Vedanta Optional Mathematics Teacher's Guide ~ 10 251 6. Prove that: a. cos4 θ + sin4 θ = 1 4 (3 + cos4θ) Solution LHS = cos4 θ + sin4 θ = (cos2 θ – sin2 θ) 2 + 2 sin2 θ . cos2 θ = (cos2θ) 2 + 1 2 (2 sinθ . cosθ) 2 = cos2 2θ + 1 2 sin2 2θ = 1 2 (2 cos2 2θ + sin2 2θ) = 1 2 (cos2 2θ + sin2 2θ + cos2 2θ) = 1 2 (1 + cos2 2θ) = 1 4 (2 + 2 cos2 2θ) = 1 4 (2 + 1 + cos4θ) = 1 4 (3 + cos4θ) ( 2 cos2 x = 1 + cos2x) = RHS proved b. cos6 θ – sin6 θ = cos2θ 1 – 1 4 sin2 2θ Solution LHS = cos6 θ – sin6 θ = (cos2 θ – sin2 θ) (cos4 θ + cos2 θ . sin2 θ + sin4 θ) = cos2θ {(cos2 θ – sin2 θ) 2 + 2 sin2 θ . cos2 θ + sin2 θ . cos2 θ} = cos2θ (cos2 2θ + 3 sin2 θ . cos2 θ) = cos2θ cos2 2θ + 3 4 (2 sinθ . cosθ) 2 = cos2θ 4 sins2 2θ + 3 sin2 2θ 4 = cos2θ 4 – 4 cos2 2θ + 3 sin2 2θ 4 = cos2θ 1 – 1 4 sin2 2θ = RHS proved
252 Vedanta Optional Mathematics Teacher's Guide ~ 10 c. sin4 θ = 1 8 (3 – 4 cos2θ + cos4θ) Solution RHS = 1 8 (3 – 4 cos2θ + cos4θ) = 1 8 {3 – 4(1 – 2 sin2 θ) + (1 – 2sin2 2θ)} = 1 8 {3 – 4 + 8 sin2 θ + 1 – 2 sin2 2θ} = 1 8 {8 sin2 θ – 2(2 sinθ . cosθ) 2 } = 1 8 {8 sin2 θ – 8 sin2 θ . cos2 θ} = 1 8 . 8 sin2 θ(1 – cos2 θ) = sin2 θ . sin2 θ = sin4 θ = LHS proved d. cos8 θ + sin8 θ = 1 – sin2 2θ + 1 8 cos4 2θ Solution LHS = cos8 θ + sin8 θ = (cos4 θ + sin4 θ) 2 – 2 sin4 θ . cos4 θ = {(cos2 θ – sin2 θ) 2 + 2 sin2 θ . cos2 θ}2 – 2 sin4 θ . cos4 θ = cos2 2θ + 1 2 sin2 2θ 2 – 1 8 sin4 2θ = cos4 2θ + cos2 2θ . sin2 2θ + 1 4 sin4 2θ – 1 8 sin4 2θ = (1 – sin2 2θ) 2 + (1 – sin2 2θ) . sin2 2θ + 1 8 sin4 2θ = 1 – 2 sin2 2θ + sin4 2θ + sin2 2θ – sin4 2θ + 1 8 sin4 2θ = 1 – sin2 2θ + 1 8 sin4 2θ = RHS proved 6. Prove that: a. 3 sin40° + 1 cos40° = 4 Solution LHS = 3 sin40° + 1 cos40°
Vedanta Optional Mathematics Teacher's Guide ~ 10 253 = 3 cos40° – sin40° sin40° . cos40° = 2 32 cos40° + 12 sin40° sin40° . cos40° = 2(cos30° . cos40° + sin30° . sin40°) sin40° . cos40° = 4 cos(40° – 30°) 2 sin40° . cos40° = 4 cos10° sin80° = 4 cos10° sin(90° – 10°) = 4 cos10° cos10° = 4 = RHS proved b. cosec10° – 3 sec10° = 4 Solution LHS = cosec10° – 3 sec10° = 1 sin10° – 3 cos10° = cos10° – 3 sin10° sin10° . cos10° = 4 12 cos10° – 32 sin10° 2 sin10° . cos10° = 4 sin30° . cos10° – cos30° . sin10° sin20° = 4 sin(30° – 10°) sin20° = 4 sin20° sin20° = 4 = RHS proved 7. Prove that: a. (2 cos θ + 1) (2 cos θ – 1) = 2 cos2 θ + 1
