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![Illustrative Problem
A horse is tied to a post by a rope. If
the horse moves along a circular path
always keeping the rope tight and
describes 88 meters when it has traced
out 72o
at the center. Find the length of
rope. [ Take π = 22/7 approx.].
Solution
P A
B
72o
Arc AB = 88 m and AP = ?
c
o 2
72 72 rad
180 5
π π
θ = = × = ÷
J002
arc AB
r AP
θ = =
l
Q
2 88 22
AP 70m [ approx.]
5 AP 7
π
= ⇒ = π =](https://image.slidesharecdn.com/07-151106161733-lva1-app6892/85/Trigonometric-ratios-and-identities-1-16-320.jpg)
![Functions Domain Range
sinθ R [-1,1]
cosθ R [-1,1]
secθ R : (2n 1)
2
π
− θ θ = +
R-(-1,1)
cosecθ { }R : n− θ θ = π R-(-1,1)
tanθ R : (2n 1)
2
π
− θ θ = +
R
cotθ
{ }R : n− θ θ = π R
Domain and Range of Trigonometric
Function
J005](https://image.slidesharecdn.com/07-151106161733-lva1-app6892/85/Trigonometric-ratios-and-identities-1-27-320.jpg)
![Transformation Formulae
• Transformation of product into sum
and difference
• 2 sinAcosB = sin(A+B) + sin(A - B)
• 2 cosAsinB = sin(A+B) - sin(A - B)
• 2 cosAcosB = cos(A+B) + cos(A - B)
Proof :- R.H.S = cos(A+B) + cos(A - B)
= cosAcosB - sinAsinB+cosAcosB+sinAsinB
= 2cosAcosB =L.H.S
• 2 sinAsinB = cos(A - B) - cos(A+B) [Note]](https://image.slidesharecdn.com/07-151106161733-lva1-app6892/85/Trigonometric-ratios-and-identities-1-41-320.jpg)
![Class Exercise - 4
Prove that
° ° ° ° =
3
sin20 sin40 sin60 sin80
16
LHS = sin20° sin40° sin60° sin80°
Solution:-
1
sin60 [2sin20 sin40 ] sin80
2
= ° × ° ° × °
3 1
[cos(40 – 20 ) – cos(40 20 )] sin80
2 2
= × ° ° ° + ° × °
3
[cos20 – cos60 ]sin80
4
= ° ° °
3 1
sin80 cos20 – sin80
4 2
= ° ° × °
[ ]
3
2sin80 cos20 – sin80
8
= ° ° °](https://image.slidesharecdn.com/07-151106161733-lva1-app6892/85/Trigonometric-ratios-and-identities-1-48-320.jpg)
![Class Exercise - 4
Prove that
° ° ° ° =
3
sin20 sin40 sin60 sin80
16
Solution:-
[ ]3
sin(80 20 ) sin(80 – 20 )– sin80
8
= ° + ° + ° ° °
[ ]
3
sin100 sin60 – sin80
8
= ° + ° °
[ ]
3
sin(180 – 80 ) sin60 – sin80
8
= ° ° + ° °
3 3
sin80 – sin80
8 2
= ° + °
3
16
=
Proved.](https://image.slidesharecdn.com/07-151106161733-lva1-app6892/85/Trigonometric-ratios-and-identities-1-49-320.jpg)
![Class Exercise - 6
The maximum value of 3 cosx + 4
sinx + 5 is
(d) None of these
(a) 5 (b) 9
(c) 7
2 23cos x 4sinx 3 4+ = +Q
Solution :-
2 2 2 2
3 4
cos x sinx
3 4 3 4
+
+ +
3 4
5 cos x sinx
5 5
= +
[ ]5 cosx cos sinx sin= α + α
3 4
Let cos sin
5 5
α = ⇒ α =
](https://image.slidesharecdn.com/07-151106161733-lva1-app6892/85/Trigonometric-ratios-and-identities-1-52-320.jpg)
![Class Exercise -10
Solution :-
16 3
1 –
25 5
= =
2 25 12
cos( – ) 1– sin ( – ) 1–
169 13
α β = α β = =
3 5
tan( ) , tan( – )
4 12
∴ α + β = α β =
tan2 tan[( ) ( – )]α = α + β + α β
3 5
tan( ) tan( – ) 564 12
3 51– tan( )tan( – ) 33
1–
4 12
+
α + β + α β
= = =
α + β α β
×
If , ,
α and β lie between0 and , then value
α of tan2α is
4
cos( )
5
α + β =
5
sin( – )
13
α β =
4
π](https://image.slidesharecdn.com/07-151106161733-lva1-app6892/85/Trigonometric-ratios-and-identities-1-62-320.jpg)
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Trigonometric ratios and identities 1
- 1.
