Trigonometry for class xi

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Trigonometry for class xi

  1. 1. WEIGHTAGE FOR CLASS ---XI One Paper Three Hours Max Marks. 100 Units Marks I. SETS AND FUNCTIONS 29 II. ALGEBRA 37 III. COORDINATE GEOMETRY 13 IV. CALCULUS 06 V. MATHEMATICAL REASONING 03 VI. STATISTICS AND PROBABILITY 12 100Fundamental Trigonometric IdentitiesBefore we start to prove trigonometric identities, we seewhere the basic identities come from.Recall the definitions of the reciprocal trigonometricfunctions, csc θ, sec θ and cot θfrom the trigonometric functions chapter:
  2. 2. After we revise the fundamental identities, we learnabout:Proving trigonometric identitiesNow, consider the following diagram where the point (x,y) defines an angle θ at the origin,and the distance from the origin to the point is r units: From the diagram, we can see that the ratios sin θand cos θ are defined as: andNow, we use these results to find an important definitionfor tan θ:
  3. 3. Now, also so we can conclude that: Also, for the values in the diagram, we can usePythagoras Theorem and obtain: y2 + x2 = r2 Dividing through by r2 gives us: so we obtain the important result:
  4. 4. sin2 θ + cos2 θ = 1 We now proceed to derive two other relatedformulas that can be used when proving trigonometricidentities. It is suggested that you remember how to find theidentities, rather than try to memorise each one. Dividing sin2θ + cos2 θ = 1 through by cos2θ gives us: so tan2 θ + 1 = sec2 θ Dividing sin2θ + cos2 θ = 1 through by sin2θ gives us: so 1 + cot2 θ = csc2 θ Trigonometric Identities Summary
  5. 5. Proving Trigonometric Identities Suggestions...1. Learn well the formulas given above (or at least, know how to find them quickly).The better you know the basic identities, the easier itwill be to recognise what is going on in the problems.2. Work on the most complex side and simplify it so that it has the same form as the simplest side.Dont assume the identity to prove the identity.This means dont work on both sides of the equalsside and try to meet in the middle.3. Start on one side and make it look like the other side.4. Many of these come out quite easily if you express everything on the most complex side in terms of sine and cosine only.5. In most examples where you see power 2 (that is, 2 ), it will involve using the identity sin2 θ + cos2 θ = 1 (or one of the other 2 formulas that we derived above).
  6. 6. Using these suggestions, you can simplify and proveexpressions involving trigonometric identities. Prove that sin y + sin y cot2 y = cosec yAnswer
  7. 7. Functio Abbreviati Descripti Identities (using radians)n on onSine sinCosine cosTangent tan (or tg)Cotange cot (or ctgnt or ctn)
  8. 8. Secant sec Cosecan csc t (or cosec) Note that these values can easily be memorized in theform but the angles are not equally spaced.
  9. 9. The values for 15°, 54° and 75° are slightly morecomplicated. [ by using formulas of sin(A-B),sin(A+B) Similarly for cosine function & tan function.] Special values in trigonometric functions There are some commonly used special values in trigonometric functions, as shown in the following table.Function s 0 1 in cos 1 0 tan 0 1 cot 1 0 sec 1 2 1 csc 2
  10. 10. The sine and cosine functions graphed on the Cartesianplane. For angles greater than 2π or less than −2π, simply continue to rotate around the circle; sine and cosine are periodic functions with period 2π: for any angle θ and any integer k. The smallest positive period of a periodic function iscalled the primitive period of the function. The primitive period of the sine or cosine is a full circle,i.e. 2π radians or 360 degrees.
