Chapter 3.1 Overview 3.1.1 The word ‘trigonometry’ is derived from the Greek words ‘trigon’ and ‘metron’ which means measuring the sides of a triangle. An angle is the amount of rotation of a revolving line with respect to a fixed line. If the rotation is in clockwise direction the angle is negative and it is positive if the rotation is in the anti-clockwise direction. Usually we follow two types of conventions for measuring angles, i.e., (i) Sexagesimal system (ii) Circular system. In sexagesimal system, the unit of measurement is degree. If the rotation from the 1 initial to terminal side is th of a revolution, the angle is said to have a measure of360 1°. The classifications in this system are as follows: 1° = 60′ 1′ =60″ In circular system of measurement, the unit of measurement is radian. One radian is the angle subtended, at the centre of a circle, by an arc equal in length to the radius of the circle. The length s of an arc PQ of a circle of radius r is given by s = rθ, where θ is the angle subtended by the arc PQ at the centre of the circle measured in terms of radians. 3.1.2 Relation between degree and radian The circumference of a circle always bears a constant ratio to its diameter. This constant 22ratio is a number denoted by π which is taken approximately as for all practical7purpose. The relationship between degree and radian measurements is as follows: 2 right angle = 180° = π radians 180° 1 radian = = 57°16′ (approx)ππ1° = radian = 0.01746 radians (approx)180 3.1.3 Trigonometric functions Trigonometric ratios are defined for acute angles as the ratio of the sides of a right angled triangle. The extension of trigonometric ratios to any angle in terms of radian measure (real numbers) are called trigonometric functions. The signs of trigonometric functions in different quadrants have been given in the following table: I II III IV sin x + + – – cos x + – – + tan x + – + – cosec x + + – – sec x + – – + cot x + – + – 3.1.4 Domain and range of trigonometric functions Functions Domain Range sine R [–1, 1] cosine R [–1, 1] tan R – {(2n + 1) π 2 : n ∈ Z} R cot R – {nπ : n ∈ Z} R sec R – {(2n + 1) π 2 : n ∈ Z} R – (–1, 1) cosec R – {nπ : n ∈ Z} R – (–1, 1) 3.1.5 Sine, cosine and tangent of some angles less than 90° 0° 15° 18° 30° 36° 45° 60° 90° sine 0 6 2 4 − 51 4 − 1 2 10 2 5 4 − 1 2 3 2 1 cosine 1 6 2 4 + 10 2 5 4 + 3 2 51 4 + 1 2 1 2 0 tan 0 2 − 3 25 10 5 5 − 1 3 52 5− 1 3 not defined nπ3.1.6 Allied or related angles The angles ±θ are called allied or related angles2 and θ ± n × 360° are called coterminal angles. For general reduction, we have the following rules. The value of any trigonometric function for (nπ±θ) is numerically2 equal to (a) the value of the same function if n is an even integer with algebaric sign of the function as per the quadrant in which angles lie. (b) corresponding cofunction of θ if n is an odd integer with algebraic sign of the function for the quadrant in which it lies. Here sine and cosine; tan and cot; sec and cosec are cofunctions of each other. 