- 2. Trigonometric ratios andIdentities Session 1
- 3. Topics Transformation of Angles CompoundAngles Definition and Domain and Range of Trigonometric Function Measurement of Angles
- 4. Measurement of Angles OA B OA InitialRay− uuur OB Ter minal Ray− uuur Angle is considered as the figure obtained by rotating initial ray about its end point. J001
- 5. Measure and Signof an Angle Measure of an Angle :- Amount of rotation from initial side to terminal side. Sign of an Angle :- O A B θ Rotation anticlockwise – Angle positive B’ − θ Rotation clockwise – Angle negative J001
- 6. Right Angle O Y X Revolving raydescribes one – quarter of a circle then we say that measure of angle is right angle J001 Angle < Right angle ⇒ Acute Angle Angle > Right angle ⇒ Obtuse Angle
- 7. Quadrants O Y Y’ X’ X II Quadrant (, )− + I Quadrant ( , )+ + IV Quadrant ( , )+ − III Quadrant ( , )− − X’OX – x - axis Y’OY – y - axis J001
- 8. System of Measurementof Angle Measurement of Angle Sexagesimal System or British System Centesimal System or French System Circular System or Radian Measure J001
- 9. System of Measurementof Angles Sexagesimal System (British System) 1 right angle = 90 degrees (=90o ) 1 degree = 60 minutes (=60’) 1 minute = 60 seconds (=60”) Centesimal System (French System) 1 right angle = 100 grades (=100g ) 1 grade = 100 minutes (=100’) 1 minute = 100 Seconds (=100”) J001 Is 1 minute of sexagesimal 1 minute of centesimal ? = NO
- 10. System of Measurementof Angle Circular System J001 O r r r A B 1c If OA = OB = arc AB c Then AOB 1radian( 1 )∠ = =
- 11. System of Measurementof Angle Circular System O A C B 1c AOC arc AC AOB arc ACB ∠ = ∠ Q 1radian r 2right angles r ⇒ = π 2right angles radian⇒ = π J001 180 radian⇒ = πo
- 12. Relation Between DegreeGrade And Radian Measure of An Angle 0 g D G 2C 90 100 = = π OR 0 g D G C 180 200 = = π J002
- 13. Illustrative Problem Find thegrade and radian measures of the angle 5o 37’30” Solution o' o 30 30 1 30" 60 60 60 120 = = = ÷ ÷ ÷× o 37 and37' 60 = ÷ o o o 37 1 45 5 37'30" 5 60 120 8 ∴ = + + = ÷ ÷ J002 We know that D G 2C 90 100 = = π
- 14. Illustrative Problem Find thegrade and radian measures of the angle 5o 37’30” g 10 G D 9 ⇒ = × ÷ ( ) g g g10 45 225 12.5 Ans 9 8 18 = × = = ÷ ÷ c and R D 180 π = × ÷ c 45 radian Ans 180 8 32 π π = × = ÷ Solution J002
- 15. Relation Between AngleSubtended by an Arc At The Center of Circle O A C 1c θ B Arc AC = r and Arc ACB = AOC arc AC AOB arc ACB ∠ = ∠ Q J002 1radian r ⇒ = θ l r ⇒ θ = l
- 16. Illustrative Problem A horseis tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 88 meters when it has traced out 72o at the center. Find the length of rope. [ Take π = 22/7 approx.]. Solution P A B 72o Arc AB = 88 m and AP = ? c o 2 72 72 rad 180 5 π π θ = = × = ÷ J002 arc AB r AP θ = = l Q 2 88 22 AP 70m [ approx.] 5 AP 7 π = ⇒ = π =
