chapter at thc cnd oflhe lesson. students

8/14/2019 Chapter at Thc Cnd Oflhe Lesson. Students

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Chapter 3 - TRIGONOI\f ETRY

At thc cnd oflhe lesson. students should be able to:

(a) understand the delinition ofthe six trigoDometric lunctions 1br angles of.rnymagnitude

(b) recall and use the exact valucs oftrigonometric lunctions of30",45" and 60'.(c) understand the complenlentify angles and supplcrncntary angles propertres(d) usc trigonometric identitir:s including cornpound anglcs, doublc anglc and lictor

formulae (listedin MF15) for. thc simplilication and cxact cvaluation ofexpressions

. solving sirnplc trigoDometric equations(e) undcrstand the use ofthc R-fornulae(1) usc thc notrlion sin l.r. ctrs I-L and tan I; to denote thc principal vahcs ofthe invcrse

tri gonolnetric relations.

l l IntroductionBy convention, anglcs ate tneas$ed liom the initial line or the r-exis with rcspcct to thc

origin.If OP is rotated drri c/ocl .risc fiom thc:r,axis, the angle so lonned is poriliyc-Bui if OP is rotated clo.tlule liom the r-axis, the angle so formed is lresdtl,e.

angle

1.2 Units of angles: Deerees & Radians

/a.., 'lAngles are measured either in degrccs or radians. What is radian?

Given a circle(as

shown ir the diagram on the right) with radius /, the angle subtended byan arc

oflength r" mcasures I radixn.Compadng the units, z rad = 180'

Positive

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1.3 Trigonomctric Ratios

The six trigonometric mtios are delined to be.vllsrnA /, cost '', lilnr--. cu5cc,

-.sec6'

t r .\ sln d co\a

srn d cos AN.'re:lso tanq - cotdcos A sin t,

1.4 Graphs of Triqonometric FunctionsIn I complete cycle of / = a sin r, amplitudc - d units, period is 2r.

In I conpletc cycle of / = o cos:r, amplitude: a units, period is 2a

In 2 complete cycles of / = a tan x , period is z

Graph of )] = atan.x

r ! al

t^n 0

P(x, y)

Graph of l, - asin -ty = a sin-r

Graphofl=acosr

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1,5 Trieonometric ratios of specinl aneles

1.6 ImDortant propcrtics

gles to dcducc values ofspccial angles.

r\"il \l[ r\-l-

We may use equilateral & isosceles trian

+l++l+

30" 45. 60" 90" 180. 2'/0" 360"

s;n 0 01

tI

J,!52

0 ,1 0

l J:)

I

T,I2

0 0

tane 0I

F I v5 0 0

ut A

Flst qu:rdrantsinl21..r-rl = sin,,os1)r r11 ,usdtJn\) 1 0t @nA

2nd quadrantltin(z a) +indcos(z d) = cos 6/ra\10t e)= rane

3rd quadrant

sin(z + A) = sinflcos(z + d) = cosd

ta\t(n + 0) = tan?

4th quadrantstlt(zft e) = .sinecos(2r - 0) = cos9 ORtan(2r e)= b\q

,4,,'i ,)= .tna'\I cos(-9) = cosB ll tan( d)= tan 97

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itt,\,^ )i ltlt. i,,. .. .)tt-


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1.7 ComDlementary angles

Tuo dngles lhat sum up lo qoo ur 7 radlans arc,alled complemenlary anglcs.

L.s.: r0" & 70". lu" & b0". or J *(, O) eomplementarv cnglc..I n 7 ,/1Rt,,tll sin 10 - cos60" - ) .'n J .n'O - 2

r,n t0 - rort0 -I- ran60 cor J0' - JJ

J:We say that sine & cosine are

!9wk!!9!l!g!t f onctlons.