3.1.7 Functions of negative angles Let θ be any angle. Then sin (–θ) = – sin θ, cos (– θ) = cos θ tan (– θ) = – tan θ, cot (–θ) = – cot θ sec (–θ) = sec θ, cosec (– θ) = – cosec θ 3.1.8 Some formulae regarding compound angles An angle made up of the sum or differences of two or more angles is called a compound angle. The basic results in this direction are called trigonometric identies as given below: (i) sin (A + B) = sin A cos B + cos A sin B (ii) sin (A – B) = sin A cos B – cos A sin B (iii) cos (A + B) = cos A cos B – sin A sin B (iv) cos (A – B) = cos A cos B + sin A sin B tan A +tan B (v) tan (A + B) = 1tanAtanB − tan A −tan B (vi) tan (A – B) = 1tanAtanB + (vii) (viii) (ix) (x) (xi) (xii) (xiii) (xiv) (xv) (xvi) (xvii) (xviii) (xix) (xx) (xxi) (xxii) (xxiii) TRIGONOMETRIC FUNCTIONS 37 cotAcotB −1 cot (A + B) = cot A +cot B cotA cotB +1cot (A – B) = cotB −cot A 2tanA sin 2A = 2 sin A cos A = 1tan 2 A+ 21–tan A cos 2A = cos2 A – sin2 A = 1 – 2 sin2A = 2 cos2A – 1 = 21+tan A 2tanA tan 2A = 21–tan A sin 3A = 3sin A – 4sin3 A cos 3A = 4cos3A – 3cos A 33tanA– tan A tan 3A = 21–3tan A A+B A–B cos A + cos B = 2cos cos 22 A+B B–A cos A – cos B = 2sin sin 22 AB+ AB−sin A+sin B = 2sin cos 22 AB A −+ B sin A– sin B = 2cos sin 22 2sin A cos B = sin (A + B) + sin (A– B) 2cos Asin B = sin (A + B) – sin (A–B) 2cos A cos B = cos (A + B) + cos (A – B) 2sin Asin B = cos (A – B) – cos (A + B) ⎡ A ⎢+if lies in quadrants I or II A1 −cosA 2sin =± ⎢22 ⎢ A – if lies in III or IV quadrants ⎢2⎣ ⎡ A ⎢+if lies in I or IV quadrants A1 +cosA 2(xxiv) cos =± ⎢22 ⎢ A – if lies in II or III quadrants ⎢2⎣ ⎡ A ⎢+if lies in I or III quadrants A1−cosA 2(xxv) tan =± ⎢21+cosA ⎢ A – if lies in II or IV quadrants ⎢2⎣ Trigonometric functions of an angle of 18° Let θ = 18°. Then 2θ = 90° – 3θ Therefore, sin 2θ = sin (90° – 3θ) = cos 3θ or sin 2θ = 4cos3 θ – 3cos θ Since, cos θ ≠ 0, we get 2sin θ = 4cos2 θ – 3 = 1 – 4sin2 θ or 4sin2 θ + 2sin θ – 1 = 0. 2−± 4 +16 −±1 5 =Hence, sin θ = 84 51−Since, θ = 18°, sin θ > 0, therefore, sin 18° = 4 2 62− 5 10 +2 5− °= Also, cos18° = 1sin 18 1 −= 16 4 Now, we can easily find cos 36° and sin 36° as follows: 625 225 − + 51+cos 36° = 1 – 2sin2 18° = 1− = = 8 84 51+Hence, cos 36° = 4 5 10 −2 52 62Also, sin 36° = 1 −cos 36°= 1 −+ = 16 4 3.1.9 Trigonometric equations Equations involving trigonometric functions of a variables are called trigonometric equations. Equations are called identities, if they are satisfied by all values of the unknown angles for which the functions are defined. The solutions of a trigonometric equations for which 0 ≤ θ < 2 π are called principal solutions. The expression involving integer n which gives all solutions of a trigonometric equation is called the general solution. General Solution of Trigonometric Equations (i) If sin θ = sin α for some angle α, then θ = nπ + (–1)nα for n ∈ Z, gives general solution of the given equation (ii) If cos θ = cos α for some angle α, then θ = 2 nπ ± α, n ∈ Z, gives general solution of the given equation (iii) If tan θ = tan α or cot θ = cot α, then θ = nπ + α, n ∈ Z, gives general solution for both equations (iv) The general value of θ satisfying any of the equations sin2 θ = sin2 α, cos2θ = cos2 α and tan2 θ = tan2 α is given by θ = nπ ± α (v) The general value of θ satisfying equations sin θ = sin α and cos θ = cos α simultaneously is given by θ = 2nπ + α, n ∈ Z. (vi) To find the solution of an equation of the form a cosθ + b sinθ = c, we put b a = r cosα and b = r sinα, so that r2 = a2 + b2 and tan α = . a Thus we find a cosθ + b sinθ = c changed into the form r (cos θ cos α + sin θ sin α) = c c or r cos (θ – α) = c and hence cos (θ – α) = . This gives the solution of the givenr equation. Maximum and Minimum values of the expression Acosθ + B sinθ are A2 + B2 and – A2 + B2 respectively, where A and B are constants. 3.2 Solved Examples Short Answer Type Example 1 A circular wire of radius 3 cm is cut and bent so as to lie along the circumference of a hoop whose radius is 48 cm. Find the angle in degrees which is subtended at the centre of hoop. Solution Given that circular wire is of radius 3 cm, so when it is cut then its length = 2π × 3 = 6π cm. Again, it is being placed along a circular hoop of radius 48 cm. Here, s = 6π cm is the length of arc and r = 48 cm is the radius of the circle. Therefore, the angle θ, in radian, subtended by the arc at the centre of the circle is given by Arc 6ππθ = = ==22.5 °. Radius 48 8 3Example 2 If A = cos2θ + sin4θ for all values of θ, then prove that ≤ A ≤ 1.4 Solution We have A=cos2 θ + sin4 θ = cos2 θ + sin2 θ sin2 θ ≤ cos2 θ + sin2 θ Therefore, A ≤ 1 Also, A = cos2 θ + sin4 θ = (1 – sin2 θ) + sin4 θ ⎛ 21 ⎞2 ⎛ 1 ⎞⎛ 21 ⎞233 = ⎜sin θ− ⎟+⎜1 −⎟ = ⎜sin θ−⎟+ ≥ ⎝ 2 ⎠⎝ 4 ⎠⎝ 2 ⎠ 44 3 ≤≤Hence, A1.4 Example 3 Find the value of 3 cosec 20° – sec 20° Solution We have 31−3 cosec 20° – sec 20° = sin 20 ° cos 20 ° ⎛ 31 ⎞cos 20 °– sin 20 °⎟⎜3cos 20 °–sin20 ° 22= = 4 ⎜⎟⎜⎟sin 20 °cos 20 °⎝ 2sin 20 °cos 20 °⎠ ⎛sin 60 °cos 20 °– cos 60 °sin 20 °⎞4= ⎜⎟⎝ sin 40 °⎠ (Why?) sin(60 °–20 ) ⎞⎛° =4 ⎜⎟ = 4 (Why?)⎝ sin 40 °⎠Example 4 If θ lies in the second quadrant, then show that 1sin −θ 1sin +θ + =− 2sec θ 1 +sin θ 1−sin θ Solution We have 1sin −θ 1sin θ sin θ sin θ+ 1 − 1 + 2+ = + = −θ2 221sin +θ 1sin 1sin −θ 1sin θ cos −θ 2 = (Since α2 = |α| for every real number α)|cos θ| Given that θ lies in the second quadrant so |cos θ| = – cos θ (since cos θ < 0). 2 Hence, the required value of the expression is = –2 secθ−cos θExample 5 Find the value of tan 9° – tan 27° – tan 63° + tan 81° Solution We have tan 9° – tan 27° – tan 63° + tan 81° = tan 9° + tan 81° – tan 27° – tan 63° = tan 9° + tan (90° – 9°) – tan 27° – tan (90° – 27°) = tan 9° + cot 9° – (tan 27° + cot 27°) (1) 12 =Also tan 9° + cot 9° = (Why?) (2)sin 9 °cos9 ° sin18 ° 1 22Similarly, tan 27° + cot 27° = = = (Why?) (3)sin 27°cos 27°sin 54 ° cos36 ° Using (2) and (3) in (1), we get 22 × 2424 × tan 9° – tan 27° – tan 63° + tan 81° = – = – =4sin18 ° cos 36 ° 51− 5 +1 sec8 θ−1 tan8 θExample 6 Prove that = sec 4 θ−1 tan 2 θ sec8 θ−1 (1 −cos8 θ) cos4 θSolution We have = sec4 θ−1 cos8 θ(1 – cos4 θ) 2sin2 4 θcos 4 θ = (Why?)2cos8 θ2sin 2 θ sin4 θ(2 sin4 θ cos 4 θ) = 22cos8 θ sin 2 θ sin 4 θ sin 8 θ = (Why?)2cos8 θ sin 2 2 θ 2sin2cos2sin8 θθθ = 2cos8 θ sin 2 2 θ tan 8 θ = (Why?)tan 2 θ Example 7 Solve the equation sin θ + sin 3θ + sin 5θ = 0 Solution We have