- 17. Definition of TrigonometricRatios J003 2 2 r x y= + xO Y X P (x,y) M θ y r y sin r x cos r y tan x θ = θ = θ = x cot y r sec x r cosec y θ = θ = θ =
- 18. Some Basic Identities sincosec 1 ; n ,n I• θ × θ = θ ≠ π ∈ 2 2 sin cos 1• θ + θ = 2 2 sec tan 1• θ − θ = 2 2 cosec cot 1• θ − θ = ( ) sin tan ; 2n 1 ,n I cos 2 θ π • θ = θ ≠ + ∈ θ ( )tan cot 1 ; 2n 1 ; n ,n I 2 π • θ × θ = θ ≠ + θ ≠ π ∈ cos cot ; n ,n I sin θ • θ = θ ≠ π ∈ θ ( )cos sec 1 ; 2n 1 ,n I 2 π • θ × θ = θ ≠ + ∈
- 19. Illustrative Problem Solution 3 sec tan.cosec= θ + θ θ 3 cosec sec 1 tan sec θ = θ + θ ÷θ ( )3 sec 1 tan cot= θ + θ θ ( )2 sec 1 tan= θ + θ ( )2 2 1 tan 1 tan= + θ + θ ( ) 3 2 21 tan= + θ ( ) 3 2 22 e Proved= − J0032 2 If tan 1 e ,prove thatθ = − ( ) 3 3 2 2sec tan .cosec 2 eθ + θ θ = − 0 2 π < θ < ÷
- 20. Signs of TrigonometricFunction In All Quadrants In First Quadrant xO Y X P (x,y) M θ y r Here x >0, y>0, 2 2 r x y= + >0 y sin 0 r θ = > x cos 0 r θ = > y tan 0 x θ = > x cot 0 y θ = > r sec 0 x θ = > r cosec 0 y θ = > J004
- 21. Signs of TrigonometricFunction In All Quadrants In Second Quadrant Here x <0, y>0, 2 2 r x y= + >0 y sin 0 r θ = > r cosec 0 y θ = > θ XX’ Y Y’ P (x,y) x y r x cos 0 r θ = < y tan 0 x θ = < x cot 0 y θ = < r sec 0 x θ = < J004
- 22. Signs of TrigonometricFunction In All Quadrants In Third Quadrant Here x <0, y<0, 2 2 r x y= + >0 r cosec 0 y θ = > r sec 0 x θ = < θ X’ X P (x,y) O Y’ Y M y sin 0 r θ = < x cos 0 r θ = < y tan 0 x θ = > x cot 0 y θ = > J004
- 23. Signs of TrigonometricFunction In All Quadrants In Fourth Quadrant Here x >0, y<0, 2 2 r x y= + >0 y sin 0 r θ = < θ XO P (x,y) Y’ M x cos 0 r θ = > y tan 0 x θ = < x cot 0 y θ = < r sec 0 x θ = > r cosec 0 y θ = < J004
- 24. Signs of TrigonometricFunction In All Quadrants I Quadrant All Positive II Quadrant sin & cosec are Positive III Quadrant tan & cot are Positive IV Quadrant cos & sec are Positive X Y’ X’ Y O J004 ASTC :- All Sin Tan Cos
- 25. Illustrative Problem θ liesin secondIf cot θ = 12 , 5 − quadrant, find the values of other five trigonometric function Solution 12 5 cot tan 5 12 θ = − ⇒ θ = −Q 2 2 2 169 sec 1 tan sec 144 θ = + θ ⇒ θ =Q ( ) 13 13 sec sec liesinsec ondquadrant 12 12 ⇒ θ = ± ⇒ θ = − θ 12 Whichgivescos 13 θ = − 13 cosec 5 ∴ θ = J004 Method : 1 5 12 5 Thensin tan cos 12 13 13 θ = θ × θ = − × − =
- 26. Illustrative Problem θ liesin secondIf cot θ = 12 , 5 − quadrant, find the values of other five trigonometric function Solution J004 Method : 2 Y θ XX’ Y’ P (-12,5) -12 5 r Here x = -12, y = 5 and r = 13 y 5 sin r 13 θ = = x 12 cos r 13 − θ = = y 5 tan x 12 θ = = − r 13 sec x 12 θ = = − r 13 cosec y 5 θ = =
- 27. Functions Domain Range sinθR [-1,1] cosθ R [-1,1] secθ R : (2n 1) 2 π − θ θ = + R-(-1,1) cosecθ { }R : n− θ θ = π R-(-1,1) tanθ R : (2n 1) 2 π − θ θ = + R cotθ { }R : n− θ θ = π R Domain and Range of Trigonometric Function J005
- 28. Illustrative problem Prove that 2 2(x y) sin 4xy + θ = is possible for real values of x and y only when x=y Solution ( ) 2 2(x y) 1 x y 4xy 4xy + ⇒ ≤ ⇒ + ≤ 2 sin 1θ ≤Q ( ) ( )2 2 x y 4xy 0 x y 0⇒ + − ≤ ⇒ − ≤ But for real values of x and y ( )2 x y− is not less than zero ( )2 x y 0 x y Pr oved∴ − = ⇒ = J005
- 29. Trigonometric Function ForAllied Angles Trig. ratio -θ 90o -θ 90o +θ 180o -θ 180o +θ 360o -θ 360o +θ cosθ cosθ sinθ - sinθ - cosθ - cosθ cosθ cosθ tanθ - tanθ cotθ - cotθ -tanθ tanθ - tanθ tanθ sinθ - sinθ cosθ cosθ sinθ - sinθ - sinθ sinθ If angle is multiple of 900 then sin ⇔ cos;tan ⇔ cot; sec ⇔ cosec If angle is multiple of 1800 then sin ⇔ sin;cos ⇔ cos; tan ⇔ tan etc.
- 30. Trigonometric Function ForAllied Angles Trig. ratio -θ 90o -θ 90o +θ 180o -θ 180o +θ 360o -θ 360o +θ secθ secθ cosecθ - cosecθ - secθ - secθ secθ secθ cosecθ - cosecθ secθ secθ cosecθ -cosecθ - cosecθ cosecθ cotθ - cotθ tanθ -tanθ -cotθ cotθ - cotθ cotθ
- 31. Periodicity of Trigonometric Function Periodicity: After certain value of x the functional values repeats itself Period of basic trigonometric functions sin (360o +θ) = sinθ ⇒ period of sinθ is 360o or 2π cos (360o +θ) = cosθ ⇒ period of cosθ is 360o or 2π tan (180o +θ) = tanθ ⇒ period of tanθ is 180o or π J005 If f(x+T) = f(x) ∀ x,then T is called period of f(x) if T is the smallest possible positive number
- 32. Trigonometric Ratio ofCompound Angle Angles of the form of A+B, A-B, A+B+C, A-B+C etc. are called compound angles (I) The Addition Formula • sin (A+B) = sinAcosB + cosAsinB • cos (A+B) = cosAcosB - sinAsinB ( ) tanA tanB tan A B 1 tanA tanB + • + = − J006 ( ) sin(A B) Pr oof: tan A B cos(A B) + − + = + Q sinA cosB cos A sinB cos A cosB sinA sinB + = −
- 33. Trigonometric Ratio ofCompound Angle J006 sinA cosB cos A sinB cos A cosB sinA sinB + = − r r Dividing N and D by cos A cosB We get tanA tanB 1 tanA tanB + = − Proved ( ) cotBcot A 1 cot A B cotB cot A − • + = +
- 34. Illustrative problem Find thevalue of (i) sin 75o (ii) tan 105o Solution (i) Sin 75o = sin (45o + 30o ) = sin 45o cos 30o + cos 45o sin 30o 1 3 1 1 3 1 2 22 2 2 2 + = × + × = ( )(ii) Ans: 2 3− +
- 35. Trigonometric Ratio of CompoundAngle (I) The Difference Formula • sin (A - B) = sinAcosB - cosAsinB • cos (A - B) = cosAcosB + sinAsinB ( ) tanA tanB tan A B 1 tanA tanB − • − = + Note :- by replacing B to -B in addition formula we get difference formula ( ) cotB cot A 1 cot A B cot A cotB + • − = − −