  11. 11. Figure 1If Q(x,y) is the point on the circle where the string ends,we may think of as being an angle by associating to itthe central angle with vertex O(0,0) and sides passingthrough the points P and Q. If instead of wrapping alength s of string around the unit circle, we decide towrap it around a circle of radius R, the angle (in radians)generated in the process will satisfy the followingrelation:
  12. 12. Observe that the length s of string gives the measure ofthe angle only when R=1.As a matter of common practice and convenience, it isuseful to measure angles in degrees, which are definedby partitioning one whole revolution into 360 equalparts, each of which is then called one degree. In thisway, one whole revolution around the unit circlemeasures radians and also 360 degrees (or ), thatis:Each degree may be further subdivided into 60 parts,called minutes, and in turn each minute may besubdivided into another 60 parts, called seconds:Angle sum identities
  13. 13. SineIllustration of the sum formula.Draw the angles α and β. Place P on the line defined by α+ β at unit distance from the origin.Let PQ be a perpendicular from P to the line defined bythe angle α. OQP is a right angle.Let QA be a perpendicular from Q to the x axis, and PBbe a perpendicular from P to the x axis. OAQ is a rightangle.Draw QR parallel to the x-axis. Now angle RPQ = α(because
  14. 14. or PQ/OP = Sin or OQ/OP = Cos (if OP ≠ 1) , so , so Or = = = . + . (∵ RB = AQ , PQ & OQ are the hyp. ) = . + . .By substituting − β for β and using Symmetry, we alsoget:
  15. 15. CosineUsing the figure above, OR PQ/OP = Sin OR OQ/OP = Cos , so , so
  16. 16. Or = = = - . (AB=RQ) = . - . .By substituting − β for β and using Symmetry, we alsoget:Also, using the complementary angle formulae,**Another simple "proof" can be given using Eulersformula known from complex analysis: Eulersformula is:Although it is more precise to say that Eulersformula entails the trigonometric identities, it followsthat for angles α and β we have:
  17. 17. ei(α + β) = cos(α + β) + isin(α + β) Also using the following properties of exponential functions: ei(α + β) = eiαeiβ = (cosα + isinα)(cosβ + isinβ) Evaluating the product: ei(α + β) = (cosαcosβ − sinαsinβ) + i(sinαcosβ + sinβcosα) This will only be equal to the previous expression we got, if the imaginary and real parts are equal respectively. Hence we get: cos(α + β) = cosαcosβ − sinαsinβ sin(α + β) = sinαcosβ + sinβcosα ] Tangent and cotangent From the sine and cosine formulae, we get Dividing both numerator and denominator by cos αcos β, we get
  18. 18. Similarly (using a division by sin α sin β), we get Double-angle identities From the angle sum identities, we get and The Pythagorean identities give the two alternativeforms for the latter of these: The angle sum identities also give
  19. 19. **It can also be proved using Eulers Formula Mulitplying the exponent by two yields But replacing the angle with its doubled version, which achieves the same result in the left side of the equation, yields It follows that By multiplying we get Because the imaginary and real parts have to be thesame, we are left with the original identities Half-angle identities The two identities giving alternative forms for cos 2θgive these:
  20. 20. One must choose the sign of the square root properly—note that if 2π is added to θ the quantities inside the square roots are unchanged, but the left- hand-sides of the equations change sign. Therefore the correct sign to use depends on the value of θ. tan function, we have If we multiply the numerator and denominator inside the square root by (1 + cos θ), and do a little manipulation using the Pythagorean identities, we get If instead we multiply the numerator anddenominator by (1 - cos θ), we get This also gives
  21. 21. Similar manipulations for the cot function giveExample. verify the identity Answer. We have which gives But and since and , we get finally Remark. In general it is good to check whether the given formula is correct. One way to do that is to substitute
  22. 22. some numbers for the variables. For example, if we take a=b = 0, we get or we may take . In this case we have Example. Find the exact value of Answer. We have Hence, using the additions formulas for the cosinefunction we get Since we get
  23. 23. Example. Find the exact value for Answer. We have Since we get Finally we have Remark. Using the addition formulas, we generate thefollowing identities
  24. 24. Double-Angle and Half-Angle formulas are very useful.For example, rational functions of sine and cosine wil bevery hard to integrate without these formulas. They areas followExample. Check the identities
  25. 25. Answer. We will check the first one. the second one isleft to the reader as an exercise. We have Hence which implies Many functions involving powers of sine and cosine are hard to integrate. The use of Double-Angle formulas help reduce the degree of difficulty. Example. Write as an expression involving the trigonometric functions with their first power. Answer. We have Hence Since , we get
  26. 26. or Example: Verify the identity Answer. We have Using the Double-Angle formulas we get Putting stuff together we get From the Double-Angle formulas, one may generateeasily the Half-Angle formulas
  27. 27. In particular, we haveExample. Use the Half-Angle formulas to findAnswer. Set . ThenUsing the above formulas, we get
  28. 28. Since , then is a positive number.Therefore, we haveSame arguments lead toExample. Check the identitiesAnswer. First note thatwhich falls from the identity. So we need to verify only one identity. For example, letus verify thatusing the Half-Angle formulas, we get
  29. 29. which reduces toProduct and Sum FormulasFrom the Addition Formulas, we derive the followingtrigonometric formulas (or identities)Remark. It is clear that the third formula and the fourthare identical (use the property to see it).The above formulas are important whenever need rises totransform the product of sine and cosine into a sum. Thisis a very useful idea in techniques of integration.