254 Vedanta Optional Mathematics Teacher's Guide ~ 10 Solution LHS = (2 cosθ + 1) (2 cosθ – 1) = 4 cos2 θ – 1 = 2(2 cos2 θ – 1) + 1 = 2 cos2θ + 1 RHS proved b. sec8θ – 1 sec4θ – 1 = tan8θ tan2θ Solution LHS = sec8θ – 1 sec4θ – 1 = 1 cos8θ – 1 1 cos4θ – 1 = 1 – cos8θ cos8θ × cos4θ 1 – cos4θ = 2 sin2 4θ . cos4θ cos8θ . 2 sin2 2θ = (2 sin4θ . cos4θ) sin4θ cos8θ . 2 sin2 2θ = sin8θ cos8θ . 2 sin2θ . cos2θ 2 sin2 2θ = tan8θ sin2θ cos2θ = tan8θ tan2θ = RHS proved c. tanθ + 2 tanθ + 4 tan4θ + 8 cot8θ = cotθ Solution Example (14) text book d. sin2 α – cos2 α . cos2β = sin2 β – cos2 β . cos2α Solution LHS = sin2 α – cos2 α . cos2β = sin2 α – cos2 α(2 cos2 β – 1) = sin2 α – 2 cos2 α . cos2 β + cos2 α = sin2 α + cos2 α – 2 cos2 β . cos2 α
Vedanta Optional Mathematics Teacher's Guide ~ 10 255 = 1 – 2 cos2 α . cos2 β = sin2 β + cos2 β – 2 cos2 α . cos2 β = sin2 β – cos2 β(2 cos2 α – 1) = sin2 β – cos2 β . cos2α = RHS proved e. 2 + 2 + 2 + 2 cos8θ= 2 cosθ Solution LHS = 2 + 2 + 2 + 2 cos8θ = 2 + 2 + 2(1 + cos8θ) = 2 + 2 + 2 . 2 cos2 4θ = 2 + 2 + 2 cos4θ = 2 + 2(1 + cos4θ) = 2 + 4 cos2 2θ = 2 + 2 cos2θ = 2(1 + cos2θ) = 2 . 2 cos2 θ = 2 cosθ = RHS proved f. sin2 α – sin2 β sinα . cosα – sinβ . cosβ = tan(α + β) Solution RHS = tan(α + β) = tanα – tanβ 1 – tanα . tanβ = tanα – tanβ 1 – tanα . tanβ = sinα cosα + sinβ cosβ 1 – sinα cosα . sinβ cosβ = sinα . cosβ + sinβ . cosα cosα . cosβ – sinα . sinβ × sinα . cosβ – sinβ . cosα sinα . cosβ – sinβ . cosa
256 Vedanta Optional Mathematics Teacher's Guide ~ 10 = sin2 α . cos2 β – sin2 β . cos2 α sinα . cosa . cos2 β – cos2 α . sinb . cosβ – sin2 α . sinb . cosβ + sin2 β . sina . cosa = sin2 α(1 – sin2 β) – sin2 β(1 – sin2 α) sinα . cosa(cos2 β + sin2 b) – sinb . cosβ(cos2 α + sin2 a) = sin2 α – sin2 β sinα . cosα – sinβ . cosβ = LHS proved 8. a) 4(cos3 10° + sin2 20°) = 3(cos10° + sin20°) Solution LHS = 4(cos3 10° + sin2 20°) = 4 cos3 10° + 4 sin2 20° = 3 cos10° + cos(3.10°) + 3 sin20° – sin(3.20°) = 3(cos10° + sin20°) + 3(cos30° – sin60°) = 3(cos10° + sin20°) + 3 3 2 – 3 2 = 3(cos10° + sin20°) + 3.0 = 3(cos10° + sin20°) = RHS proved b) sin3 10° + cos3 20° = 3 4 (cos20° + sin10°) Solution LHS = sin3 10° + cos3 20° By using formula, cos3 θ = 1 4 (3 cosθ + cos3θ) sin3 θ = 1 4 (3 sinθ – sin3θ) LHS = 1 4 [3 sin10° – sin(3.10°) + 3 cos20° + cos(3.20°)] = 1 4 [3 sin10° – sin30° + 3 cos20° + cos60°] = 3 4 (sin10° + cos20°) + 1 4 – 1 2 + 1 2 = 3 4 (sin10° + cos20°) = RHS proved 9. Prove that: a) cot(A + 45°) – tan(A – 45°) = 2 cos2A 1 + sin2A