- 36. Illustrative problem If tan(θ+α) = a and tan (θ - α) = b Prove that a b tan2 1 ab − α = − Solution ( ) ( ){ }tan2 tanα = θ + α − θ − α ( ) ( ) ( ) ( ) tan tan 1 tan tan θ + α − θ − α = + θ + α θ − α a b 1 ab − = +
- 37. Some Important Deductions •sin (A+B) sin (A-B) = sin2 A - sin2 B = cos2 B - cos2 A • cos (A+B) cos (A-B) = cos2 A - sin2 B = cos2 B - sin2 A ( ) tanA tanB tanC tanA tanB tanC tan A B C 1 tanA tanB tanB tanC tanC tanA + + − • + + = − − −
- 38. To Express acosθ+ bsinθ in the form kcosφ or λsinψ acosθ +bsinθ 2 2 2 2 2 2 a b a b cos sin a b a b = + θ + θ ÷ + + 2 2 2 2 a b Let cos ,thensin a b a b α = α = + + ( )2 2 acos b sin a b cos cos sin sin∴ θ + θ = + θ α + θ α ( )2 2 a b cos= + θ − α Similarly we get acosθ + bsinθ = λsinψ 2 2 k cos ,where k a b ,= φ = + φ = θ − α
- 39. Illustrative problem 7cosθ +24sinθ Findthe maximum and minimum values of 7cosθ + 24sinθ Solution 2 2 2 2 2 2 7 24 7 24 cos sin 7 24 7 24 = + θ + θ ÷ + + 7 24 25 cos sin 25 25 = θ + θ ÷ 7 24 Let cos sin 25 25 α = ⇒ α = ( )7cos 24sin 25 cos cos sin sin∴ θ + θ = θ α + θ α
- 40. Illustrative problem Find themaximum and minimum value of 7cosθ + 24sinθ Solution ( )25cos 25cos where= θ − α = φ φ = θ − α 1 cos 1− ≤ φ ≤Q 25 25cos 25⇒ − ≤ φ ≤ ∴ Max. value =25, Min. value = -25 Ans.
- 41. Transformation Formulae • Transformationof product into sum and difference • 2 sinAcosB = sin(A+B) + sin(A - B) • 2 cosAsinB = sin(A+B) - sin(A - B) • 2 cosAcosB = cos(A+B) + cos(A - B) Proof :- R.H.S = cos(A+B) + cos(A - B) = cosAcosB - sinAsinB+cosAcosB+sinAsinB = 2cosAcosB =L.H.S • 2 sinAsinB = cos(A - B) - cos(A+B) [Note]
- 42. Transformation Formulae • Transformationof sums or difference into products C D C D sinC sinD 2sin cos 2 2 + − • + = C D C D sinC sinD 2cos sin 2 2 + − • − = C D C D cosC cosD 2cos cos 2 2 + − • + = C D C D cosC cosD 2sin sin 2 2 + − • − = − C D D C cosC cosD 2sin sin 2 2 + − • − = or Note By putting A+B = C and A-B = D in the previous formula we get this result
- 43. Illustrative problem Prove that cos8xcos6x cos 4x cot 6x sin8x sin6x sin4x + + = + + Solution (cos8x cos 4x) cos6x L.H.S (sin8x sin4x) sin6x + + = + + 8x 4x 8x 4x 2cos cos cos6x 2 2 8x 4x 8x 4x 2sin cos sin6x 2 2 + − + = + − + 2cos6x cos2x cos6x 2sin6x sin2x sin6x + = + 2cos6x(cos2x 1) 2sin6x(cos2x 1) + = + cot 6x= Proved
- 44. Class Exercise -1 If the angular diameter of the moon be 30´, how far from the eye can a coin of diameter 2.2 cm be kept to hide the moon? (Take p = approximately) 22 7 A B E ( E y e ) r M o o n