  30. 30. Example. Express the product as asum of trigonometric functions.Answer. We havewhich givesNote that the above formulas may be used to transform asum into a product via the identitiesExample. Express as a product.Answer. We have
  31. 31. Note that we used .Example. Verify the formulaAnswer. We haveandHencewhich clearly impliesExample. Find the real number x such that andAnswer. Many ways may be used to tackle this problem.Let us use the above formulas. We have
  32. 32. HenceSince , the equation gives and the equation gives .Therefore, the solutions to the equationareExample. Verify the identityAnswer. We haveUsing the above formulas we getHence
  33. 33. which implies Since , we get TRIGONOMETRIC EQUATIONS Example : Solve for x in the following equation.There are an infinite number of solutions to this problem. Tosolve for x, you must first isolate the tangent term.
  34. 34. = tan(± ) General solution: X = nπ ± ∀ n Є Z(integers) Example : Solve for x in the following equation. There are an infinite number of solutions to this problem.To solve for x, set the equation equal to zero and factor.
  35. 35. then when when , and when when andThis is impossible because The exact value solutions are and
  36. 36. Example : Solve for x in the following equation. There are an infinite number of solutions to this problem. Isolate the sine term. To do this, rewrite the left side ofthe equation in an equivalent factored form. The product of two factors equals zero if at least one of the factors equals zeros. This means that if or We just transformed a difficult problem into two easier problems. To find the solutions to the original equation, , we find the solutions to the equations OR
  37. 37. Sinx = sin(-π/6) x= nπ + (-1) (π/6) ∀ nЄ Z n OR sinx = sin(π/2) x = nπ + (-1)n (π/2) ∀ nЄ Z.Example : Solve for x in the following equation.(general solution)There are an infinite number of solutions to this problem.
  38. 38. Cos(3x-1) = cos(π/2) 3x-1 = 2nπ ± π/2 3x = 2nπ ± π/2 +1 x = 1/3(2nπ ± π/2 +1) ∀ n Є Z. 1. Solve the trigonometric equation analytically 4 tan x − sec2 x = 0 (for 0 ≤ x < 2π) Answer 4 tan x − sec2 x = 0 In 0 ≤ x < 2π, we need to find values of 2x such that0 ≤ 2x < 4π.
