Trigonometry Tricks. a. Sin θ= ऱम ब / णण, cosec θ = णण / ऱम ब b. cos θ= आध र / णण, sec θ= णण / आध र c. tan θ = ऱम ब / आध र, cot θ = आध र/ ऱम ब
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1 Trigonometry Tricks 1. क स भ सम ण (Right angle) लऱय स त र (formula) णण2 = ऱम ब2 + आध र2 2. अब य द रख य LAL/KKA, (ऱ ऱ/ क ) L- ऱम ब, A- आध र, K- णण 3. अब इन क रम sin θ, cos θ, tan θ, तथ cot θ, sec θ, cosec θ इन ठ उल ट ह त ह a. Sin θ= ऱम ब / णण, cosec θ = णण / ऱम ब b. cos θ= आध र / णण, sec θ= णण / आध र c. tan θ = ऱम ब / आध र, cot θ = आध र/ ऱम ब
2 Pythagorean Identities sin 2 θ + cos 2 θ = 1 tan 2 θ + 1 = sec 2 θ cot 2 θ + 1 = csc 2 θ Negative of a Function sin ( x) = sin x cos ( x) = cos x tan ( x) = tan x csc ( x) = csc x sec ( x) = sec x cot ( x) = cot x If A + B = 90 o, Then Sin A = Cos B Sin 2 A + Sin 2 B = Cos 2 A + Cos 2 B = 1 Tan A = Cot B Sec A = Csc B
3 For example: If tan (x+y) tan (x-y) = 1, then find tan (2x/3)? Solution: Tan A = Cot B, Tan A*Tan B = 1 So, A +B = 90 o (x+y)+(x-y) = 90 o, 2x = 90 o, x = 45 o Tan (2x/3) = tan 30 o = 1/ 3 If A - B = 90 o, (A B) Then Sin A = Cos B Cos A = - Sin B Tan A = - Cot B If A ± B = 180 o, then Sin A = Sin B Cos A = - Cos B If A + B = 180 o Then, tan A = - tan B If A - B = 180 o Then, tan A = tan B For example: Find the Value of tan 80 o + tan 100 o? Solution:Since = 180 Therefore, tan 80 o + tan 100 o = 1
4 If A + B + C = 180 o, then Tan A + Tan B +Tan C = Tan A * Tan B *Tan C sin θ * sin 2θ * sin 4θ = ¼ sin 3θ cos θ * cos 2θ * cos 4θ = ¼ cos 3θ For Example:What is the value of cos 20 o cos 40 o cos 60 o cos 80 o? Solution: We know cos θ * cos 2θ * cos 4θ = ¼ cos 3θ Now, (cos 20 o cos 40 o cos 80 o ) cos 60 o ¼ (Cos 3*20) * cos 60 o ¼ Cos 2 60 o = ¼ * (½) 2 = 1/16 If a sin θ + b cos θ = m & a cos θ - b sin θ = n then a 2 + b 2 = m 2 + n 2 For Example: If 4 sin θ + 3 cos θ = 2, then find the value of 4 cos θ - 3 sin θ: Solution: Let 2 cos θ - 3 sin θ = x By using formulae a 2 + b 2 = m 2 + n = x = 4 + x 2 X = 21
5 If sin θ + cos θ = p & csc θ - sec θ = q then P (1/p) = 2/q For Example: If sin θ + cos θ = 2, then find the value of csc θ - sec θ: Solution: By using formulae: P (1/p) = 2/q 2-(1/2) = 3/2 = 2/q Q = 4/3 or csc θ - sec θ = 4/3 If a cot θ + b csc θ = m & a csc θ + b cot θ = n then b 2 - a 2 = m 2 - n 2 If cot θ + cos θ = x & cot θ - cos θ = y then x 2 - y 2 = 4 xy If tan θ + sin θ = x & tan θ - sin θ = y then x 2 - y 2 = 4 xy
6 If y = a 2 sin 2 x + b 2 csc 2 x + c y = a 2 cos 2 x + b 2 sec 2 x + c y = a 2 tan 2 x + b 2 cot 2 x + c then, y min = 2ab + c y max = not defined For Example: If y = 9 sin 2 x + 16 csc 2 x +4 then y min is: Solution: For, y min = 2* 9 * = 2*3* = = 28 If y = a sin x + b cos x + c y = a tan x + b cot x + c y = a sec x + b csc x + c then, y min = + [ (a 2 +b 2 )] + c y max = - [ (a 2 +b 2 )] + c
