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( Basic Angular Systems/4 Siiat wafeat ) Ex, exagesimal System/Wtza 4 1 - 30 of a right angle = 1° 1 — =} Ue SAS FOF = 1 @o MRT min.] ° 1 , rst g5 =r Fre] 1 %0 of 1'= 1" [1 sec.] Sov Ul aee] . Lar Convert degrees, minutes to decimal degrees. 941° 27'8 ; fest, frre at aerrera feat Hage” 941° 27° 8". , HINTS: cop grcoase s 2248 wogtes Me 941° 27° 8" = 941° + 60+ eox60 ~ 20° 450 = 941° + 0.45° + 0.0022°= 941.452° Convert the decimal degrees to dgree, minutes and seconds, 48.63°, wend agi wi fet, fre sik Gave A aerii 48.63°. HINTS: 48.63°> D°M'S" D° = 48° M' = 0.63° x 60 = 37.8 M'=37' S" = 0.8 60 = 48" D°M'S" = 48.63° = 48°37'48" Ex, = ~x|radians x degrees 180 (= x x| degrees xradians r 1e(s -2) 57°16'22" 7 1°c (x = 3.141592...) 57°17'21" a 0.01746 radian t=10 Where [= length of arc r = radius ofcircle @ = angle on centre (9 is always in radians Convert the radian measure to degree measure/ # un al feat ag A agi 531 10 HINTS: —53n 10 We know that; 180° Tv 1 radian = ~53x 180° 10 “= 18% (+53) = -954° Radian System /tfzat Circumference - Diameter =n (Irrational number) = Fixed Number = | radian (14 circle #1 radius % @Wat FH arc centre W 1 radian 7 angle VET AB =L=2erx.—0 360° 50 L=r* Tee 180° L=rx@ | Where, @=6°* A rail road curve is to be laid out on a circle. What radius should be used if the track is to change direction by 25° a distance of 40 meters. te Fah UH ta We wad faorn oem 2) aie Se i 25°, 40 et al ga A fem seer 2 ot fea fen me Seam fee ST azar HINTS: Given, Length of the arc = 40 m Ed Sa . @=25°= 25 => radian “180 36 U T We know that, 9=- => r=— r 8 = 40 _40x36*7 _ 9) 64m Sx 5x22 36 So, radius of track should be 91.64m Trigonometric Ratios /faanrmtuta am) Consider an acute angle zYAX =6 with initial side AX and terminal side AY. Let P be any point on the terminal Side AY. PM perpendicular from P on AX to get the right angled trianlge AMP in which ZPAM = 0. THUG Spr _ZVAX = O21 fara wrfines ST AX Ta PTH 3 AYE pat Ay wid fay PEL PM fay PA A AX OH "Oar 2 fe gin AMP aaein # fae ZPAM = 021 a ¥ Pp h ip A 5 OM * In the figure, p = perpendicular/74 b = base/ Aur h = hypotaneous/#71 By the ‘pythagoras theorem p+ b? = h? Note: h is always greater than that of p and b. | | Perpendicular / 77 _ p fi) Sin@= “Hypotenuse / = h _ Base/27%_ _b (i) Cos0= Hypotenuse /71 "A Perpendicular / 74 _ p (iii) tané = Base / a7 b iy (Eotcg Hypotenuse / #71 ab liv) Cosec® = perpendicular / 74 p _ Hypotenuse / #7 _h () SecO= So se/ amb F Base / Tum _b vi) Cot0 = perpendicular / 3 ~ P — Important Pythagorean triplets. B45) 6.12.13) 5 815,17) (11, 60, 61) (12, 35, 37) (13, 89, 85) (16, 63, 65) (28, 45,53) [(33.56,65)_[ (36.77.85) | _(99,50,65 (65, 72, 97) (20, 99, 101) (60, 91, 109) (15, 112, 123) (88, 105, 137) | (17, 134, 145) @ Relations Between Trigonometric Ratios fratnfifa enue & dre gee (a) sind = cosec@ (b) cosecd = = (c) sin® . cosecO = 1 sin® = V1-cos*0 cos @= v1~sin’ 0 sec’ 0 = 1+ tan’?