Introduction to Trigonometry Shobhit Nirwan

TRIGONOMETRY
HANDWRITTEN NOTES

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Designed with ta
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Shobhit Nisman

[standard 3¥ Basic- fifth strategy 5th Notes at video # alt E ) ②

k3B Some important terms related to Triangles ( ETI Ite HIGHT It that ):p
,
as a greek letter (Piatt Angle AT # IT #TvseeHFD
perpendicula→r Impotence

y 01 B

C TBase
② Pythagoras Theorem : (Hypotenuse)I(Perpendicular)'t (Base) '

for ex.in above Bg # B) ' = #C) 2+4302
*③** tHfypToHteTnuIsTe,
Base 3¥ Perpendicular depend that # That angle # Respect

HT Fl
→ Slant aint side EIHT Hypotenuse EITI
→ III Angle # 9¥ # I said HH# dint side perpendicular eiditl
→ att GT side is Base

redfin! wEEEn¥E
← PCeHrpITeTnIdidciutltayr ← Base

10 f r

Base Pe(rHpeINndTictuhlart)

TRIGONOMETRIC RATIOS

A

④ ABBE AAI Aff }Baf→IP) ratios of sides of ad
T , ,, - - - - - .

to r C 3¥ 344T # ratios hF4HT If # off trigonometric Ratios FEIT
B f, YoppIIriff
Hstlypotenvse Sino - cost -tfTopp tano ,Iz Iott
(B) -
B- Base
p → Perpendicular
tfcoset - HgSeco - tfCoto -_
-

Sino -- 1ta-no-ocost0fg=S-e1cosfnfcose.CcOost

org cosec E- d- org Seco =L tangorg Coto = = COSI
Sina
Sino cost

K3B ① sino # Csinlxlo ) ( product ietf ! ]
G-Pronounced as " sin of angle O " not
" sin into o- " X
egg :- SPNCATB) t sin At sin B

②

② sin-A = (sin A)2

COS ZA I do SA) 2

:

I
'

③ (sin A) ' t sin-TA

G (sin AT' means Hsin A but sin L- A is called sin inverse A
.

④ sino tf value tfL- s Fatt tf et etat I

mathematically , sino EEL , y - channel tf video lectures

3¥ SHEET the AHHH EM

LI: Given ;DsinA = , calculate all other trigonometric ratios
.

Lol:- T.IN#:3-HITerftTsftAptT trigonometric ratio fit valve yet I # EIGHT
death 2A yet AT THAT I
3

given, sin A=q3 I :3 I . 10=3 and tf 4 µm p
.

by pythag HE 132-1102 A ya C

1412=132 + (3) 2 BI F
B = +Ft
LG - 9 = 132 ⑤yq )or
(because side cannot be Ove .
%lB=FT

-

314T Hg Bgp IHF HT T - Att et ITH WHT !

¥ GINow , sin A = = Igcosec A- =

tf IIcos A = = see A = I

Iz ¥tan A = = Ft

cot A = II

3

K'B Ex - 8.I ¥5419115 questions Htt et Fg yet tMfTAHttrmiglon#omsettorirceyrat i os Atik
T diff
q tf then IIT 2A expression at value yet et ' A - cotta ggioonnaete. 34

uestion tf further like sin'At cost Ag sin

siie . mply value put the FT

right angledIP :- In BOPQ , at P g OP = 7cm and 00 PO- - Lem . Determine
values of Sind
Concept : * HH tf BUT NHAT AT HIT -

and Cosa . tf Pythaddtdttl

tf question

0 ③

set: Given : OP= Perp - tem tem I
. sp
00 - PQ = Lcm
100=41002-10
aH

Totnd : Sind , cos Q

Now by pytha g 002=0102 top -
O ' = (7)' t ( OPP
( Lt POI ' - 49 + Cpa) ' ( from ②)
t.tl#t2lPQI=49tlp/Q122PQ=48/PQ=24T
putting in ② g 00=1+24--25

% Perp (P) = 7cm
H = 25

** 13=24

LII- ooo sin 0=17--215 4 cos0=13--2215

If LB and LO are acute angles such that sin B= since , then prove that 43=20
.