- 45. Class Exercise -1 If the angular diameter of the moon be 30´, how far from the eye can a coin of diameter 2.2 cm be kept to hide the moon? (Take p = approximately) 22 7 A B E ( E y e ) r M o o n Solution :- Let the coin is kept at a distance r from the eye to hide the moon completely. Let AB = Diameter of the coin. Then arc AB = Diameter AB = 2.2 cm c c 30 1 30´ 60 2 180 360 π π θ = = = × = o arc 2.2 radius 360 r π ∴ θ = ⇒ = 360 2.2 360 2.2 7 r 252 cm 22 × × × = = = π
- 46. Class Exercise -2 Solution :- Prove that tan3A tan2A tanA = tan3A – tan2A – tanA. We have 3A = 2A + A ⇒ tan3A = tan(2A + A) ⇒ tan3A = tan2A tanA 1– tan2A tanA + ⇒ tan3A – tan3A tan2A tanA = tan2A + tanA ⇒ tan3A – tan2A – tanA = tan3A tan2A tanA (Proved)
- 47. Class Exercise -3 If sinα = sinβ and cosα = cosβ, then – sin 0 2 α β =(c) – cos 0 2 α β =(d) sin 0 2 α + β =(a) cos 0 2 α + β =(b) Solution :- α = βQsin sin α = βcos cosand ⇒ α β = α β =sin – sin 0 and cos – cos 0 – – 2sin cos 0 and – 2sin sin 0 2 2 2 2 α β α + β α β α + β ⇒ = = – sin 0 2 α β ⇒ =
- 48. Class Exercise -4 Prove that ° ° ° ° = 3 sin20 sin40 sin60 sin80 16 LHS = sin20° sin40° sin60° sin80° Solution:- 1 sin60 [2sin20 sin40 ] sin80 2 = ° × ° ° × ° 3 1 [cos(40 – 20 ) – cos(40 20 )] sin80 2 2 = × ° ° ° + ° × ° 3 [cos20 – cos60 ]sin80 4 = ° ° ° 3 1 sin80 cos20 – sin80 4 2 = ° ° × ° [ ] 3 2sin80 cos20 – sin80 8 = ° ° °
- 49. Class Exercise -4 Prove that ° ° ° ° = 3 sin20 sin40 sin60 sin80 16 Solution:- [ ]3 sin(80 20 ) sin(80 – 20 )– sin80 8 = ° + ° + ° ° ° [ ] 3 sin100 sin60 – sin80 8 = ° + ° ° [ ] 3 sin(180 – 80 ) sin60 – sin80 8 = ° ° + ° ° 3 3 sin80 – sin80 8 2 = ° + ° 3 16 = Proved.
- 50. Class Exercise -5 Prove that n n n cos A cosB sin A sinB sin A – sinB cos A – cosB A B 2 cot , if n is even 2 0, if n is odd + + + + = Solution :- n n cos A cosB sinA sinB LHS sinA – sinB cos A – cosB + + = + n n A B A – B A B A – B 2cos cos 2sin cos 2 2 2 2 A B A – B A B A – B 2cos sin –2sin sin 2 2 2 2 + + = + + + n n A– B A – B cot –cot 2 2 = +
- 51. Class Exercise -5 Solution :- n n nA – B A – B cot (–1) cot 2 2 = + n n n A – B 2cot , if n is evenA – B cot 1 (–1) 2 2 0, if n is odd = + = Prove that n n n cos A cosB sin A sinB sin A – sinB cos A – cosB A B 2 cot , if n is even 2 0, if n is odd + + + + =
- 52. Class Exercise -6 The maximum value of 3 cosx + 4 sinx + 5 is (d) None of these (a) 5 (b) 9 (c) 7 2 23cos x 4sinx 3 4+ = +Q Solution :- 2 2 2 2 3 4 cos x sinx 3 4 3 4 + + + 3 4 5 cos x sinx 5 5 = + [ ]5 cosx cos sinx sin= α + α 3 4 Let cos sin 5 5 α = ⇒ α =
- 53. Class Exercise -6 The maximum value of 3 cosx + 4 sinx + 5 is Solution :- 3 4 Let cos sin 5 5 α = ⇒ α = 5cos(x – )= α –1 cos(x – ) 1≤ α ≤Q –5 5cos(x – ) 5⇒ ≤ α ≤ –5 5 5cos(x – ) 5 10⇒ + ≤ α + ≤ 0 3cosx 4sinx 5 10⇒ ≤ + + ≤ ∴ Maximum value of the given expression = 10.