7 For Example: If y = 1/(12sin x + 5 cos x +20) then y max is: Solution: For, y max = 1/x min = 1/- ( ) +20 = 1/(-13+20) = 1/7 Sin 2 θ, maxima value = 1, minima value = 0 Cos 2 θ, maxima value = 1, minima value = 0
8 Trigonometric Function
9 Trigonometric Functions (Right Triangle)
10 Special Angles
11 Trigonometric Function Values in Quadrants II, III, and IV Examples:
12 Example2:
13 Example: 3: Unit Circle
14 Addition Formulas: cos(x+y) = cosxcoxy sinxsiny cos(x-y) = cosxcoxy + sinxsiny sin(x+y) = sinxcoxy + cosxsin sin(x-y) = sinxcoxy cosxsiny tan(x+y) = [tanx+tany]/ [1 tanxtany] tan(x-y) = [tanx-tany]/ [1+ tanxtany] cot(x+y) = [cotx+coty-1]/ [cotx+coty] cot(x-y) = [cotx+coty+1]/ [cotx-coty] Sum to Product Formulas: cosx + cosy = 2cos [(X+Y) / 2] cos[(x-y)/2] sinx + siny = 2sin [(X+Y) / 2] cos[(x-y)/2] Difference to Product Formulas cosx - cosy = - 2sin [(X+Y) / 2] sin[(x-y)/2] sinx + siny = 2cos [(X+Y) / 2] sin[(x-y)/2] Product to Sum/Difference Formulas cosxcosy = (1/2) [cos (x-y) + cos (X+Y)] sinxcoxy = (1/2) [sin (x+y) + sin (X-Y)] cosxsiny = (1/2) [sin (x+y) + sin (X-Y)] sinxsiny = (1/2) [cos (x-y) + cos (X+Y)] Double Angle Formulas sin (2X) = 2 sin X cos X cos (2X) = 1 2sin 2 X= 2cos 2 X 1 tan(2x) = 2tanX/[1-tan 2 X]
15 Multiple Angle Formulas More half-angle formulas Law of Sines a/sina = b/sinb= c/sinc Law of Cosines a 2 = b 2 +c 2 2bcCosA b 2 = a 2 + c 2 2ac CosB c 2 = a 2 + b 2 2abCosC Pythagorean Identities a. sin 2 X + cos 2 X = 1 b. 1 + tan 2 X = cec 2 X a. 1 + cot 2 X = csc 2 X
16 Given Three Sides and no Angles (SSS) Given three segment lengths and no angle measures, do the following: Use the Law of Cosines to determine the measure of one angle. Use the Law of Sines to determine the measure of one of the two remaining angles. Subtract the sum of the measures of the two known angles from 180 to obtain the measure of the remaining angle. Given Two Sides and the Angle between Them (SAS) Given two segment lengths and the measure of the angle that is between them, do the following: Use the Law of Cosines to determine the length of the remaining leg. Use the Law of Sines to determine the measure of one of the two remaining angles. Subtract the sum of the measures of the two known angles from 180 to obtain the measure of the remaining angle. Given One Side and Two Angles (ASA or AAS) Given one segment length and the measures of two angles, do the following: Subtract the sum of the measures of the two known angles from 180 to obtain the measure of the remaining angle. Use the Law of Sines to determine the lengths of the two remaining legs.