Given : LB LO - acute angles
- ,
sin B=sTnQ

Preet To Iprtove : 43=20 are two right angled triangle.
AABC & DPOR
:

AP

B) cc QI R

sin D= AI Sind - PI

AB Pd

A-7dg sin B= sing

Ffs - pigs

p#aerpend¥icular k¥7 = ApBTeIf¥¥tfgpo et #

Hint → AT tret tf Base # Ratio sit suit ratio tf K

BC - I¥Ac2 4 QR= SER' ( by pytha) =

IAp¥AyNowy Bfg = T = Intel = kJ #

11¥22 EX

④

Sog ,A÷=FBq=Bfp
do by similarity of Ds concept :
so proofedDACBNDPRQ
✓AM 43=20
LO 00 300 450 Hence
600 900

Sino o 42 452 Bk I

also I 53/2 HE 42 0 Very
tano o
453 LB n.gg. hhfapobyetant

cosec O ND. 2 52 453 I # Rata Macedo

Seco I 453 52 2 N 'D.

Coto N 'D 53 I 453 O

27 BUT standard results
valve direct questions tf
zfetet gvalves.it#answerYI3tTlLPI-InBABCgoPght[#ZHPIgHhTtI HKT angles # Htt Hitt

for example - ex- 8.2Mt 040249

angled at BgAB=5cm,LACB=3O°. Determine the length
of sides BC and Ac .
¥g3# standard result use that liked value
soft concept : if # AT
tltilttfg LIT question then ④4dTdIoTl
A
given
:-p - 5cm
-

5cm LAC 13=300 or 300

c) 30' Toted : BC 4Ac
¥ppg
NtOaWk,ing C=3O° sides ✓ why Isin t2IH4T ? ?
sin both ans :
. P→lIH# stilted
sine = sin sintfgtantf
30 ' SHIFT sin that I

= -1 Han 2A Faith # D

,2

H = Plz) (5) (2)

' fH=LOT
-

-

by pythia g H'=P ' -1132 1y13=551
100=25+132

⑤

LI:- In DPQR , right angled at Q , PO - 3cm and PR- 6cm . Determine
-

sod :- LQPR and LPRO . tilt , Itt UTST Gut 1941-1 P

concept : feet 196¥

Given :- H- 6cm (for both angle) - Gm
-
3cm

for LP g B - 3cm
-

for LR g P =3 CM 7 lR

Q

Totind , LP & LR

conceit : - yet tis HT angle mast , let say P g. Now 134M than 3M¥ ?

for P we have g base & Hypo cos It

To cos D= -13 o

H

#12cos P = g or cos P= cos 60
:o/P=6Izcos P=

Now, 40+42+20=180 (sum of all angles of D)
60't LRT 900=1800
yM
/LR=3Q
Complementary Angles
-

G ett at angles HTT sum 90 Eft
A

→ for a right angled Dg At Btc = 1800

A -190 t C = 180
1AtC=90
bet yet ¥1

.
some9mp.to#aes:sPnl90-o-)=cosO/tanC90-O-7=cotOcotC9O-o-I-complementaryB 7CItTrigorighttfaFnegletdtD tf IIHF 3¥

Right angled

cos Cao - of = Sino fancy / sect 90 - ok coseco
f)cosec 190 - = Seco

313=7 & complementary concept AT tht tht It ?
* sit tits ett angle Hf standard angles I
tf HUT Edf
f- tht steal
net # tf def
g

⑥

LIE Evaluate tan . ?

