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Mathematics

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**Trigonometric ratios and Identities**

Session 1 Trigonometric ratios and Identities

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**Topics Measurement of Angles Definition and Domain and Range**

of Trigonometric Function Compound Angles Transformation of Angles

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**Measurement of Angles J001 B O A**

Angle is considered as the figure obtained by rotating initial ray about its end point.

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**Measure and Sign of an Angle**

J001 Measure of an Angle :- Amount of rotation from initial side to terminal side. Sign of an Angle :- B Rotation anticlockwise – Angle positive O A B’ Rotation clockwise – Angle negative

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Right Angle O Y X J001 Revolving ray describes one – quarter of a circle then we say that measure of angle is right angle Angle < Right angle Acute Angle Angle > Right angle Obtuse Angle

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**Quadrants J001 Y O X’ X X’OX – x - axis Y’OY – y - axis Y’ II Quadrant**

III Quadrant IV Quadrant X’OX – x - axis Y’OY – y - axis

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**System of Measurement of Angle**

J001 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**

J001 Sexagesimal System (British System) 1 right angle = 90 degrees (=90o) 1 degree = 60 minutes (=60’) 1 minute = 60 seconds (=60”) Is 1 minute of sexagesimal 1 minute of centesimal ? = Centesimal System (French System) 1 right angle = 100 grades (=100g) 1 grade = 100 minutes (=100’) 1 minute = 100 Seconds (=100”) NO

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**System of Measurement of Angle**

J001 Circular System O r A B 1c If OA = OB = arc AB

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**System of Measurement of Angle**

J001 Circular System O A C B 1c

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**Relation Between Degree Grade And Radian Measure of An Angle**

J002 OR

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**Illustrative Problem J002**

Find the grade and radian measures of the angle 5o37’30” Solution

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**Illustrative Problem J002 Solution**

Find the grade and radian measures of the angle 5o37’30” Solution

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**Relation Between Angle Subtended by an Arc At The Center of Circle**

B J002 Arc AC = r and Arc ACB =

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**Illustrative Problem J002 Solution B Arc AB = 88 m and AP = ? 72o P A**

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 = ?

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**Definition of Trigonometric Ratios**

x O Y X P (x,y) M y r J003

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Some Basic Identities

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Illustrative Problem J003 Solution

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**Signs of Trigonometric Function In All Quadrants**

J004 In First Quadrant x O Y X P (x,y) M y r Here x >0, y>0, >0

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**Signs of Trigonometric Function In All Quadrants**

J004 In Second Quadrant X X’ Y Y’ P (x,y) x y r Here x <0, y>0, >0

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**Signs of Trigonometric Function In All Quadrants**

J004 X’ X P (x,y) O Y’ Y M In Third Quadrant Here x <0, y<0, >0

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**Signs of Trigonometric Function In All Quadrants**

J004 In Fourth Quadrant X O P (x,y) Y’ M Here x >0, y<0, >0

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**Signs of Trigonometric Function In All Quadrants**

J004 X Y’ X’ Y O II Quadrant sin & cosec are Positive I Quadrant All Positive III Quadrant tan & cot are Positive IV Quadrant cos & sec are Positive ASTC :- All Sin Tan Cos

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**Illustrative Problem J004 lies in second If cot =**

quadrant, find the values of other five trigonometric function Solution Method : 1

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**Illustrative Problem J004 Solution Y P (-12,5) r 5 -12 X X’ Y’**

lies in second If cot = quadrant, find the values of other five trigonometric function Solution Method : 2 Y X X’ Y’ P (-12,5) -12 5 r Here x = -12, y = 5 and r = 13

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**Domain and Range of Trigonometric Function**

J005 Functions Domain Range sin R [-1,1] cos R [-1,1] tan R cot R Faculty should explain domain of any one of trigonometric function in the class with the help off funda book. sec R-(-1,1) cosec R-(-1,1)

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**Illustrative problem Prove that J005**

is possible for real values of x and y only when x=y J005 Solution But for real values of x and y is not less than zero

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**Trigonometric Function For Allied Angles**

If angle is multiple of 900 then sin cos;tan cot; sec cosec If angle is multiple of 1800 then sin sin;cos cos; tan tan etc. Trig. ratio - 90o- 90o+ 180o- 180o+ 360o- 360o+ sin - sin cos cos sin - sin - cos tan - tan cot - cot -tan

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**Trigonometric Function For Allied Angles**

Trig. ratio - 90o- 90o+ 180o- 180o+ 360o- 360o+ cot - cot tan -tan -cot sec cosec - cosec - sec cosec - cosec sec -cosec