sin θ + sin 3θ + sin 5θ = 0 or (sin θ + sin 5θ) + sin 3θ = 0 or 2 sin 3θ cos 2θ + sin 3θ = 0 (Why?) or sin 3θ (2 cos 2θ + 1) = 0 1 or sin 3θ = 0 or cos 2θ = – 2 n π When sin 3θ = 0, then 3θ = nπ or θ = 3 12π 2ππWhen cos 2θ = – = cos , then 2θ = 2nπ ± or θ = nπ ±23 33 ππ which gives θ = (3n + 1) or θ = (3n – 1)33 n πAll these values of θ are contained inθ = , n ∈ Z. Hence, the required solution set3 n πis given by {θ : θ = , n ∈ Z}3 Example 8 Solve 2 tan2 x + sec2 x = 2 for 0 ≤ x ≤ 2π Solution Here, 2 tan2 x + sec2 x = 2 1 which gives tan x = ± 3 TRIGONOMETRIC FUNCTIONS 43 1 π 7π If we take tan x = , then x = or (Why?)3 66 −15π 11 π,then x = or Again, if we take tan x = (Why?)3 66 Therefore, the possible solutions of above equations are π 5π7π11πx = , , and where 0 ≤ x ≤ 2π666 6 Long Answer Type ⎛ π⎞⎛ 3π⎞⎛ 5π⎞⎛ 7π⎞Example 9 Find the value of ⎜1cos ⎟⎜ + ⎟⎜ 1cos ⎟⎜ + cos ⎟+ 1cos + 1⎝ 8 ⎠⎝ 8 ⎠⎝ 8 ⎠⎝ 8 ⎠ ⎛ π⎞⎛ 3π⎞⎛ 5π⎞⎛ 7π⎞++ cos Solution Write ⎜1cos ⎟⎜ 1cos + ⎟⎜ 1cos + ⎟⎜ 1 ⎟⎝ 8 ⎠⎝ 8 ⎠⎝ 8 ⎠⎝ 8 ⎠ ⎛ π⎞⎛ 3π⎞⎛ ⎛ 3π⎞⎞⎛ ⎛ π⎞⎞ + 1cos + π− = ⎜1cos + ⎟⎜ 1cos ⎟⎜+ ⎜π− ⎟⎟⎜ 1cos ⎜ ⎟⎟⎝ 8 ⎠⎝ 8 ⎠⎝ ⎝ 8 ⎠⎠⎝ ⎝ 8 ⎠⎠ ⎛ 2 π⎞⎛ 23π⎞ = ⎜1 − cos ⎟⎜ 1− cos ⎟ (Why?)⎝ 8 ⎠⎝ 8 ⎠ π 23π2 = sin sin 88 1 ⎛ π⎞⎛ 3π⎞ = ⎜1cos − −⎟ (Why?)⎟⎜ 1cos 4 ⎝ 4 ⎠⎝ 4 ⎠ 1 ⎛ π⎞⎛ π⎞+= ⎜1cos − ⎟⎜ 1cos ⎟ (Why?)4 ⎝ 4 ⎠⎝ 4 ⎠ 12 π⎞ 1 ⎛ 1 ⎞ 1 = ⎜1cos ⎠⎟ = ⎜1 ⎟= ⎛− − 4 ⎝ 44 ⎝ 2 ⎠ 8 2π 4π Example 10 If x cos θ = y cos (θ + ) = z cos ( θ + ), then find the value of33 xy + yz + zx. ⎛111 ⎞Solution Note that xy + yz + zx = xyz ++ .⎜⎟xyz⎝⎠ 2ππIf we put x cos θ = y cos (θ + ) = z cos ⎛⎜θ+4 ⎟⎞ = k (say).3 ⎝ 3 ⎠kk k Then x = θ, y = and z = cos 2π⎞ 4π⎞ cos ⎛⎜θ+ ⎟ cos ⎛⎜θ+ ⎟⎝ 3 ⎠⎝ 3 ⎠ 111 1 ⎡⎛ 2π⎞ ⎛ 4π⎞⎤so that ++ = cos θ+cos ⎜θ+ ⎟+cos ⎜θ+ ⎟⎢ ⎥xyz k ⎣⎝ 3 ⎠⎝ 3 ⎠⎦ 12π 2π[cos θ+cos θcos −sin θsin = k 33 4π 4π + cos θcos −sin θsin ]33 1 −131 3 = [cos θ+cos θ()− sin θ− cos θ+ sin θ] (Why?)k 22 22 1 = 00×= k Hence, xy + yz + zx = 0 Example 11 If α and β are the solutions of the equation a tan θ + b sec θ = c, 2ac then show that tan (α + β) =2 2. a −c Solution Given that atanθ + bsecθ = c or asinθ + b = c cos θ Using the identities, θ 2 θ2 tan 1 −tan 22and cos θ=sin θ = 2 θ 2 θ1tan + 1tan + 2 2 ⎛θ⎞ ⎛ 2 θ⎞a⎜2tan ⎟ c⎜1−tan ⎟⎝ 2 ⎠⎝ 2 ⎠We have, b+=θθ1tan + 2 1tan 2+22 θθor (b+ c) tan 2 2 + 2atan 2 + b– c= 0 θαβAbove equation is quadratic in tan and hence tan and tan are the roots of this 2 22 αβ−2a β−α bc equation (Why?). Therefore, tan + tan = and tan tan = (Why?)bc 2 bc22 +2 +αβtan +tan ⎛α β⎞ 22Using the identity tan ⎜ +⎟ = ⎝22 ⎠αβ1tan tan − 22 −2a ⎛αβ⎞+ −2a −abcWe have, tan ⎜ +⎟= = = ... (1)⎝ bc22 ⎠− 2cc1−bc+Again, using another identity α+β2tan ⎛α+β⎞ tan 2 ⎜ ⎟⎠ = 2,⎝ 22 α+β1tan − 2 ⎛ a⎞2⎜− ⎟⎝ c⎠ 2ac We have tan (α+β )= 2 = 2 − 2 [From (1)]a ac1−2c Alternatively, given that atanθ + bsecθ = c ⇒ (a tanθ – c)2 = b2(1 + tan2θ) ⇒ a2 tan2θ – 2ac tanθ + c2 = b2 + b2 tan2θ ⇒ (a2 – b2) tan2θ – 2ac tanθ + c2 – b2 = 0 ... (1) Since α and β are the roots of the equation (1), so b2 2 2 2ac c tanα + tanβ = and tanα tanβ = 22 2 − b−−baa −tan α+tan 1tan βTherefore, tan (α + β) =αβtan 2ac 2 b22ac− −− a = = 2 b2 a2 −c2 b2 c 2a Example 12 Show that 2 sin2 β + 4 cos (α + β) sin α sin β + cos 