  39. 39. So So or x = 0.2618, 1.309, 3.403, 4.451 2. Solve the trigonometric equation analytically for 0≤ x < 2π: sin 2x cos x − cos 2x sin x = 0 Answer We recognise the left hand side to be in the form: sin(a − b) = sin a cos b − cos a sin b, where a = 2x and b = x. So sin 2x cos x − cos 2x sin x = sin(2x − x) = sin x Now, we know the solutions of sin x = 0 to be:
  40. 40. x = 0, π. 3. Solve the given trigonometric equationanalytically and by graphical method (for 0 ≤ x < 2π): sin 4x − cos 2x = 0 Answer sin 4x − cos 2x = 0 2sin 2x cos 2x − cos 2x = 0 cos2x (2sin 2x - 1) = 0 EITHER cos 2x = 0 OR sin 2x = 1/2
  41. 41. Question Solve the equation tan 2θ − cot 2θ = 0 for0 ≤ θ < 2π. Answer tan2 2θ = 1 tan 2θ = ± 1 Since 0 ≤ θ < 2π , we need to consider values of 2θ such that 0 ≤ 2θ < 4π. Hence, solving the above equation, we have: So
  42. 42. Question Solve the equationfor 0 ≤ θ < 2π. Answer By using the half angle formula for andthen squaring both sides, we get: = 1 + cos x So we have: 2 cos2 x + 3 cos x + 1 = 0 (2 cos x + 1)(cos x + 1) = 0 Solving, we get cos x = − 0.5 or cos x = − 1
  43. 43. Now gives . However, on checking in the original equation, wenote that but So the only solution for this part is Also, cos x = − 1 gives x = π. So the solutions for the equation are GENERAL SOLUTIONS OF TRIGONOMETRICEQUATIONS 1. If sinx = 0 ⇨ x = nπ, ∀ n є Z
  44. 44. 2. If cosx = 0 ⇨ x = (2n+1) π/2, ∀ n Є Z 3. If tanx = 0 ⇨ x = nπ, n Є Z 4. If sinx = six ⇨ x = nπ + (-1)n , ∀ n Є Z 5. If cosx = cos ⇨ x = 2nπ ± , ∀ n Є Z 6. If tanx = tan ⇨ x = nπ+ , ∀ n Є Z. EXAMPLE: Solve the equation: sin3θ + cos2θ = 0 SOLUTION: cos2θ = - sin3θ ⇨ cos2θ = cos (+3θ) ⇨ 2θ = 2nπ± (+3θ) , ∀ nЄZ -θ = nπ+ and5θ = nπ- , ∀ nєz. Question –1 If cos(A+B)=4/5 , sin(A-B)=5/13 andA,B lie between 0 and π/4 , prove that tan 2A = 56/33. Answer: Since A-B Є (-π/4 , π/4) and A+B Є(0,π/2) both are positive sin(A+B)=3/5 , cos(A-B)=12/13
  45. 45. tan(A+B)=3/4 , tan(A-B) = 5/12 thentan2A = tan(A+B+A-B) = =56/33 Question – 2 If cos (A-B)+cos(B-C)+cos(C-A) = -3/2 ,Prove that CosA+cosB+cosC = sinA+sinB+sinC = 0 Answer : From given result we get2cosAcosB+2sinAsinB+2cosBcosC+2sinBsinC+2cosAcosC+2sinAsinC +3 =02cosAcosB+2sinAsinB+2cosBcosC+2sinBsinC+2cosAcosC+2sinAsinC + sin2A+cos2A+sin2B+cos2B +sin2C+cos2C = 0. ( 3=1+1+1 and 1 can bewritten as sin2x+cos2x) (sin2A+sin2B+sin2C+2sinAsinB+2sinBsinC+2sinAsinC) +(cos2A+cos2B+cos2C +2cosAcosB+2cosBcosC+2cosAcosC) = 0 (sinA+sinB+sinC)2+ (cosA+cosB+cosC)2 = 0. Question – 3 Prove that (1+cos ) (1+ cos ) (1+ cos ) (1+ cos )= .