E,g.: (i) sin40" - cos50'3tr it(xl ) tan = cot-',(

l{

,91yp!g1g91141t fwtctions. Also. langenl & colangcnl are

1.8 Basic Angle

The basic atrgle is defined to be the positiv, acute angle between the line OP & its projection onthe ).-axis.For any general angle, there is a basic angle associated with it.Let (,t be the basic angle, 0" < d < 90"

3n )t(rr ) cos- = srn-

(iv) cot 35" = tan 55'

a = l8O" 0 a=0-18O"

5n(rr) cos-

( 2n\trrrt tanl : J

Examplc ISimplify (i) sin 210'

Solution:

(i) sin 210"

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5n(1t) cos- :

rii,r t-(-4) -

I2

Example 2lf sin:r - 0.6, cos"r = 0.8, find (i)

Solution:(D

sin(3a x)= sin(2tr + zr x)= sin(z -x)

sin(32 -.r) , (ii) cos(42 + r) .

(iD

cos(42 + rc)

= cos(2n + x)

= 0.8

Example 3Sol\ e tlre lbllowing equarions:

ru.)*in1l-t a) ! "1"." o. 0 - n. (b)2 J2Solution:

(a) Since sinlZ-8) is positive, it is in the (b)1'1or 2d quadrant.

Basic ansle- a=1"4

SrrJt''2"4""4-5n3n- H=- or H=-44

Since0!d(z-3n.. H=

4

ExerciseSolve tle equation sec(9 + 30') = 2

cos(28-250) -0.8 *here 0" < d - 180'

Since cos(2d + 25") is negative, it is iothe 2nd or 3rd quadrant

Basic angle, a =36.87"-'- 20 +25" =180" -36.8'7' or

20 + 25" =180" + 36.87'

Hence d = 59.1' or 95.9"

Answer: d = 30",270'

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l2.t li:r5ic ldcnriric\ \ ., )..- --

2.2 lnlportant Fornlulac

Example 4

It is given that tan I = 3,01A {zr. Find, without using calculator,(D the exact value of tana, given that tan(d+,4) = 5,(ir) the exact value of tanL giventhat s1n(6 + A)=2cos(d- A),(iiD the exact value of sin 2.4 .

(l) ,;12A1so"2n.1

Dividing (1) throughout by cos2 A.

(l) tan2;\ I I sec2 A

Dividing (l) throughout by sin2 A,

(r) l+rotl{ rn\.rl \

(l ) Compound Angle lrormuhc (givcn in MIr 15)sin[4tB) = sinlcos.BlcoslsilB

cos(liB) = cos,.lcos,B+sinlsill,R

tinl / + B)= trnlltunBli t.,n ,1 r.,n B

(2) Doublc.\nglc Fornrulac (givcn in t\'1F15)sin 2,1 : 2 sin

Icos

Icos 2-,4 : cos2l sin2,4^ J, ,

: I 2 sin2 A

tat2A= 2tat A1 ta\'\2 A

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Solution:(i) Given ran (A+,4):5 and tan I : -3, = lrr g *+t..[+(a lb + l+1

. tcn9 31+3taDA

5+15tan0 =tan0 3

.. tun0 = -!7(ii) Given sin (l +,1)=).cos(i l)ardtanl=-3,sin f cos I + cos Qsitt A =2lcos/ cosl + sinl sinll

(rh 0 i ."' / hi A = ).14f + n4 lar Hsin I 3cos Q :2(cos /-3sin()Tsrn(=5cos/

t"nd:l7(iii) Giventan,.1 : 3 and O< A


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Solution:, 16

Since sin' r.1 :25

1

5

Butl is obtuse,

s

cosA= -25

{i) cas4A- l' IJrn-){: i - r( Li^Ac.rAJ]

=, ,f/1ll lll'| \54 5/./

=t 21 4\'I 2s ,/52762s

(ii)


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Solution:(i) cos 3,1 =


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Some points to note:ln future topics you may need to simplify liacliol1s involving trigoDometric identities in thedenominator.

tllSome eramDlcs: .rnd

| (u\v I I cnc 46l ) )\rt' H

which fonnulac can you use to simplity the dcnominator so that you can get rid ofthe constant?