2 (α + β) = cos 2α Solution LHS = 2 sin2 β + 4 cos (α + β) sin α sin β + cos 2(α + β) = 2 sin2 β + 4 (cos α cos β – sin α sin β) sin α sin β + (cos 2α cos 2β – sin 2α sin 2β) = 2 sin2 β + 4 sin α cos α sin β cos β – 4 sin2 α sin2 β + cos 2α cos 2β – sin 2α sin 2β = 2 sin2 β + sin 2α sin 2β – 4 sin2 α sin2 β + cos 2α cos 2β – sin 2α sin 2β = (1 – cos 2β) – (2 sin2 α) (2 sin2 β) + cos 2α cos 2β (Why?) = (1 – cos 2β) – (1 – cos 2α) (1 – cos 2β) + cos 2α cos 2β = cos 2α (Why?) Example 13 If angle θ is divided into two parts such that the tangent of one part is k times the tangent of other, and φ is their difference, then show that k+ −1 1 sin φsin θ = k Solution Let θ = α + β. Then tan α = k tan β tan α k or = tan β1 Applying componendo and dividendo, we have tan α+tan β k+1 β = tan α−tan k−1 sin αcos β+cos αsin β k+1 or = (Why?)sin αcos β−cos αsin βk−1 sin ( α+β) k+1i.e., = (Why?)sin ( α−β) k−1 Given that α – β = φ and α + β = θ. Therefore, sin θ k+1 k+1 φ = or sin θ = sin φsin k–1 k−1 Example 14 Solve 3 cos θ + sin θ = 2 Solution Divide the given equation by 2 to get 31 1 ππ π cos θ+ sin θ= or cos cos θ+sin sin θ=cos 22266 4 ⎛π⎞π ⎛π⎞ π cos = or cos ⎝6 ⎠ 4 ⎝ 6 ⎠ 4or ⎜ −θ⎟ cos ⎜θ− ⎟=cos (Why?) ππ Thus, the solution are given by, i.e., θ = 2mπ ± + 46 Hence, the solution are ππ ππ5ππ θ = 2mπ + + and 2mπ – + , i.e., θ = 2mπ + and θ = 2mπ – 4646 12 12 Objective Type Questions Choose the correct answer from the given four options against each of the Examples 15 to 19 −4 Example 15 If tan θ = , then sinθ is3 −4 4 −44(A) but not (B) or55 55 44(C) but not − (D) None of these55 4Solution Correct choice is B. Since tan θ = − is negative, θ lies either in second3 4 quadrant or in fourth quadrant. Thus sin θ = if θ lies in the second quadrant or54sin θ = − , if θ lies in the fourth quadrant.5 Example 16 If sin θ and cos θ are the roots of the equation ax2 – bx + c = 0, then a, b and c satisfy the relation. (A) a2 + b2 + 2ac = 0 (B) a2 – b2 + 2ac = 0 (C) a2 + c2 + 2ab = 0 (D) a2 – b2 – 2ac = 0 Solution The correct choice is (B). Given that sin θ and cos θ are the roots of the bc equation ax2 – bx + c = 0, so sin θ + cos θ = and sin θ cos θ = (Why?)aaUsing the identity (sinθ + cos θ)2 = sin2θ + cos2θ + 2 sin θ cos θ, we have 2b 2c =+1 or a2 – b2 + 2ac = 02 aa Example 17 The greatest value of sin x cos x is 1 (A) 1 (B)2 (C) (D) 2 Solution (D) is the correct choice, since 11 sinx cosx= sin 2x ≤ , since |sin 2x | ≤ 1.22 Eaxmple 18 The value of sin 20° sin 40° sin 60° sin 80° is −35 3 1 (A) (B) (C) (D)1616 16 16 Solution Correct choice is (C). Indeed sin 20° sin 40° sin 60° sin 80°. 33 = sin 20° sin (60° – 20°) sin (60° + 20°) (since sin 60° = )22 3 = sin 20° [sin2 60° – sin2 20°] (Why?)2 33 = sin 20° [ – sin2 20°]2431 =× [3sin 20° – 4sin3 20°]24 31 =× (sin 60°) (Why?)24 31 33 = ×× = 242 16 π2π4π8πExample 19 The value of cos cos cos cos is55 55 1 −1 −1 (A) (B) 0 (C) (D)16 816 Solution (D) is the correct answer. We have π2π4π8π cos cos cos cos55 55 1 ππ 2π 4π 8π = 2sin cos cos cos cos π 555 552sin 5 12π 2π 4π 8π = sin cos cos cos π 5 555 (Why?)2sin 5 14π 4π 8π = sin cos cos π 555 (Why?)4sin 5 18π 8π = sin cos 