Vedanta Optional Mathematics Teacher's Guide ~ 10 257 Solution LHS = cot(A + 45°) – tan(A – 45°) = 1 tan (A + 45°) – tan(A – 45°) = 1 tanA + tan45° 1 – tanA . tan45° – tanA – tan45° 1 + tanA . tan45° = 1 – tanA 1 + tan A – tanA – 1 1 + tan A = 1 – tanA – tanA + 1 1 + tan A = 2(1 – tanA) 1 + tan A = 2 1 – sinA cosA 1 + sinA cosA = 2 cosA – sinA cosA + sinA × cosA + sinA cosA + sinA = 2 cos 2 A – sin 2 A cos 2 A + sin 2 A + 2 sinA . cosA = 2 cos2A 1 + sin2A = RHS proved b) tan(A + 45° ) + tan(A – 45°) = 2 tan2A Solution LHS = tan(A + 45°) + tan(A – 45°) = tanA + tan45° 1 – tanA . tan45° + tanA – tan45° 1 + tanA . tan45° = tanA + 1 1 – tan A + tanA – 1 1 + tan A = (1 + tanA) 2 + (tanA – 1) (1 – tnaA) 1 – tan 2 A = 1 + 2 tanA + tan 2 A + tanA – tan 2 A – 1 + tanA 1 – tan 2 A = 4 tanA 1 – sin 2 A cos 2 A = 4 sinA cos A × cos 2 A cos 2 A – sin 2 A
258 Vedanta Optional Mathematics Teacher's Guide ~ 10 = 2 2 sinA cosA cos2A = 2 sin2A cos2A = 2 tan2A = RHS proved c) tan(A + 45°) – tan(A – 45°) = 2 sec2A Solution LHS = tan(A + 45°) – tan(A – 45°) = tanA + tan45° 1 – tanA . tan45° – tanA – tan45° 1 + tanA . tan45° = tanA + 1 1 – tanA – tanA – 1 1 + tanA = tanA + 1 1 – tanA + 1 – tanA 1 + tanA = (1 + tanA)2 + (1 – tanA)2 1 – tan2 A = 1 + 2 tanA + tan2 A + 1 – 2 tanA – tan2 A 1 – tan2 A = 2 (1 + tan2 A) 1 – tan2 A = 2 1 1 – tan2 A 1 + tan2 A = 2 cos2A = 2 sec2A = RHS proved d) tanA + tan π 3 + A – tan π 3 – A = 3 tan3A Solution LHS = tanA + tan π 3 + A – tan π 3 – A = tanA + tan2 π 3 + tanA 1 – tan π 3 . tanA – tan π 3 – tanA 1 + tan π 3 . tanA = tanA + 3 + tanA 1 – 3 tanA – 3 – tanA 1 + 3 tanA = tanA + ( 3 + tanA) (1 + 3 tanA) – ( 3 – 3tanA) (1 – 3 tanA) 1 – 3 tan2 A
Vedanta Optional Mathematics Teacher's Guide ~ 10 259 = tanA + 3 + 3 tanA + tanA + 3 tan2 A – 3 + tanA + tanA – 3 tan2 A 1 – 3 tan2 A = tanA + 8 tanA 1 – 3tan2 A = tanA – 3 tan3 A + 8 tanA 1 – 3 tan2 A = 9 tanA – 3 tan3 A 1 – 3 tan2 A = 3(3 tanA – tan3 A) 1 – 3 tan2 A = 3 tan3A = RHS proved 10.a) If 2 tanα = 3 tanβ, prove that : tan(α – β) = sin2β 5 – cos2β Solution Given, 2 tanα = 3 tanβ or, tanα = 3 2 tanβ LHS = tan(α – β) = tanα – tanβ 1 + tanα . tanβ = 3 2 tanβ – tanβ 1 + 3 2 tanβ . tanβ = 3 tanb – 2 tanβ 2 + 3 tan2 b = sinβ cosβ 2 + 3 sin2 β cos2 β = sinβ cosb × cos2 β 2 cos2 b + 3 sin2 b = 2 sinβ . cosβ 4 cos2 b + 6 sin2 b = sin2β 4 – 4sin2 b + 6 sin2 b = sin2β 4 + 2 sin2 b = sin2β 5 – 1 + 2 sin2 b