@ sec*O-tan*0=1 (sec 0+ tan 0)(sec6 ~ tan§) = 1 Note: If (cosecO— cot) Be l+cor’o O-corO=1 (cose cO~cor 0}(cosecO + cotG) = 1 Pp, then (cos psi cosec 0 = 2p & cotd= note: If (sec + tan0) = k, then (seco - mo =1/k IfsecO + tan6 = FS Kel Ke atb = sec0 = & tan = —— then, cosee@ + cot = => * (luigonomenis Ratios in Terms of other Trigonometric Ratios TaD Rit arqura qut franfafe arqunit da sin@ | cos6 tan@ coté sec@ | cosecd | fe { sin@ sia —cos?6 tan@ 1 sec? 9-1 1 | v1l+tan?@ vi+cot?9 secO cosecd | ; / cos0 fe sin? cosé 1 cotO 1 cosec’6 —1 | Vi+tan?@ | fi+cot?9 sec@ cosecd sin® v1-cos?6 1 | tan@ | ———— | ——— tand — Vsec?@—1 | —————_ | 1-sin’9 cos@ cote vcosec?8-1 | v1-sin?6 co% | cot 5 80 Z cote |——1__} Jeosec*@—1 sin® | Vi-cos?6| _tan@ vsec?6-1 1 1 z, | ¥l+cot?@ cosecO sec8 | ————— — 1+tan*@ | ———— sec@ > 1-sin?6 cos@ cot veosec? -1 ¥1+ tan? cosec@ A 1 tan’@ V1+cot?6 _Sec8 cosec8 sind ¥1-cos?@ tand vsec’ 0-1 Sign of Trigonometric Ratios () First Quadrant/sa4 "gata FECAMICIGE arqarat a foe All trigonometric ratios are positive/ast fratihfa sy7R : wae ed Zi 90" (ii) Second Quadrant/fgia agaiz AXIS ¥ ; dent sin@ and cosec®@ are positive./sinO 3% cosecO rant. s II quadrant. a ae (90° + x) Py (ii) Third Quadrant/qdta agate {180° - x) (90° - x) 7 tan@ and cot@ are positive./tand 3X cot TARAS ta a Only sin and cosec All ratios are positive (iv) Fourth Quadrant/agel aqufst ti iti ° . nating are posiive 0, 360 cos@ and sec® are positive./cos@ 3k secO WAH Ba Bt oe MI quadrant IV quadrant => 90°, 270° ....... (will convert to an odd multiple of 90°) T G => 0°, 180°, 360° ...... (Will not convert to an even multiple (180° +29 (270° +x) of 90°) (270° — x) (360° - 4 Remember: Only tan and cot Only cos and sec In Quadrant: 1 Wm IV ratios itive ratios are positive are post T-ratios: All sin tan cos + +Ve 270° -y-axis Trigonometric Ratios of Some Special Anaes| wf tan2G, tan70 © _ Sah & tral wmifaa ae = q HINTS: sin(90° - 0) = cos0 | cosec(90°~0)=sec0 —_ - cos(90°-@) = sind | sec(90°— 1) = cosecd =8= 10 [tan(90°-0)=cote | cot(90*-0) = tand Now, sin30 + cos120 sin(-6) = -sin®d | cosec(-6)= -cosec0® sin30° + cos120 oe cos(-8) = cos6 sec(-8) = sc A, _ lig tan(-6) = -tane cot(-0) = -cotu | : -373 7 B. IfA+B= 180°, sin(90°+6)=cos@ | cosec(90°+6) = secd ; : S cos(90° + 8) = —sinO | sec(90° + 6) = —cosec 8 sinA cosecB = tan(90°+6)=~-cot0| cot(90°+ 6) = -tand C. IfA+B=45° = 225°, (i) (1 + tanA) (1 + tanB) = 2 - sin(180°+ 6) =-sin0 | cosec(270°+6) = -secd Cos(180°+ 8) = —cos6 | sec(270°+ 6) = cosec 0 7 y tan(180°-6)=-tané| _cot{180°+ 6) = coto Ee [rstan{ za vx-y|fstan{2at ty-x|] sin(270°+ 0).cos(360°+ @).tan(170°+ 0). sin60° cos(180°+ 6).sin(270°— 6).cot(260°+ 4). cos 30° HINTS: (ii) (1 — cotA) (1 — cotB) = 2 HINTS: $ 1 Here, 22% +x-yt 225 ty—x= 45° ~coseos 9tan(180°s 0—10°)%2 Hence, 7 costeos 0cot(270"+ 0-107) ¥2 [rssan{ 203 soy) stan(208 yx] =2 ~1xtan(@-10°) = Ex. Find the value of (1 + tan11°) (1 + tan 21°) (1 + tan 249 tan(@-10°) (1 + tan 34°) (1 + tan 45°) Ex. Find the value of cosec (1020°) (1 + tan11°) (1 + tan 21°) (1 + tan 249) (1 + tan 34°) cosec (1020°) #1 4M et il ; (1+ tan 45°) st Am ara ei HINTS: HINTS: = cosec(3 x 360° — 60°) S 5 (1 + tan 11°) (1 + tan 219) (1 + tan 249) (1 + tan 34) 2 (1 + tan 45%) =—cosec60° = 7 v3 =2x2=4 When sum of two angles are given D. IfA+B=135°, wa at aot at direct frat et . (i (1 - tana) (1 - tanB) = 2 A. IfA+ B= 90° 25 - , (i) (1+ Cota) (1 + CotB) = 2 (i) sinA secB = 1 : (ii) cosA cosecB =1 Ex. Ifx+y = 135° then find ( + tanS](1+ tan! 3 3)" (iii) tanA tanB = 1 (iv) cotA cotB = 1 aft x+y = 135° 4 (2+s0n8}(1+tan2] ama (v) sin?A + sin?B = 1 2 2 HINTS: Ex. If tan26. tan76 = 1 then find the value of sin30 + cos128. ATQ, x+y = 135 Transformation of sum or difference into product Results to find fen al valueg | (am sreen arert ar or afters) Stat A Wd A r ufrom c& | Fy asin@+ beos0 = : ; . (S22 | sinC+sinD= 2sin| bsinO-acos@=d ‘or’ acos6-bsin@=q Jeos{ cP) =altb?=c?+d? C2D) ia c-D (2) asec0 + btan0 =c, bsecO Fa tan@=d =>a?-b? =¢c?-d? +D c-D (3) acosec 0+ beot8 = c, bcosec 8@F acotG =d cos 2 42.242 =a’-b’=c’-d | sinc - sinD = 2eo0s| cosC+cosD = me cos C -cosD = 2sin(S2?}sin D-c Ex. IfpsecO—qtan6 = 10 and ptané - qsec# = 8 then fing 4 2 2 value of p? - q? + 2. afe psec - qtand = 10 4M ptand - qsec9 = 8% Tp Some Important Results q? + 21 AA aa Hil (eo vecrqet after) HINTS: (1) sin A.sin2A.sin4A = sd oin3a Using foppuiee 4 -q? = 10?- (2) cos A.cos2A.cos4A = [00834 >Pp- (3) tanA.tanA.tan4A = tan3A 1: Adding both the sides '2' (4) sin(60° - A).sin A.sin(60°+ A) = jsin3a => p?-q?+2=38 [eon Facts Useful for Solving Question on Trigonometry (5) cos(60°- A).cos A. cos(60°+ A) = Zc0s3A fratofaf & wei a aa wet Hwee SA aT —_ (6) tan(60°—- A).tan A.tan(60° + A)= tan3A (i) (sin? A+cos? A - sin Acos A)(sinA + cos A) = sin? A+ cos’ (7) sin?(60° — A)+sin? A +sin?(60°+ A) = = 2 ; ey (iatans] tanta 3 1-coté (8) cos?