  46. 46. Answer : cos = cos( π - ) = - cos , cos = cos ( π - ) = - cos L.H.S. (1+cos ) (1+ cos ) (1+ cos ) (1+ cos ) = (1 – cos2 ) (1 – cos2 ) = sin2 . sin2 = (2sin2 ). (2sin2 ) = (1 - cos ) ( 1 - cos )= .Question – 4 Prove that cos cos cos cos = .Answer : L.H.S. ⇨ - cos cos cos cos[∵cos = cos(π - )] = -[ ] , where A= [∵ all angels are in G.P. , short-cut Method] = -[ ]= - = = . [ ∵ sin (π + ) lies in 3rd quadrant] OR
  47. 47. L.H.S. ⇨ - (2 sin cos ) cos cos cos = - ( 2sin cos )cos cos =- ( 2sin cos )cos = - (2 sin .cos ) , now you can apply above result. Similarly you can prove cosAcos2Acos4Acos8A =sin16A/16sinA.Question – 5 (i) Prove that sin200 sin400 sin600 sin800 = Answer : L.H.S. sin200 sin400 sin600 sin800 ( ⇨ (2sin200 sin400 sin800) ⇨ [(cos200 – cos600) sin800 ] ⇨ [cos200 . sin800– cos600. sin800 ]
  48. 48. (1/2) ⇨ [( cos200 .sin800 – sin800 ]⇨ [(2 cos200 .sin800 –sin800 ] ⇨ [(sin1000+sin600 – sin800 ] =( ∵ sin1000 lies in 2nd quadrant) [sin(1800 - 800)] = sin800 (ii) Prove that: cos cos cos =- . [Hint: let x = , then (2sinx cosx cos2x cos4x)] Sin2x(iii) Prove that: tan200 tan400 tan800 = tan600.[hint: L.H.S. solve as above method.] Question – 6 Solve : 2sinx + cosx = 1+ sinxAnswer : sinx + cosx = 1 cosx + sinx = [ dividing by ,where a = and b = 1] cos . cosx + .sinx = cos
  49. 49. ⇨ cos(x - ) = cos ⇨ x- = 2nπ ± ∀ n Є Z (integers) ⇨ x = 2nπ ± + = 2nπ + , 2nπ - ∀ n Є Z (integers).Question – 7 Solve: 3cos2x - 2 sinx .cosx – 3sin2x =0. Answer: 3 cos2x - 3 sinx .cosx + sinx .cosx – 3sin2x = 0 ⇨ 3cosx ( cosx - sinx) + sinx (cosx - sinx) = 0 ⇨ (3cosx + sinx) (cosx - sinx) = 0 ⇨ (3cosx + sinx) = 0 or (cosx - sinx) = 0 ⇨ tanx = - = tan(- ) or tanx = =tan( ) ⇨ x = nπ+(- ) ∀ n Є Z (integers) or x = nπ + Question – 8 If , are the acute angles and cos2 = , show that tan = tan .
  50. 50. Answer: According to required result , we have to convert given part into tangent function By using cos2 = ∴ we will get = = = By C & D = ⇨ = 2 ⇨ =2 ⇨tan = tan . Question – 9 Prove that =cot4A.
  51. 51. Answer : = [∵ sinA+sinB = 2sin(A+B)/2. cos(A-B)/2 & cosA-cosB = - 2sin(A+B)/2.sin(A-B)/2 ] = = cot 4A. Question – 10 Prove that = 2COSθ. Answer: = = == 2cosθ. [By using = 2cos2θ] Question – 11 Find the general solution of thefollowing equation:
  52. 52. 4sinxcosx+2sinx+2cosx+1 = 0 Answer: Above equation can be written as(4sinxcosx+2sinx) + (2cosx+1 ) = 0 ⇨ 2sinx (2cosx+1) +(2cosx+1) = 0 ⇨ (2sinx+1) (2cosx+1)=0 ⇨ (2sinx+1) = 0 or2cosx+1 = 0 ⇨ sinx = -1/2 orcosx = -1/2 ⇨ sinx = sin(π + )or cosx = cos (π - ) ⇨ x = nπ +(-1)nor x = 2nπ ± , ∀ n Є Z (Integers). Question – 12 If cosx = - and π <x < , find thevalue of sin3x and cos3x. Answer: Since x lies in 3rd quadrant ∴ sinx isnegative.
