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Trigonometry Tricks
1. किसी भी समिोण (Right angle) िे लऱये सूत्र (formula) – िणण2 = ऱम्ब2 + आधार2
2. अब याद रखिये LAL/KKA, (ऱाऱ/ िक्िा) L- ऱम्ब, A- आधार, K- िणण 3. अब इनिा क्रम sin θ , cos θ, tan θ, तथा cot θ, sec θ, cosec θ इनिे ठीि उल्टे होते हैं a. Sin θ= ऱम्ब / िणण, cosec θ = िणण / ऱम्ब
b. cos θ= आधार / िणण, sec θ= िणण / आधार
c. tan θ = ऱम्ब / आधार , cot θ = आधार/ ऱम्ब
Pythagorean Identities
sin2 θ + cos2 θ = 1 tan2 θ + 1 = sec2 θ cot2 θ + 1 = csc2 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 = 90o, Then
Sin A = Cos B Sin2A + Sin2B = Cos2A + Cos2B = 1 Tan A = Cot B Sec A = Csc B
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 = 90o
(x+y)+(x-y) = 90o, 2x = 90o , x = 45o
Tan (2x/3) = tan 30o = 1/√3
If A - B = 90o, (A › B) Then
Sin A = Cos B Cos A = - Sin B Tan A = - Cot B
If A ± B = 180o, then
Sin A = Sin B Cos A = - Cos B
If A + B = 180o
Then, tan A = - tan B
If A - B = 180o
Then, tan A = tan B
For example:
Find the Value of tan 80o + tan 100o ?
Solution:Since 80 + 100 = 180
Therefore, tan 80o + tan 100o = 1
If A + B + C = 180o, 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 20o cos 40o cos 60o cos 80o?
Solution: We know cos θ * cos 2θ * cos 4θ = ¼ cos 3θ
Now, (cos 20o cos 40o cos 80o ) cos 60o
¼ (Cos 3*20) * cos 60o
¼ Cos2 60o = ¼ * (½)2 = 1/16
If a sin θ + b cos θ = m & a cos θ - b sin θ = n
then a2 + b2 = m2 + n2
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 a2 + b2 = m2 + n2
42 + 32 = 22 + x2
16 + 9 = 4 + x2
X = √21
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 b2 - a2 = m2 - n2
If
cot θ + cos θ = x & cot θ - cos θ = y
then x2 - y2 = 4 √xy
If
tan θ + sin θ = x & tan θ - sin θ = y
then x2 - y2 = 4 √xy
If
y = a2 sin2x + b2 csc2x + c
y = a2 cos2x + b2 sec2x + c
y = a2 tan2x + b2 cot2x + c
then,
ymin = 2ab + c
ymax = not defined
For Example:
If y = 9 sin2 x + 16 csc2 x +4 then ymin is:
Solution:
For, y min = 2* √9 * √16 + 4
= 2*3*4 + 20 = 24 + 4 = 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, ymin = + [√(a2+b2)] + c
ymax = - [√(a2+b2)] + c
For Example:
If y = 1/(12sin x + 5 cos x +20) then ymax is:
Solution:
For, y max = 1/x min
= 1/- (√122 +52) +20 = 1/(-13+20) = 1/7
Sin2 θ, maxima value = 1, minima value = 0
Cos2 θ, maxima value = 1, minima value = 0
Trigonometric Function
Trigonometric Functions (Right Triangle)
Special Angles
Trigonometric Function Values in Quadrants II, III, and IV
Examples:
Example2:
Example: 3:
Unit Circle
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 – 2sin2X= 2cos2X – 1
tan(2X) = 2tanX/[1-tan2X]
Multiple Angle Formulas
More half-angle formulas
Law of Sines
a/sinA = b/sinB= c/sinC
Law of Cosines
a2 = b2 +c2 – 2bcCosA
b2 = a2 + c2 – 2ac CosB
c2 = a2 + b2 – 2abCosC
Pythagorean Identities
a. sin2 X + cos2 X = 1
b. 1 + tan2 X = cec2 X
a. 1 + cot2 X = csc2 X
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.
Some Important Tricks
Remember Useful Point :
tan1. tan2. ……… tan89 = 1
cot1. cot2 ……. Cot890 = 1
cos10.cos20…… cos900 = 0
cos10.cos20…… to (greater than cos900) = 0
sin10.sin20.sin30 ……… sin1800 = 0
sin10. sin20 sin30 ….. to (greater than sin1800) = 0
प्रश्न संख्या -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
= 1+1 – 2
= 0 उत्तर
प्रश्न संख्या -2
1+ 1 – sec227 + 1 – cosec227
cot263 sin263
= 1 + tan263 – sec2(90- 63) + cosec263 – cosec2(90-63)
= sec263 – cosec263 + cosec263 – sec263
= 0 उत्तर
प्रश्न संख्या- 3
यदद x = cos@ तो 1- sin@
cos@ िा मान क्या होगा ??
1+sin@
हऱ: x = cos@
1- sin@
1 = 1- sin@
x cos @
= (1- sin@) (1+ sin@)
cos @ (1+ sin@)
= 1 – sin2@
cos @ (1+ sin@)
= cos2@
cos @ (1+ sin@)
= cos@ उत्तर
(1+ sin@)
प्रश्न संख्या 4-
यदद tan @ + cot @ = 2 तो @ िा मान क्या होगा?
इसे हऱ िरने िे लऱये हम @ िा एि ऐसा मान सोचेंगे जिसिा मान tan तथा cot िे लऱये 1 हो और ऐसा 45 ऩर सम्भव है
tan 45 + cot 45 = 1
अत: @ = 1
प्रश्न संख्या -5
tan2@+3 = 3 sec @ तो @ िा मान क्या होगा ?
हऱ:
tan2@+3= 3sec@
sec2@ – 1 +3 = 3 sec @
sec2@ – 3sec@ +2 = 0
गुणनिण्ड िरने ऩर
sec2@ – 2sec@ – sec @ +2 = 0
sec@(sec@-2) -1(sec@-2)=0
(sec@-1)(sec@-2) = 0 अत: sec@ =1 तथा @= 0
या sec@=2 तथा @ = 60
प्रश्न संख्या-6
sin265+sin225+cos235+ cos255 िा मान क्या होगा?
हऱ: sin265+sin2(90-65)+cos2(90-55)+ cos255)
= sin265+cos265+sin255+ cos255
= 1+1
= 2
प्रश्न संख्या-7
1-2sin2Q+sin4Q िा मान ज्ञात िीजिये ?
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 – 1 + 1- 2cos2Q + cos4Q
= cos4Q उत्तर
प्रश्न संख्या-8
त्रत्रभुि ABC में Sin (A+B)/2 िा मान किसिे बराबर है?
हऱ:
किसी त्रत्रभुि ABC में –
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 उत्तर
प्रश्न संख्या-9
tan Q+Cot Q = /3
तो tan3 Q + cot3 Q िा मान क्या होगा ?
हऱ:
दोनों ओर घन िरने 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 उत्तर
प्रश्न संख्या-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 – 1
2
= 1/2