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7.1 Trigonometric Ratios & Identities

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Published by LIM TENG KEE KM-Pensyarah, 2023-10-01 23:50:55

SM015 Chapter 7 (Lecture 1 of 4)

7.1 Trigonometric Ratios & Identities

7.0 TRIGONOMETRIC FUNCTIONS LECTURE 1 OF 4 7.1 Trigonometric Ratios And Identities


Learning Outcomes (a) State trigonometric ratios of sin θ ,cos θ, tan θ , cosec θ , sec θ and cot θ. sinθ (b) Express tanθ = , sin(90°- θ) = cos θ, cosθ cos (90° -θ) = sin θ and tan(90°- θ) = cotθ (c) Use some special angles. (d) Find the angle of trigonometric equations. 2 2 2 2 2 2 sin + cos =1 1 + cot = cosec 1 + tan = sec (e) Use the basic trigonometric identities :


TRIGONOMETRIC RATIOS For any acute angle θ, there are six trigonometry ratios, each of which is define be referring to a right angle triangle containing θ. x y r θ y x 90


From the diagram tan θ = = adjacent opposite y x cos θ = = hypotenuse adjacent r x y sin θ = = opposite hypotenuse r x r θ x y y cot θ = 1 tan θ = y x sec θ = 1 cos θ = r x cosec θ = 1 sin θ = y r


Example 1: If and θ is acute angle, find cos θ, tan θ, cosec θ, sec θ, and cot θ. 3 sin 5 Solution: 4 cos 5 θ 4 3 5 3 tan = 4 5 cosec = 3 4 cot 3 5 sec = 4


Example 2: 1 If tan θ = and θ is in the fourth quadrant 2 find the sec θ and cosec θ. Solution: 1 3 sec = cos 2 1 cosec = 3 sin θ 1 2 3


Relation of sinθ,cosθ and tanθ r y θ x 90 x = y x (i) sin(90 ) r cos sin(90 ) cos y = y (ii) cos(90 ) r sin r cos(90 ) sin x = x (iii) tan(90 ) y cot y tan(90 ) cot


TRIGONOMETRIC RATIOS FOR SPECIAL ANGLES 3 2 1 60 30 1 2 45 45 1 30 45 60 sin 1 2 1 2 3 2 cos 1 2 3 2 1 2 tan 1 3 1 3


Graph of sin , cos and tan 90 180 270 360 0 1 1 y sin y 0 90 180 270 360 1 1 y cos y


Note 60 = 60 180 = rad 3 4 4 180 rad 5 5 144 DEGREE RADIAN RADIAN DEGREE


2 2 1- tan 30 Prove that cos60 = 1+ tan 30 2 2 2 2 1 1 1 tan 30 3 : 1 tan 30 1 1 3 RHS Example 3: Solution: 1 1 3 1 1 3 2 3 3 4 1 2 cos60 LHS


Example 4: 2 1 3 1 3 2 3 2 3 2 Solution: LHS : sec 30 tan 60 + sin 45 cosec 45 + c os 30 cot 60 1 2 1 2 7 2 RHS 7 sec 30 tan 60 + sin 45 cosec 45 +cos 30 cot 60 = 2 Prove that


Types of Angle Acute angle Right angle Obtuse Angle Reflex angle TRIGONOMETRIC RATIOS FOR ANY ANGLE 0 < < 90 = 90 90 < < 180 180 < < 360


Tangent positive Basic Angle (α) α = α α α Sine positive All positive Cosine positive II I III IV II =180 - III =360 - IV = -180 α III II IV I I 0/ 360 90 180 270


Positive and Negative Angles θ -θ S T A C sin (- ) = - sin cos (- ) = cos tan (- ) = - tan


Example 5: (a) sin(-330 ) (b) cos(-225 ) Solution: (a) sin(-330 in(330 ) ) [ sin(360 330 )] sin30 1 2 s


(b) cos(-225 (225 225 ) cos ) cos( 180 ) cos45 1 2 Solution:


Remember r P(x,y) y x y Phytagoras Theorem : (1) 2 2 2 r x y ......


2 2 2 2 2 2 r r r r x y 2 2 x y 1 r r 2 2 1 cos sin ........... (2) Divide ( ) by : 1 2 r


2 2 2 2 2 cos cos 1 cos si cos n 2 2 1 sin 1 cos cos 2 2 sec 1 tan (2) 2 Divide by cos : 2 2 2 2 2 sin sin 1 cos sin sin 2 2 1 cos 1 sin sin (2) 2 Divide bysin : 2 2 cosec cot 1 2 2 sin cos 1 2 2 1 cot cosec 2 2 tan 1 sec Trigonometric Identities


Prove the following identities. ( ) cos tan cos 2 2 2 b ec ec 1 1 Example 6: 2 2 (a) sin cos sin cos 2


2 2 a sin cos sin cos 2 LHS : 2 2 sin cos sin cos 2 2 2 2 sin 2sin cos cos sin 2sin cos cos 2 2 2 2 sin cos sin cos 1 1 2 RHS Solution:


( ) cos tan cos 2 2 2 b ec ec 1 1 cos 2 ec RHS : cos tan 2 2 LHS ec 1 1 cot (sec ) 2 2 cos sin cos 2 2 2 1 sin 2 1


RHS: cos 1 sin cos 1 1 sin sin 1 sin 2 1 sin cos cos sin 2 2 cos cos sin cos cos cos sin cos 1 sec tan (LHS) cos cos sin cos 2 Prove the identity cos sec tan 1 sin Example 7: Solution:


CONCLUS IONS 2 2 sin cos 1 2 2 1 cot cosec 2 2 tan 1 sec Trigonometric Identities sin (- ) = - sin cos (- ) = cos tan (- ) = - tan + & - Angles 3 2 1 60 30 1 2 45 45 1 TRIGONOMETRIC RATIOS FOR ANY ANGLE


Graph of sin , cos and tan 90 180 270 360 0 1 1 y sin y 0 90 180 270 360 1 1 y cos y


Tangent positive Basic Angle (α) α = α α α Sine positive All positive Cosine positive II I III IV II =180 - III =360 - IV = -180 α III II IV I I 0/ 360 90 180 270


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