  53. 53. Sinx = – = - then sin3x =3sinx – sin3x = - Cos3x = 4cos3x – 3cosx = -[by putting the values of sinx & cosx] Question – 13 (i) If cos(A+B) sin(C-D) = cos(A-B)sin(C+D) , then show that tanA tanB tanC + tanD = 0 (ii) If sinθ = n sin(θ+2 ), prove thattan(θ+ ) = tan . Answer: We can write above given result as = By C & D = = -cotA.cotB = tanC.cotD - . = tanC. ⇨ - tanD = tanA tanB tanC ⇨ tanA tanB tanC + tanD = 0. (ii) = , by C & D =
  54. 54. ⇨ = ⇨ tan(θ+ ) = tan .Question – 14 Prove that (i) cos 520+cos 680+cos 1720 = 0 (ii) sinA .sin(600 – A).sin(600 + A) = sin3A.(it lies in 2nd quad.) Answer: (i) L.H.S. cos 520+(cos 680+cos 1720) =cos520 + 2 cos1200 cos520 =cos520 + 2 ) cos520 [∵ cos1200 = cos(1800-600)] =0 (ii) L.H.S. sinA.[sin(600 – A).sin(600 +A)] = sinA [sin2 600 – sin2 A] =sinA [ - sin2 A] = [3sinA – 4sin3A]= sin3A. [We know that sin2A-sin2B = sin(A+B)sin(A-B)& cos2A – sin2B = cos(A+B)cos(A-B)]
  55. 55. Question – 15 If sinx + siny = a and cosx + cosy = b, findthe values (i) (ii) . Answer: sinx + siny = a ⇨ 2sin( cos )=a..............1 cosx + cosy = b ⇨ 2cos( cos )=b..............2 (i) By squaring & adding above results, we get 4 cos ) [sin²( + cos²( ] = a2+b2 (1) Sec2 = ⇨ tan2 = -1⇨tan = .(ii) dividing 1 by 2, we get tan = .Question – 16 Prove that (i) tan700 = 2 tan500 + tan200 . (ii) Prove that: tan300+ tan150+ tan300. tan150 =1. (iii) cos2 A + cos2 B – 2cosAcosBcos(A+B) = sin2 (A+B). Solution: (i) We have tan700 = tan(500 + 200) =
  56. 56. ⇨ tan700 [ ]= ⇨ tan700 - tan700 = ⇨ tan700 – tan(900-200) = (cot200 = ) ⇨ tan700 – tan500 = ⇨ tan700 = 2 tan500 + tan200. (ii) [hint: take tan450 = tan(300+150) (iii) L.H.S. cos2 A + cos2 B –(2cosAcosB)cos(A+B) = cos2 A + cos2 B – (cos(A+B)+cos(A-B))cos(A+B) cos2 A + cos2 B – [cos2(A+B)+cos(A-B)cos(A+B)] cos2 A + cos2 B – [cos2(A+B)+cos2(A) – sin2(B] cos2 A + cos2 B - cos2(A+B)-cos2(A) + sin2(B] 1 - cos2(A+B) = sin2 (A+B). Question – 17 If x+y = , prove that (i)(1+tanx)(1+tany) = 2 (ii) (cotx – 1)(coty – 1) = 2.Answer: (i) tan(x+y) = tan ⇨ =1 ⇨ tanx+tany+tanx.tany = 1
  57. 57. ⇨ (1+ tanx) + tany(1 + tanx) = 1+1 ⇨ (1+tanx)(1+tany) = 2(ii) Similarly for second part by using cot(x+y) = .Question – 18 Prove that (i) tan1890 =(ii) Find the value of tan220 30’. Answer: (i) we can take both sides to prove aboveresult L.H.S. tan1890 = tan(1800 +90) = tan9° =tan(45°-36°) == = . (ii) Let x= 220 30’ then 2x = 450 We know that tanx = ⇨tanx = = = tan220 30’ = [∵ it lies in 1st quad.] Question: If tanx+tany = a and cotx+coty = b, prove that1/a - 1/b = cot(x+y).
  58. 58. Answer: L.H.S. = - = {after simplification}. Question – 19 Prove that cos2 x + cos2 (x+ ) + cos2 (x- )= . Answer: L.H.S. cos2 x + cos2 (x+ ) +[1 - sin2 (x- ) ] ⇨ 1+ cos2 x + [ cos2 (x+ ) - sin2 (x- ) ] ⇨ 1+ cos2 x + [ cos (x+ ) cos(x+ -x+ ) ] ⇨ 1+ cos2 x + [cos (2x) .cos( )] ( ∵cos(π - it lies in 2nd quad.) ⇨ 1+ cos2 x + [cos (2x) )] ⇨ = = = .Question – 20 Prove that (a) sin3x + sin3( +x) + sin3( +x) = - sin3x. [Hint: Use sin3A = 3sinA – 4sin3A ⇨ 4sin3A = 3sinA -sin3A ⇨ sin3A = ¼[3sinA - sin3A]] (b) If tanx + tan(x+ ) + tan +x) = 3 , thenshow that tan3x = 1. [ use formula of tan(x+y)]