Answer: By applying appropriately thc doublc anglc fbnnulaeI I r r*c rdlI cu'9 , r-:..^'fdJ ,.o" l'l ' L 'r' ) )) ' )ltlt

^ -5(-ldI .os49 t ,lt."J tO I )cos' '0 'tl

By applying appropriatl3 basic identities

llll' )srn rr )l I .;n-Bt zcoJa l"'"\/

(3) 'IheR-lbrmula (Rcvisc on your orvn, it's in'O'lc!el Additiirnal Maihs syllabus)4sind l bcosd = Rsin(d ! a)4cosd tbsin P = Rcos(P+ a)

where R = , R>0 tar\ a :b

a0--.a

1t

2

Derivalion of R-formulaLet's look at the case wherea sin d + rcosd = R sin(0 + d).a sin g+ 6cos I = Rsin(d + rr)

-R sin d cos a + 1l cos 9sin a

By comparing the coefficients of sin 9, we have Rcosa = a ------- (1)

By comparing the coefficients of cos 9, we have ,Rsin(l: , ---- (2)Solving the simultaneous cquations,

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(2).(r) '

You may derive other forms

.l 1 r \

\iR2 =n2 +b2 = n=

b

(t)2 r(2)2 :

ofR-formula using the sinrilar approach as shown above.

Examplc 7Express 3 sin r + 4 cos jr in the f-olrn ,R sin (,r + rr.Solution:3sin x+ 4cosr = Rsin(.:r+a)

n=J:'+A = "[s =:43

a = 53.13'. - 3sin r+ 4cosr = 5 sin(r. + 53. 13" )

ExcrciscExpress 3sinr cos-r in the lbrrn of Rsin(r a) where 0


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By lettin9 P: A + B arld Q: A B, thus we obtain:

(l) + (2): sin P+sin O =2"i^!g r g!"o"f,@ -Q)(r) -(2): sin p s;ne=zcos)@ +e)"*;(p q)

(3)+ (4): cosp+ cosg =z*"1{r t4*"}P O)(3) (4) : cosP-cosO = z"i"l{rr O)"""f,Q' - O)

Example 8Show that(i) cosd + cos3d + cos50 = cos3A(4cos'? 9 l)(iD sind + sin 3d +sin 5A + sin 7A = l6sinAcos' g

"o"220

Solution:(D cosA+cos3d+cos5A: (.actqc + -tt) + .cr3.

=2cos30cos20 + cns30

= ..tlf ( )arlc +t_ ..ttc ( I(16rle {) +,

= cos3d(4cos'z a l)

(iD LHS: Sinf + Ji^jt + Ji^tl + Ji^'lt

= (ri"e + li^lfl+ ( sincc t3;n1";

= 2stn20 cos4 +2sin60 cosq

= )


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=2cose 2sin4e cos20

= 8cos, 2sinAcosBcos' 2B= l6 sin A cos' A cos' 2,: RHS

Exercisesin9 )sin )0 ' srn30 , gProvc lhal - lansin d + 2 sin 29+ sin 30 2

Example 9Express each ofthe followings as a sum or difference of two sine or two cosine:(a) sin5dcos39 (b) cosT9cos29 (c) sin5rsinr( ln order to integrate these three fiulctions, it is required to express them in terms of a singlefunction ofsine or cosine by using the factor formulae)

(a) sin5dcos3g= *f rln15"+tc) + ri^(5c -3d)] = !"ingg *!"rn26

(b) cosTpcos2d: i1t or('ra.1a1 + ..r (). -r5.)] =1"o"98*1"o.58

(c) sinsrsin.r: t[ - (sr,n J - ..r (\-! ] =1"o.4t-f "o"6"

ExerciseExpress 2sin 6d sin 4A as a difference of two cosine.2sin60 sin4e = -[cos(69++a)-cos(ao- ae)] =cos29 cosl0d

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3 Principal Value and Principal Range

Consider the eqLration sin d = 0.5.

sin d

Thc graph above shows that without restriction on d, therc arc infinitely many solutions. Howcvcr,

, I, O -? tl'cnd \in Ll.' lJkc. c uniqu" r'u|r". Ls. rr = sin 6., I *1'cr*1 I.4,'s

the principal range, and the angle thal lies within the principle range is thc principnl valuc.

When solving a trigonometric equation in ,ts pdncipal range, wc have the fbllowing:

']l:0.5

Function Notation Domain Principal Range

l!-r(1 '-l +'tllnvctse cosine I < -t


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My Notes:

chapter at thc cnd oflhe lesson. students

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