Y’
X’ X
II Quadrant( , )
I Quadrant( , )
IV Quadrant( , )
III Quadrant( , )
X’OX – x - axis
Y’OY – y - axis
J001
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System of Measurement of Angle
Measurement of Angle
Sexagesimal System or British System
Centesimal System or French System
Circular System or Radian Measure
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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
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System of Measurement of Angle
Circular System
J001
O
r
r
r
A
B
1c
If OA = OB = arc ABcThen AOB 1radian( 1 )
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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
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Relation Between Degree Grade And Radian Measure of An Angle
0 gD G 2C
90 100
OR
0 gD G C
180 200
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Method : 1
5 12 5Thensin tan cos 12 13 13
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Illustrative Problem
lies in secondIf cot = 12 ,5
quadrant, find the values of other five trigonometric function
Solution
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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Illustrative problem
If tan (+) = a and tan ( - ) = b
Prove that a btan2 1 ab
Solution
tan2 tan
tan tan1 tan tan
a b1 ab
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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
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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 ,
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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
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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.
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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]
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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
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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
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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
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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
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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)
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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
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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
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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.
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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
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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
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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
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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.
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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),
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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).
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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
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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)
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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
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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)
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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
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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 ( )
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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