8sin π 5 5 (Why?) 5 16π ⎛π⎞sin sin 3π+⎜⎟5 ⎝ 5 ⎠= = ππ16sin 16sin 55 π−sin 5= π (Why?)16 sin 5 1 = − 16 Fill in the blank : Example 20 If 3 tan (θ – 15°) = tan (θ + 15°), 0° < θ < 90°, then θ = _________ Solution Given that 3 tan (θ – 15°) = tan (θ + 15°) which can be rewritten as tan( θ+15°) 3 tan( θ−15°) = 1. tan ( θ+15°)+tan ( θ – 15°) = tan ( θ+15°) −tan ( θ –15°) Applying componendo and Dividendo; we get 2 sin ( θ+ 15°) cos ( θ−15°)+sin ( θ−15°) cos ( θ+ 15°) ⇒ = 2 sin ( θ+15°) cos ( θ−15°) −sin ( θ− 15°) cos( θ+15°) sin 2 θ⇒ = 2 i.e., sin 2θ = 1 (Why?)sin 30 °πgiving θ= 4 State whether the following statement is True or False. Justify your answer 1 −Example 21 “The inequality 2sinθ + 2cosθ≥ 1 holds for all real values of θ” 2 Solution True. Since 2sinθ and 2cosθ are positive real numbers, so A.M. (Arithmetic Mean) of these two numbers is greater or equal to their G.M. (Geometric Mean) and hence sinθ cos θ2 +2 ≥ 2 sin θ+cos θ 1 ⎛ 11 ⎞sin θ+ cos θ⎜ ⎟2 ⎠22 ⎝ 2≥ 2 = 2 1 sin ⎛π+θ⎞⎟≥ 22 ⎝⎜ 4 ⎠ ⎛π ⎞ Since, –1 ≤ sin ⎜ +θ⎟≤ 1, we have⎝4 ⎠sin θ cos θ−1 − 112 +2 2 sin θ cos θ≥ 2 ⇒ 2 + 2 ≥ 22 2 Match each item given under the column C to its correct answer given under column C12 Example 22 C1 C2 1cos − xx (a) (i) cot2 sin x 2 1cos xx+ (b) (ii) cot 1cos x 2− 1cos x+ (c) (iii) cos x +sin xsin x x (d) 1sin2x (iv) tan + 2 Solution 2sin2 x 1cos −xx2(a) = = tan .sin x xx 22sin cos 22 Hence (a) matches with (iv) denoted by (a) ↔ (iv) 21cos x 2sin x+ 22 x(b) = = cot . Hence (b) matches with (i) i.e., (b) ↔ (i)1cos x 2sin 2 x 2−2 x +2cos2 1cos xx2(c) = = cot .sin x xx 22sin cos 22 Hence (c) matches with (ii) i.e., (c) ↔ (ii) 22(d) 1sin 2x= sin x+cos x+2 sin +xcos x = (sin x+cos x)2 = (sin x+cos x) . Hence (d) matches with (iii), i.e., (d) ↔ (iii) Short Answer Type tanA +secA–1 1 +sinA 1. Prove that = tan A −secA +1 cos A 2sin α −α+ sin 1cos α2. If = y, then prove that is also equal to y. 1cos + α+ sin α+1sin α1cos α+ sin α − α+ sin α+ α+ sin ⎡− 1cos 1cos α⎤Hint :Express = .⎢⎥1sin α 1+sin α 1 +cos α+ sin +α⎣⎦ mn+ 3. If msin θ = nsin (θ + 2α), then prove that tan (θ + α) cot α = mn− sin( 2 mθ+ α )[Hint: Express = and apply componendo and dividendo]sin θn 45 π 4. If cos (α + β) = and sin (α – β) = , where α lie between 0 and , find the513 4 value of tan2α [Hint: Express tan 2 α as tan (α + β + α – β] ab+ ab−b +5. If tan x= , then find the value of a ab ab−+ θ 9θ6. Prove that cosθ cos – cos3θ cos = sin 7θ sin 8θ.2 2 1 θ 9θ[Hint: Express L.H.S. = [2cosθ cos – 2 cos3θ cos ]22 2 7. If acos θ + bsin θ = mand asin θ – bcos θ = n, then show that a2 + b2 = m2 + n2 8. Find the value of tan 22°30 ′ . θ θθsin 2sin cos θ sin θ2 22[Hint: Let θ = 45°, use tan == = ]2 θ 2 θ 1 +cos θcos 2 cos 22 9. Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A. 10. If tanθ + sinθ = mand tanθ – sinθ = n, then prove that m2 – n2 = 4sinθ tanθ [Hint: m+ n= 2tanθ, m– n= 2 sinθ, then use m2 – n2 = (m+ n) (m– n)] p+q11. If tan (A + B) = p, tan (A – B) = q, then show that tan 2 A = 1−pq [Hint: Use 2A = (A + B) + (A – B)] 12. If cosα + cosβ = 0 = sinα + sinβ, then prove that cos 2α + cos 2β = – 2cos (α + β). [Hint: (cosα + cosβ)2 – (sinα + sinβ)2 = 0] sin ( x+y) ab + tan xa 13. If = , then show that = [Hint: Use Componendo andsin ( x−y) ab − tan yb Dividendo]. sin α−cos α 14. If tanθ = , then show that sinα + cosα = 2 cosθ.sin α+cos αππ [Hint: Express tanθ = tan (α – ) ⇒ θ = α – ]44 15. If sinθ + cosθ = 1, then find the general value of θ. 16. Find the most general value of θ satisfying the equation tanθ = –1 and 1 cosθ = 17. If cotθ + tanθ = 2 cosecθ, then find the general value of θ. 18. If 2sin2θ = 3cosθ, where 0 ≤ θ ≤ 2π, then find the value of θ. π 19. If secx cos5x + 1 = 0, where 0 < x ≤ , then find the value of x.2 Long Answer Type 20. If sin (θ + α) = a and sin (θ + β) = b, then prove that cos 2(α – β) – 4ab cos (α – β) = 1 – 2a2 – 2b2 [Hint: Express cos (α – β) = cos ((θ + α) – (θ + β))] 1−m21. If cos (θ + φ) = m cos (θ – φ), then prove that tan θ= cot φ. 1+m cos ( θ+φ ) m[Hint: Express = and apply Componendo and Dividendo]cos ( θ−φ)1 22. Find the value of the expression 3ππ 3 [sin4 ( 2 −α ) + sin4 (3π + α)] – 2 {sin6 (2 + α) + sin6 (5π – α)] 23. If a cos 2θ + b sin 2θ = c has α and β as its roots, then prove that 2b tanα + tan β = . ac+ 1tan 2 θ θ− 2tan [Hint: Use the identities cos 2θ = 2 and sin 2θ = 2].+ +1tan θ1tan θ 24. If x = sec φ – tan φ and y = cosec φ + cot φ then show that xy + x – y + 1 = 0 [Hint: Find xy + 1 and then show that x – y = – (xy + 1)] 8 25. If θ lies in the first quadrant and cosθ = , then find the value of17 cos (30° + θ) + cos (45° – θ) + cos (120° – θ). 4 π 43π 45π 47π26. Find the value of the expression cos +cos +cos +cos 88 8 8 4 π 43π cos [Hint: Simplify the expression to 2 ( cos + )88 ⎡ 2 ⎤⎛ 2 π 23π⎞ 2 π 23π =2cos +cos −2 cos cos ⎢⎜⎟ ⎥ ⎢⎝ 88 ⎠ 88 ⎥⎣ ⎦ 27. Find the general solution of the equation 5cos2θ + 7sin2θ – 6 = 0 28. Find the general solution of the equation sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x 29. Find the general solution of the equation ( 3 – 1) cosθ + ( 3+ 1) sinθ = 2 ππ[Hint: Put 3 – 1= r sinα, 3 + 1 = r cosα which gives tanα = tan ( – )46 π ⇒ α = ]12 Objective Type Questions Choose the correct answer from the given four options in the Exercises 30 to 59 (M.C.Q.). 30. If sin θ + cosec θ = 2, then sin2 θ + cosec2 θ is equal to (A) 1 (B) 4 (C) 2 (D) None of these 31. If f(x) = cos2 x + sec2 x, then (A) f (x) < 1 (B) f (x) = 1 (C) 2 < f (x) < 1 (D) f(x) ≥ 2 [Hint: A.M ≥ G.M.] 11 32. If tan θ = and tan φ = , then the value of θ + φ is23 π π (A) (B) π (C) 0 (D) 6 4 33. Which of the following is not correct? 1 (A) sin θ = – (B) cos θ = 15 1 (C) sec θ = (D) tan θ = 202 34. The value of tan 1° tan 2° tan 3° ... tan 89° is (A) 0 (B) 1 1 (C) (D) Not defined2 21tan 15 °− 35. The value of is21tan 15 °+ 3 (A)1 (B) 3 (C) (D)22 36. The value of cos 1° cos 2° cos 3° ... cos 179° is 1 (A) (B)0 (C)1 (D) –1 2 37. If tan θ = 3 and θ lies in third quadrant, then the value of sin θ is (A) 1 10 (B) − 1 10 (C) 3 10 − (D) 3 10 38. The value of tan 75° – cot 75° is equal to (A) 23 (B) 2 + 3 (C) 2 − 3 (D) 1 39. Which of the following is correct? (A) sin1° > sin 1 (B) sin 1° < sin 1 (C) sin 1° = sin 1 (D) sin 1° = 18 π ° sin 1 [Hint: 1 radian = 180 ° π = 57 30 ° ′ approx] 40. If