260 Vedanta Optional Mathematics Teacher's Guide ~ 10 = sin2β 5 – (1 – 2 sin2 b) = sin2β 5 – cos2β = RHS proved b) tanθ = 1 7 and tanb = 1 3, prove that : cos2θ = sin4b Solution Here, tanθ = 1 7 , tanb = 1 3 LHS = cos2θ = 1 – tan2 q 1 + tan2 q = 1 – 1 7 2 1 + 1 7 2 = 1 – 1 49 1 + 1 49 = 48 49 × 49 50 = 24 25 RHS = sin4b = sin2(2b) = 2 sin2b . cos2b = 2 2 tanb 1 + tan2 b × 1 – tan2 b 1 + tan2 b = 4 × 1 3 1 + 1 9 × 1 – 1 9 1 + 1 9 = 4 3 × 9 10 × 8 9 × 9 10 = 24 25 LHS = RHS proved 11. Prove that : a) cosA – 1 + sin2A sinA – 1 + sin2A = tanA
Vedanta Optional Mathematics Teacher's Guide ~ 10 261 Solution LHS = cosA – 1 + sin2A sinA – 1 + sin2A = cosA – sin 2 A + cos 2 A + 2 sinA cosA sinA – sin 2 A + cos 2 A + 2 sinA cosA = cosA – (sinA + cosA) 2 sinA – (sinA + cosA) 2 = cosA – sinA – cosA sinA – sinA – cosA = –sinA –cosA = tanA = RHS proved b. 1 tan3 θ + tan θ – 1 cot3 θ + cot θ= cot4 θ Solution LHS = 1 tan3 θ + tan θ – 1 cot3 θ + cot θ = 1 tan3 θ + tan θ – 1 1 tan3 θ + 1 tan θ = 1 tan3 θ + tan θ – tan θ . tan3θ tan θ + tan3 θ = 1 – tan θ . tan3 θ tan3 θ + tan θ = 1 tan3 θ+ tan θ 1 – tan θ . tan3 θ = 1 tan(3 θ + θ ) = cot4 θ = RHS proved c. cotA cotA – cot3A – tanA tan3A – tanA = 1 Solution LHS = cotA cotA – cot3A – tanA tan3A – tanA = 1 tanA 1 tanA – 1 tan3A – tanA tan3A – tanA
262 Vedanta Optional Mathematics Teacher's Guide ~ 10 = 1 tanA tanA . tan3A tan3A – tanA – tanA tan3A – tanA = tanA tan3A – tanA – tanA tan3A – tanA = tan3A – tanA tan3A – tanA = 1 = RHS proved 12. Prove that: a) 8 1 + sin π 8 1 + sin 3π 8 1 – sin 5π 8 1 – sin 7π 8 = 1 Solution LHS = 8 1 + sin π 8 1 + sin 3π 8 1 – sin 5π 8 1 – sin 7π 8 = 8 1 + sin π 8 1 + sin 3π 8 1 – sin π – 3π 8 1 – sin π – π 8 = 8 1 + sin π 8 1 + sin 3π 8 1 – sin 3π 8 . 1 – sin π 8 = 8 1 – sin2 π 8 . 1 – sin2 3π 8 = 8 cos2 π 8 . cos2 3π 8 ∴cos2 3π 8 = cos2 π 2 – π 8 = sin2 π 8 = 8 cos2 π 8 . sin2 π 8 = 2 4 sin2 π 8 . cos2 π 8 = 2 2 sin π 8 . cos π 8 2 = 2 sin2 π 4 = 2 1 2 2 = 1 = RHS proved b) sin4 π 8 + sin4 3π 8 + sin4 5π 8 + sin2 7π 8 = 3 2 Solution LHS = sin4 π 8 + sin4 3π 8 + sin4 5π 8 + sin2 7π 8 = sin4 π 8 + sin4 3π 8 + sin4 π 2 + π 8 + sin4 π 2 + 3π 8 = sin4 π 8 + sin4 3π 8 + cos4 π 8 + cos4 3π 8 = sin4 π 8 + sin4 π 2 – π 8 + cos4 π 8 + cos4 π 2 – π 8
Vedanta Optional Mathematics Teacher's Guide ~ 10 263 = sin4 π 8 + cos4 π 8 + cos4 π 8 + sin4 π 8 = 2 sin4 π 8 + cos4 π 8 = 2 sin2 π 8 + cos2 π 8 2 – 2 sin2 π 8 . cos2 π 8 = 2 1 – 1 2 4 sin2 π 8 . cos2 π 8 = 2 1 – 1 2 sin2 π 4 = 2 1 – 1 2 . 1 2 = 2 1 – 1 4 = 2 . 