*ACt oncept :- Yett 26

Toto tf feted
Fatal tf mist at standard
o: Taft at ett tf complementary angle af I !
MT concept
soft = tot

tanto = tan 190-64 )
d-

= cot 64 eTxeshricamitspetlfiIfystef2t1a1ttAfM19co4qn9cueeTpsttI iodndsH3E#etthFenAI
t.az#--gottg4y }:o
±

LIE c Sa me t ype Q → o f " e Ex 2,7- 8.3 → I
,
con eIpfts:i-nsHo2infAt 3=tcAroigs=mionpncleo(om+smC9eAen0-tt-ar2i9rcy60)t+ria,t-tf3icioAnos)ndc④AepH.t④dT of Rfef 3 ¥ t NHI
Asadi et *
Idf :- ata I ( IIHF e t tf change AT # t)

.

I f ]: sin 190 A- ) = COSA
= Sin ( 90 - 190-3AD )cos 190 - 3A

of )sin 3A = cos CA - 26
)cos (90-3 A) = cos CA -26

do 90 - 3A = A -26

[ /A=zgt
-

Same concept YT Ex-8.3 → 03,4g 5,6

TRIGONOMETRIC IDENTITIES

hit A 44T exercise lionfact that questions * 3-Hat 3-HIT ) at St
3 formula es 4T A based # or *i fsin'

k ¥ 3 :- 35A LI TA1T3I¥2hAoHteIIlTmtpTfoyprtHeasnHtFAqzuiTtesEatiotn s see O t C 050=1
Este f
- tan20=1
1947 E-
MKT IF left Type of question tf
CO see O - cot 20 =L
¥42 44T click TEH TEI

B⑧ IT 2A Hitt Question # 3kt TBH # HIT Types tf recall that I
CB4obodg
TYPE # I ° Direct formula the :.
.

IPI sin 263 t sin ' 27 Evaluate.

¥7tf 9

0=1 ⑦

Sdf:- Sin ' 63 + Isin 190-6335 = sin263 t cos263 I②
- costs

cos217 t (cos 190 - HD2 -

Ex - 8.4 → 03 → fig } Sin ' Ot cos 2-0=1
On → ④ Civ) questions
05 → lvii ) ( x) Type # I .

Type #2 : converting or Expressing in terms of one trigonometric
Ratio :.
Express the trigonometric ratio sin A g Seca and
LIE tan A in term of Cota
.

**Concept :- find otrfigtorignoonmometerticric ratio tf convert that ¥9272117

that # ratio treat formulae # that # I

forego here → cot A MYTH formulae# 19M¥ ?
cosec2A - cotta =L
Isaf Haff yet
use MTT l

/sinA=¥aµ⑤SOIcosec'A =Lt cot A ②-
cosec A- -
Eta

spina = Iota

Now, sin ' At COSA =L COSA = Ttt F-Lt coth

costa FEAnow,
seeA = =

secA=g#⑤( t - ¥ot2A ②④ &⑤
- ,

and ftana-co.FI ⑦- transfer
I

-

Al** Cota of sin AgcoSA # loathe that :-
tan Alcott # sin AlcoSA hose CAI see A
TYPE # 3 : Tan # sin A , cosA tf convert NHI th terms
→concept : II Ht tqhuaetsEtioFn#ctoanntAaiKnointga

prove

II:- proore that taffnoj.si?noIo=sseefof#

÷÷÷¥s;÷series : . sina.io#.:::s.I
.

..
÷

⑧

= s Cdtcosf)
-
SPIN (L - Cosa

= It 4secf sect -1L sec

- se Seco - I

I - theca sect - RHI

EPI 73¥ + = Ltsecocoseco SEE
no

Sql: sino cosofspno

:I÷÷¥÷÷÷¥ Sino - Cosa + toss
-
cost

scions - sine + Eto . so
¥t*5¥00
e.cm#tfxfaei
minus common tat etat't
:::¥%÷o::n¥÷.

sina.co#olsIonsoI-csoEnoI

1 - sTn30-cc
(Sino - Cosa) (Sino cost)