- 54. Class Exercise -7 If a and b are the solutions of a cosθ + b sinθ = c, then show that 2 2 2 2 a – b cos( ) a b α+β = + Solution :- We have … (i)acos bsin cθ + θ = acos c – bsin⇒ θ = θ 2 2 2 a cos (c – bsin )⇒ θ = θ ( )2 2 2 2 2 a 1– sin c –2bcsin b sin⇒ θ = θ + θ ( )2 2 2 2 2a b sin – 2bcsin (c – a ) 0⇒ + θ θ + = ∴αβare roots of equatoin (i),
- 55. Class Exercise -7 If a and b are the solutions of acosθ + bsinθ = c, then show that α + β = + 2 2 2 2 a – b cos( ) a b Solution :- 2 2 2 2 c – a sin sin a b α β = + Hence Again from (i),bsin c – acosθ = θ 2 2 2b sin (c – acos )⇒ θ = θ 2 2 2 2 2b (1– cos ) c a cos –2cacos⇒ θ = + θ θ 2 2 2 2 2(a b )cos – 2ac cos c – b 0⇒ + θ θ + = ∴sinα and sinβ are roots of equ. (ii).
- 56. Class Exercise -7 If a and b are the solutions of acosθ + bsinθ = c, then show that α + β = + 2 2 2 2 a – b cos( ) a b Solution :- (iv) ∴α and β be the roots of equation (i), ∴cosα and cosβ are the roots of equation (iv). 2 2 2 2 c – b cos cos a b α β = + 2 2 2 2 2 2 2 2 2 2 2 2 c – b c – a a – b – a b a b a b = = + + + cos( ) cos cos – sin sinα + β = α β α βNow
- 57. Class Exercise -8 If a seca – c tana = d and b seca + d tana = c, then (a) a2 + b2 = c2 + d2 + cd (c) a2 + b2 = c2 + d2 (d) ab = cd + = +2 2 2 2 1 1 a b c d (b)
- 58. Class Exercise -8 If a seca – c tana = d and b seca + d tana = c, then asec – c tan dα α =Q a c sin – d cos cos α ⇒ = α α Solution :- b sec dtan cα + α =QAgain b dsin c cos cos α ⇒ + = α α ⇒ = α αb ccos – dsin ….. ii a csin dcos⇒ = α + α ….. (I) Squaring and adding (i) and (ii), we get 2 2 2 2 2 2 2 2a b c (sin cos ) d (cos sin )+ = α + α + α + α 2cd sin cos 2cd sin cos+ α α − α α 2 2c d= +
- 59. Class Exercise -9 22A A sin – sin – 8 2 8 2 π π + The value of (a) 2 sinA (c) 2 cosA 1 sinA 2 (b) 1 cosA 2 (d)
- 60. Class Exercise -9 22A A sin – sin – 8 2 8 2 π π + Q Solution :- A A A A sin – sin – – 8 2 8 2 8 2 8 2 π π π π = + + + 1 sin sin A sin A 4 2 π = ⋅ = 2 2 sin A – sin B sin(A B)sin(A – B) = + Q 2 2A A sin – sin – 8 2 8 2 π π + The value of
- 61. Class Exercise -10 If, , α and β lie between0 and , then value α of tan2α is 4 cos( ) 5 α + β = 5 sin( – ) 13 α β = 4 π (a) 1 (c) 0 (b) 56 33 (d) 33 56 Solution :- ∴α and βbetween 0 and , π 4 – – and 0 4 4 2 π π π ∴ < α β < < α + β < Consequently, cos(α − β) and sin(α + β) are positive. 2 sin( ) 1– cos ( )α + β = α + β
- 62. Class Exercise -10 Solution:- 16 3 1 – 25 5 = = 2 25 12 cos( – ) 1– sin ( – ) 1– 169 13 α β = α β = = 3 5 tan( ) , tan( – ) 4 12 ∴ α + β = α β = tan2 tan[( ) ( – )]α = α + β + α β 3 5 tan( ) tan( – ) 564 12 3 51– tan( )tan( – ) 33 1– 4 12 + α + β + α β = = = α + β α β × If , , α and β lie between0 and , then value α of tan2α is 4 cos( ) 5 α + β = 5 sin( – ) 13 α β = 4 π
Editor's Notes
- #28 Faculty should explain domain of any one of trigonometric function in the class with the help off funda book.
- #34 Ask the students to attempt cot(A+B) themselves
- #38 Ask the students to find the expressions for Sin ( A+B+C) and Cos ( A+B+C)
- #41 Ask them to try using the Sin (A+B ) form
- #42 Ask student to do 2 sinAsinB = cos(A - B) - cos(A+B)