(60° - A)+ cos? A + cos7(60° + A) = = 2 l-sind _1-sin@ ®)\iysind~coso 7 S°Co- tame (" Some Standard identities in triangle = I-cosd 1 (fas at ao ares wdaftrent) (a) fAES088 _ 140088 seco + cot L ‘T+eos® sind (A) IfA+B+C= 180° or x, men (5) Sosec zi 7 Sec + tand (1) tanA + tanB+tanC = tanAtanBtanC | A, BB. CC, A 6 jee (2) tan .tanz ttan5-tan> +tan> tan5=1 (8) \ ecco tang = Seco tane i i i A B c icosec 8+ coté (3) sinA+sinB+sinC = 4cos-> cos > cos > (7 [Eee OEE cosccoscotd C-1+4sin4 sinB sin€ (4) cos A +cosB+cosC = i a) (8) (Cosec 6 ~ Cot 6)?= Tretet 070 < 90" (5) sin2A +sin2B+sin2C = 4sinAsinBsinC +Cos0 9 tan@, +cot@, ——__tan8,+coto, () tan(90-6,)+cot(90 26 7 7 t2n0,,cot 0, i (B) A+ B+ C= 90% or 5 them (10) tand(1 + sec26)(1 + sec 40)(1 + sec 80) = tan80 cotA + cotB + cotC = cotA.cotB.cotC (11) sec*0-tan‘o = = sec’@+tan?Q =1+2tan’0 a —— 4 Cot7O = 14 Qeot?g = 2cosec? § ~ j b+ tand fx 4 eee Tig! y-tant mgt) if cosecO-sin0= mand secO-cos8=n then (m'n)” +(mn") =] if cotO# tan@ = x and sec0'-cos@=y then 243 Pata 2 (sY" Gy =1 6 I sinO+sin’O=1 then sin@=cos"6 and cos’@+cos'8=1 and cos""8+3cos!"8+3c08"0+c0s"8+ 2cos‘0+2cos8-2= 1 +7) If tan@=l-a? then secO+tan*6cos®cosecd=(2-a* na (17) : (18) If sin@+cos6=/3 then sin2@ = 2 {19] I cosec8 + cot = mand cosec® - cot® = n then mn = 1 70) 1 Xcos0 + Xsin@ = 1and ~siné - Zcos@ = 1 then a b a b xeiy? eee : . ar 2/8 (1) If x=acos’eand y = bsin*@ then (< (3) =1l. (22) FoosA+cos?A =1 then sin? A+sin* A=1 (23) tfsind+2cos8=1 then 2sin0-cos0=2 — (24) If coseex -sinx = a’ and secx-cosx=b” then a’b?(a? +b?) =1 25) If cotx(l+sinx)=4m and cot x{i-sinx) =4n then m?~n?=mn 26) If cosec@-sind =1 and sec@-cos0 =m, then Pm? (P +m? +3) =1 , ; a? _b? 2--,41 Q7) x= asing andy = btand then 42" y’ 2 28) x=rsinAcosC, y =rsinAsinC® and z=reosA then r?2x?4y? #2" 29) If tanossind=m and tand~sind =n; then m?—n? = 4vmn sin 80° = 0 @ [a) sind°.sin2°,sin3°.sin4?... {b) Sin}® sin2*.sin?.sind®. se 180° or, (Greater'than sin 180°) = 9 {i} (aj co (6) cos }?,cos2”.....cns90” or igreaier vee tanBy? = 1 S312 190 - 14 + Ie Sin (ili) (a) van” .tan2 (iv) Sines + Sint (9 + 24 + Tn alterms 2 (90 - 6) = Sin? 5° + Sin® 10° + Sin? 15° Sin? 907 Total terms = ~~ _ Totalterms 18 9 Sum = 2 a 2. Sin?5° + Sin? 6? + eer 85-5 1 +1=81 Total terms = Totalterms 81, = =—=40 sum 2 2 ie Max/Min Values of Trigonometric Functions | fafa wert & afer apras AT | A. msin@+ncos@ - ) Maximum value/aiftred# WA = Jim? sn? (i) Minimum value/=—0" 4H = —Jm? +n" Ex: Find the maximum and minimum value of 9sin® + 40cos0. Qsin® + 40cos0 Ht siftenan ak =AaH AM sa afr HINTS: : 9sind + 40cos®. Maximum value = (9) +(40) = J81+1600 ~ 41 Minimum value = =/(97 +(40 = —Ji681 =- 41 B. asin*0 + bcos?6 {) Maximum value/aifttmda AM = Max fa, b] (i) Minimum value /2—07 AH = Min fa, bj Ex. Find the minimum value.of 12sin*6 + 23cos*6 “12sin?@ + 23c0s"0 I ATH AR ara afere HINTS: s 12sin?6 + 23cos76 “Maximum value = 23 Minimum value = 12+0 +12 