[ as - b3 = (a-b) Casetbtab))

ftp.sin20-tcos20-tslnf.cos#sinsfCsTnO-cosoC)sPno.cosQ

Ltsinocososinocoso

teenaged#join . +t

Ltsecocoseco Rts

LI:-(seeAtlanta) G- sinA) = cosec A
(A) SECA (B) Sina (D) COSA
El (
%n#)k )+ -sina

( E)Lt sina ( L - sina) [ ]sin2Ateos2A=L
costa
trgggffa . :L- s9n2A=cos2A
COSAS
COSA (D) V

I

⑨

LIE )(cosec O -
- Coto tcoso
=

It cost
Ets:- (sine - go.ms#-7k-cos-07sPn2O-
( )5h20 -1 cos
f)( 2 5h20 =L- cooee

= L - cos A

7*0

(= L - cos

t.IE#oELespeeneePnrred.

TYPE # 4 : Rationalise :-

stintII : Ey = see A + tanA that's
seal: t.es : Essien An

.

Fante u.si#n- cotsatsinsna

I + tan A

see A + tan A - RE

HomeportedTYPE # 5 : LHS 7¥ at solve # HIT if RHS ht simplify htt :

LIE Lt SECA = sin 2 A

SETA FCOSA

¥ Pooove that ¥344 sinks It 41¥ see A # Faisal etat I

sat: HII : L+÷A_ = c0{¥T¥

TOSA Yagi7¥ 3¥49 solve # yet # pus

②COSA + I - - ,

RHS (42- cossa ¥AHL#A)

= Tosa (L-C0SA#

⑤It cos A -

ProofedWe can see ②=④
Fe. LHS = RHS Hence

LI:-(cosec A - sin A) CsecA - COSA)= L ④

34¥ u - sin A. COSA

soft - , set tan At Cota ④

.

a0

IN

Sina

-
( ÷f ) (÷A )LHS
COSA

-

rising kiosks)
gqsntf sCiOsStAa sina - Cosa ②-

Rts I 1- sinoan.FI#I

ta¥tA sinsttatcgfnaa

tlencepnrred②= Fe
. LHS -_ RHS

TYPE # 6 : mlnlplq type of Question :-

LI:- If tanftsinom 4 tano - sino - n g show m2- n2=4Tmn
-

set: LES: mm fnfatnn.oIT#+tano-sinoI.Ctanof+sino-ta/no-+sinot
2 tano - Lsino
→ 54¥ 3M¥ solve that grind :. RHS theft.
4 tan # Sino
Fmsifnsoon'HE ''
+since Isin
tano CITCOSQ) qq.no#.=tanoy
gionoa-
sino sino -146%1)

-

tano ( L- cost)

Now, 45mm = 4 Ittanotldtoso ) Hana) CL- cost)
-

-4= tan 20 (L- cos20)

Is sin 20

Proved= 4 tanf Sino
= LHS HencIe

TYPE # 7 : Miscellaneous :-

LI:- If Isin F - 4050=0 , find tanto
Seco
SEL: Hino #so
- this was main Pmp. starting step.
1tano=I

④

0=3NOW , tan -P
→B

by pythag H2 = 2+45

1H=T5T
Hq IISecoa

so
- =

-

Now, tangy Idgaf Iz - Rationalistng fExT¥
⑤e
II : Prove that cos40 - cos20 = gonna - spree .

set : firstly making the question as: Same power a# same

COSY f - sin't ⑦ = coff - sin 20 → side , main starting step.

Nowg HIS ( Q'cos'- Cspnzojz

4050+820) )( cos20 -sino
-

=L Rtlsgyeneeproored

cos 20 - sin 't

I

LEFT + tosspot - 2 sea

concept : GET AT GHAT HIT 3-14 adding # equations # um et Faf!

HIS CeosA) CoosA) HustonAT c05AtLtsPn2At2sPn#

Ithaca ( Lt sin A) COSA 2cusin#

pyas of this v i d e o . £tp2¥n÷ WHOS A

Etowahchdaepstcerriptionalsooifns a m e 2 Seca teRnHuSpAnrwed