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**Periodicity of Trigonometric Function**

J005 If f(x+T) = f(x) x,then T is called period of f(x) if T is the smallest possible positive number 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 2 cos (360o+) = cos period of cos is 360o or 2 tan (180o+) = tan period of tan is 180o or

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**Trigonometric Ratio of Compound Angle**

J006 Angles 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

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**Trigonometric Ratio of Compound Angle**

J006 We get Proved Ask the students to attempt cot(A+B) themselves

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**Illustrative problem Find the value of (i) sin 75o (ii) tan 105o**

Solution (i) Sin 75o = sin (45o + 30o) = sin 45o cos 30o + cos 45o sin 30o

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**Trigonometric Ratio of Compound Angle**

(I) The Difference Formula sin (A - B) = sinAcosB - cosAsinB cos (A - B) = cosAcosB + sinAsinB Note :- by replacing B to -B in addition formula we get difference formula

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**Illustrative problem If tan (+) = a and tan ( - ) = b Prove that**

Solution

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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 Ask the students to find the expressions for Sin ( A+B+C) and Cos ( A+B+C)

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**To Express acos + bsin in the form kcos or sin**

Similarly we get acos + bsin = sin

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Illustrative problem Find the maximum and minimum values of 7cos + 24sin Solution 7cos +24sin

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**Illustrative problem Solution Max. value =25, Min. value = -25 Ans.**

Find the maximum and minimum value of 7cos + 24sin Solution Max. value =25, Min. value = Ans. Ask them to try using the Sin (A+B ) form

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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 Ask student to do 2 sinAsinB = cos(A - B) - cos(A+B) 2 sinAsinB = cos(A - B) - cos(A+B) [Note]

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**Transformation Formulae**

Transformation of sums or difference into products By putting A+B = C and A-B = D in the previous formula we get this result Note or

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Illustrative problem Prove that Solution Proved

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Class Exercise - 1 If 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)

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**Class Exercise - 1 Solution :-**

If 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) 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

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**Class Exercise - 2 Prove that tan3A tan2A tanA = tan3A – tan2A – tanA.**

Solution :- We have 3A = 2A + A Þ tan3A = tan(2A + A) Þ tan3A = Þ tan3A – tan3A tan2A tanA = tan2A + tanA Þ tan3A – tan2A – tanA = tan3A tan2A tanA (Proved)

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**Class Exercise - 3 If sina = sinb and cosa = cosb, then (b) (a) (d)**

Solution :- and

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**Class Exercise - 4 Prove that Solution:-**

LHS = sin20° sin40° sin60° sin80°

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Class Exercise - 4 Prove that Solution:- Proved.

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Class Exercise - 5 Prove that Solution :-

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Class Exercise - 5 Prove that Solution :-

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**Class Exercise - 6 The maximum value of 3 cosx + 4 sinx + 5 is**

(d) None of these (a) 5 (b) 9 (c) 7 Solution :-

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**Class Exercise - 6 Solution :-**

The maximum value of 3 cosx + 4 sinx + 5 is Solution :- \ Maximum value of the given expression = 10.

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**Class Exercise - 7 If a and b are the solutions of a**

cosq + b sinq = c, then show that Solution :- We have … (i) are roots of equatoin (i),

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**Class Exercise - 7 Solution :- sin and sin are roots of equ. (ii).**

If a and b are the solutions of acosq + bsinq = c, then show that Solution :- sin and sin are roots of equ. (ii). Hence Again from (i),

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**Class Exercise - 7 Solution :- (iv)**

If a and b are the solutions of acosq + bsinq = c, then show that Solution :- (iv) and be the roots of equation (i), cos and cos are the roots of equation (iv). Now

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Class Exercise - 8 If a seca – c tana = d and b seca + d tana = c, then (a) a2 + b2 = c2 + d2 + cd (b) (c) a2 + b2 = c2 + d2 (d) ab = cd

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Class Exercise - 8 If a seca – c tana = d and b seca + d tana = c, then Solution :- ….. (I) Again Squaring and adding (i) and (ii), we get ….. ii

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Class Exercise -9 The value of (a) 2 sinA (c) 2 cosA (b) (d)

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Class Exercise -9 The value of Solution :-

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**Class Exercise -10 If , , and b lie between0 and , then value**

of tan2a is (a) 1 (c) 0 (b) (d) and between 0 and , Solution :- Consequently, cos(a - b) and sin(a + b) are positive.

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**Class Exercise -10 Solution :- If , ,**

and b lie between0 and , then value of tan2a is Solution :-

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