17 Some Important Tricks
18
19
20 Remember Useful Point : tan1. tan2. tan89 = 1 cot1. cot2. Cot89 0 = 1 cos1 0.cos2 0 cos90 0 = 0 cos1 0.cos2 0 to (greater than cos90 0 ) = 0 sin1 0.sin2 0.sin3 0 sin180 0 = 0 sin1 0. sin2 0 sin3 0.. to (greater than sin180 0 ) = 0
21 प रश न स ख य -1 sin 43 + cos 19 8cos260 cos 47 sin 71 हऱ :- sin 43 + cos 19 8cos260 cos (90-43) sin (90-19) = sin 43 + cos 19 8(1/2)2 sin 43 cos 19 = = 0 उत तर प रश न स ख य sec cosec227 cot263 sin263 = 1 + tan263 sec2(90-63) + cosec263 cosec2(90-63) = sec263 cosec263 + cosec263 sec263 = 0 उत तर
22 प रश न स ख य - 3 यदद x = cos@ त 1- sin@ cos@ म न क य ह ग?? 1+sin@ हऱ: x = cos@ 1- sin@ 1 = 1- sin@ x = (1- sin@) (1+ sin@) (1+ sin@) = 1 sin2@ (1+ sin@) = cos2@ (1+ sin@) = cos@ उत तर (1+ sin@)
23 प रश न स ख य 4- यदद + = 2 म न क य ह ग? इस हऱ रन लऱय ए ऐस म न स च ग ज स म न tan तथ cot लऱय 1 ह और ऐस 45 ऩर सम भव ह tan 45 + cot 45 = 1 = 1 प रश न स ख य -5 tan2@+3 = 3 म न क य ह ग? हऱ: tan2@+3= 3sec@ sec2@ 1 +3 = 3 sec2@ 3sec@ +2 = 0 ग णन ण ड रन ऩर sec2@ 2sec@ +2 = 0 sec@(sec@-2) -1(sec@-2)=0 (sec@-1)(sec@-2) = 0 sec@ =1 0 य sec@=2 = 60 अत:
24 प रश न स ख य -6 sin265+sin225+cos235+ cos255 म न क य ह ग? हऱ: sin265+sin2(90-65)+cos2(90-55)+ cos255) = sin265+cos265+sin255+ cos255 = 1+1 = 2
25 प रश न स ख य sin2Q+sin4Q म न ज ञ त ज य? हऱ: 1-2sin2Q+sin4Q = 1 2sin2Q+sin2Q.sin2Q = 1- sin2q sin2q+sin2q.sin2q = cos2q sin2q + (1- cos2q.1- cos2q) = cos2q (1-cos2Q) + (1- cos2q.1- cos2q) = cos2q 1 + cos2q + (1-2cos2Q + cos4q) = cos2q 1 + cos2q + 1-2cos2Q + cos4q = 2cos2Q cos2Q + cos4q = cos4q उत तर
26 प रश न स ख य -8 त रत रभ ABC म Sin (A+B)/2 म न क स बर बर ह? हऱ: क स त रत रभ ABC म A+B+C = 180 A+B+C = 90 2 A/2 + B/2 + C/2 = 90 A/2 + B /2 = 90 C/2 A+B = 90- C/2 2 Sin (A+B) = Sin 90- C/2 2 = Cos C/2 उत तर
27 प रश न स ख य -9 tan Q+Cot Q = /3 त tan3 Q + cot3 Q म न क य ह ग? हऱ: द न ओर घन रन ऩर (tan Q + cot Q)3 = (/3)3 tan3 Q + cot3 Q+ 3 tanq.cotq(tan Q+Cot Q) = 3./3 tan3 Q + cot3 Q+ 3.(/3) = 3./3 tan3 Q + cot3 Q = 3./3 3./3 tan3 Q + cot3 Q= 0 उत तर
28 प रश न स ख य -10 Cot 40 1 { cos 35} tan 50 2 {sec 55} हऱ: Cot 40 1 { cos 35} tan (90-40) 2 {sec (90-35)} Cot 40 1 { cos 35} Cot 40 2 {cos 35} = 1/2
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