tan α = 1 m m + , tan β = 1 2 1m + , then α + β is equal to (A) 2 π (B) 3 π (C) 6 π (D) 4 π 41. The minimum value of 3 cosx + 4 sinx + 8 is (A) 5 (B) 9 (C) 7 (D) 3 42. The value of tan 3A – tan 2A – tan A is equal to (A) tan 3A tan 2A tan A (B) – tan 3A tan 2A tan A (C) tan A tan 2A – tan 2A tan 3A – tan 3A tan A (D) None of these π 51. If α + β = , then the value of (1 + tan α) (1 + tan β) is4 (A)1 (B) 2 (C) – 2 (D) Not defined −4 θ52. If sin θ = and θ lies in third quadrant then the value of cos is5 2 1 111 – –(A) (B) (C) (D)5 10 510 53. Number of solutions of the equation tan x + sec x = 2 cosx lying in the interval [0, 2π] is (A) 0 (B)1 (C)2 (D) 3 ππ 2π 5π54. The value of sin + sin + sin +sin is given by18 9 9 18 7π 4π (A) sin + sin (B) 118 9 π 3π ππ (C) cos + cos (D) cos + sin 67 99 55. If A lies in the second quadrant and 3 tan A + 4 = 0, then the value of 2 cotA – 5 cos A + sin Ais equal to −5323 37 7 (A) (B) (C) (D)1010 10 10 56. The value of cos2 48° – sin2 12° is 51 51+ −(A) (B)8 8 +51 51+(C) (D)5 22 [Hint: Use cos2A – sin2 B = cos (A+ B) cos (A – B)] TRIGONOMETRIC FUNCTIONS 59 11 57. If tan α = , tan β = , then cos 2α is equal to73 (A) sin 2β (B) sin 4β (C) sin 3β (D) cos 2β a 58. If tan θ = b , then b cos 2θ + a sin 2θ is equal to a (A) a (B) b (C) (D) Noneb 1 59. If for real values of x, cos θ = x + , then x (A) θ is an acute angle (B) θ is right angle (C) θ is an obtuse angle (D) No value of θ is possible Fill in the blanks in Exercises 60 to 67 : sin 50 ° 60. The value of is _______ .sin 130 °π 5π 7π⎛⎞ ⎛⎞ ⎛⎞ 61. If k = sin ⎜⎟ sin sin ⎜⎟ , then the numerical value of k is _______.⎜⎟ ⎝⎠ 18 ⎝⎠ 18 ⎝⎠18 1cos B − 62. If tan A = , then tan 2A = _______. sin B 63. If sin x + cos x = a, then (i) sin6 x + cos6 x = _______ (ii) | sin x – cos x | = _______. 64. In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is _______. 2 [Hint: A + B = 90° ⇒ tan Atan B = 1 and tan A + tan B = ]sin 2A 65. 3 (sin x – cos x)4 + 6 (sin x + cos x)2 + 4 (sin6 x + cos6 x) = _______. 66. Given x > 0, the values of f(x) = – 3 cos ++23 xx lie in the interval _______. 67. The maximum distance of a point on the graph of the function y = 3 sin x + cos x from x-axis is _______. In each of the Exercises 68 to 75, state whether the statements is True or False? Also give justification. 1– cosB 68. If tan A= , then tan 2A = tan B sin B 69. The equality sin A + sin 2A + sin 3A = 3 holds for some real value of A. 70. sin 10° is greater than cos 10°. 2π 4π 8π 16 π 1 71. cos cos cos cos = 15 15 15 15 16 72. One value of θ which satisfies the equation sin4 θ – 2sin2 θ – 1 lies between 0 and 2π. π 73. If cosec x = 1 + cot x then x = 2nπ, 2nπ + 2 nπ π74. If tan θ + tan 2θ + 3 tan θ tan 2θ = 3 , then θ= + 39 1⎛ π⎞75. If tan (π cosθ) = cot (π sinθ), then cos ⎜θ– ⎟ = ± ⎝ 4 ⎠22 76. In the following match each item given under the column C1 to its correct answer given under the column C2 : (a) sin (x + y) sin (x – y) (i) cos2 x – sin2 y 1tan θ− (b) cos (x + y) cos (x – y) (ii) 1tan θ+ ⎛π 1tan θ+ (c) cot ⎜ +θ⎞⎟ (iii)⎝4 ⎠ 1tan θ− (d) tan ⎜⎛π+θ⎟⎞ (iv) sin2 x – sin2 y⎝4 ⎠

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