Mathematics

Trigonometric ratios and Identities Session 1 Trigonometric ratios and Identities

Topics Measurement of Angles Definition and Domain and Range of Trigonometric Function Compound Angles Transformation of Angles

Measurement of Angles J001 B O A Angle is considered as the figure obtained by rotating initial ray about its end point.

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

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

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

System of Measurement of Angle J001 Measurement of Angle Sexagesimal System or British System Centesimal System or French System Circular System or Radian Measure

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

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

System of Measurement of Angle J001 Circular System O A C B 1c

Relation Between Degree Grade And Radian Measure of An Angle J002 OR

Illustrative Problem J002 Find the grade and radian measures of the angle 5o37’30” Solution

Illustrative Problem J002 Solution Find the grade and radian measures of the angle 5o37’30” Solution

Relation Between Angle Subtended by an Arc At The Center of Circle  B J002 Arc AC = r and Arc ACB = 

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 = ?

Definition of Trigonometric Ratios x O Y X P (x,y) M  y r J003

Some Basic Identities

Illustrative Problem J003 Solution

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

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

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

Signs of Trigonometric Function In All Quadrants J004 In Fourth Quadrant  X O P (x,y) Y’ M Here x >0, y<0, >0

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

Illustrative Problem J004  lies in second If cot  = quadrant, find the values of other five trigonometric function Solution Method : 1

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

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)

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

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

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

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 

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

Trigonometric Ratio of Compound Angle J006 We get Proved Ask the students to attempt cot(A+B) themselves

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

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

Illustrative problem If tan (+) = a and tan ( - ) = b Prove that Solution

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)

To Express acos + bsin in the form kcos or sin Similarly we get acos + bsin = sin

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

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 = -25 Ans. Ask them to try using the Sin (A+B ) form

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]

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

Illustrative problem Prove that Solution Proved

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)

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

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)

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

Class Exercise - 4 Prove that Solution:- LHS = sin20° sin40° sin60° sin80°

Class Exercise - 4 Prove that Solution:- Proved.

Class Exercise - 5 Prove that Solution :-

Class Exercise - 5 Prove that Solution :-

Class Exercise - 6 The maximum value of 3 cosx + 4 sinx + 5 is (d) None of these (a) 5 (b) 9 (c) 7 Solution :-

Class Exercise - 6 Solution :- The maximum value of 3 cosx + 4 sinx + 5 is Solution :- \ Maximum value of the given expression = 10.

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),

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),

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

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

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

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

Class Exercise -9 The value of Solution :-

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.

Class Exercise -10 Solution :- If , , and b lie between0 and , then value of tan2a is Solution :-