  59. 59. (c) Find in degrees and radians the anglesubtended b/w the hour hand and the minute hand Of a clock at half past three. [answer is 750 ,5 /12] (d) - = cot 2A. [Hintput cot3A = 1/tan3A & cotA = 1/tanA then take l.c.m.and use formula tan(A-B).] (e) If , are the distinct roots of acosθ +bsinθ = c, prove that sin( + ) = . Solution of (e) If , are the distinct roots of acosθ +bsinθ = c, then acos + bsin = c & acos + bsin = cBy subtracting , we get a(cos – cos ) + b(sin – sin ) = 0 ⇨ a(cos – cos ) = b(sin – sin ) ⇨ 2a sin sin = 2b cos sin ⇨ tan = ⇨ sin( + ) = ⇨sin( + ) = .Question: If sin2A = k sin2B, prove that =
  60. 60. [ Hint by c & d = , use formula ofsinx+siny= 2sin( cos )]Question: If cos(x+2A) = n cosx, show that cotA =tan(x+A).Question: Prove that tan( = , if 2tan =3tan .[Hint: L.H.S. = , put the value of tan andsimplify it].Question: If sinx + siny = a and cosx + cosy = b,find the valueof cos(x-y). Answer: squaring and adding above results, we will getcos(x-y) = ½[a2+b2 – 2]Question: Show that - =4[Hint: 2 [ - ]=2[ - ]Question: Prove that : = tan(x/2) [ Hint: use 1 – cosx = 2sin2x/2 , 1+cosx = 2cos2x/2 and sinx= 2sinx/2.cosx/2] ** Question Solve: sec∅ + tan∅ = 1
  61. 61. Solution + sin∅ = cos∅ ⇨ cos∅ - sin∅ =dividing by = ∵ a=1,b=1 (cos∅ - sin∅ )/ = 1 ⇨ cos(п/4) cos∅ -sin(п/4) sin∅ =1 ⇨ cos(∅+п/4) = cos00 ∅+п/4 = 2nп±0, n∈Z ⇨ ∅ = 2nп – п/4. ** Question: If tan2A = 2tan2B + 1, prove that cos2A +sin2B =0. Answer: L.H.S. + sin2B = + sin2B ,by putting above result and simplify it. ** Question: Find the maximum and minimum values ofsinx+cosx. Answer: maximum value of (asinx+bcosx) = = Minimum value of (asinx+bcosx) = - =- Or [ sinx + cosx] = sin(x+п/4), as -1 ≤sin(x+п/4) ≤1 ∀ x **Question: Find the minimum value of 3cosx+4sinx+8. [answer is 3 as above result]
  62. 62. **Question: ∀ x in (0,п/2), show that cos(sinx) >sin(cosx). Answer: п/2> [∵п/2 = 22/7=1.57 and =1.4] We know that п/2 > ≥ sinx+cosx ⇨ п/2 -sinx > cosx ⇨ cos(sinx) > sin(cosx). ** Question: If A = cos2x + sin4x ∀ x, prove that ¾ ≤ A ≤1 Answer: A = cos2x + sin2x. sin2x ≤ cos2x + sin2x ⇨ A ≤ 1 ,A = (1 - sin2x) + sin4x = [sin2x – (½)]2 + (¾) ≥ ¾. ** Question: (i) find the greatest value of sinx.cosx [½.(2sinx.cosx) ≤ ½] (ii) If sinx and cosx are the roots of ax2 – bx+c= 0, then a2 – b2 +2ac = 0. [ Hint: sum and product of roots ⇨ (sinx+cosx)2 =1+2(c/a)] ** Question : If f(x) = cos2x+sec2x , then find which iscorrect f(x)<1, f(x)=1, 2< f(x)<1, f(x)≥2. [A.M. ≥ G.M.⇨ f(x)/2= (cos2x+sec2x)/2 ≥ (cos2x.sec2x) ,ANSWER IS f(x)≥2]. ** Question: solve = 28. [hint: use 2cos2x = 1+ cos2x , 2sin2x = 1 – cos2x and letsin2x+cos2x = y
  63. 63. Above equation becomes = 28. Thenput t = ⇨ 3t2 – 28t +9=0 ⇨ t=1/3 , 9 When y=-1 then sin2x+cos2x = -1 ⇨ cosx = 0 or tanx = -1 When y = 2 then sin2x+cos2x =2 ⇨ sin2x+ cos2x = ⇨ sin(2x+ >1 which is not possible

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