3 4 = 3 2 = RHS proved Questions for practice 1. If cosθ = 1 2, then find the values of sin2θ, cos2θ and tan2θ. 2. Prove that cotθ = ± 1+cos2θ 1– cos2θ 3. Prove the following : (a) 2sin2 π 4 – A = (1– sinA) (b) tanα + cotα = 2cosec2α (c) tan π 4 + θ = cos2θ 1–sin2θ (d) 1–tan2 π 4 – θ 1 + tan2 π 4 – θ = sin2θ (e) 1+sin2A 1–sin2A = cotA+1 cotA–1 2 4. Prove the following: (a) cos2 θ + sin2 θ.cos2ß = cos2 ß + sin2 ß . cos2θ (b) cosec 20° + cot 40° = cot 10° – cosec 40° (c) 4 cosec2θ . cot2θ = cosec2 θ – sin2 θ 5. Prove that : tanθ + 2tan2θ + 4cot4θ = cotθ 6. Prove that : tanθ + tan π 3 + θ + tan 2π 3 – θ = 3 tan3θ.
264 Vedanta Optional Mathematics Teacher's Guide ~ 10 Sub multiple angles Estimated Periods: 7 1. Objectives S,N. Level Objectives (i) Knowledge(k) To define sub-multiple angles of an angle. To tell the formulae of trigonometric ratios of sub-multiple angles. (ii) Understanding(U) To explain to derive the formulae of sub-multiple angles by using compound angle formulae. (ii) Application(A) To solve problems of trigonometric identities of sub-multiple angles. (iv) Higher Ability (HA) To solve very long question of trigonometric ratios multiple angles. 2. Teaching Materials Formula chart of trigonometric ratios sub-multiple angles. Teaching Strategies – Review the formulae of trigonometric ratios of compound angles and multiple angles. – Define sub-multiple angles of q as q 2 , q 3 , q 4 etc. – Show how to derive the formulae of trigonometric ratios of sub-multiple angles by using compound angle formulae. – Compare formula of multiple and sub-multiple angle formulae of rigonometry like sin2q = 2 sinq . cosq sinq = sin2 q 2 = 2 sin q 2 . cos q 2 . cos2q = cos2 q – sin2 q = 2 cos2 q – 1 = 1 – tan2 q 1 + tan2 q cosq = cos2 q 2 – cos2q 2 – sin2q 2 = 1 – tan2 q 2 1 + tan2q 2 5. Discuss how to evaluate values of sinq, cosq, tanq if sinq = 1 2 . 6. Discuss to solve problems related to trigonometric ratios of sub-multiple questions given in exercises. List of formulae: 1. sinθ = 2 sin q 2 . cos q 2 = 2 tan q 2 1 – tan2q 2
Vedanta Optional Mathematics Teacher's Guide ~ 10 265 2. cos θ = cos 2 q2 – sin 2 q2 = 2 cos 2 q2 – 1 = 1 – 2 sin 2 q2 = 1 – tan2 q2 1 + tan 2 q2 3. sin θ = 3 sin q3 – 4 sin 3 q3 cos θ = 4 cos 3 q3 – 3 cos q3 4. tan θ = 3 tan q3 – tan3 q3 1 – 3 tan 2 q3 5. cot θ = cot2 q2 – 1 2 cot q2 6. cot θ = cot3 q3 – 3 cot q3 3 cot 2 q3 – 1 Some solved problems Prove the following : 1. cotx = cot3 x3 – 3 cot x3 3 cot2 x3 – 1 Solution LHS = cotx = cot x3 + cot 2x3 = cot x3 . cot 2x3 – 1 cot 2x3 + cot x3 = cot x3 . cot2 x3 – 1 2 cot x3 – 1 = cot2 x3 – 1 2 cot x3 + cot x3 = cot3 x3 – 3 cot x3 3 cot2 x3 – 1
266 Vedanta Optional Mathematics Teacher's Guide ~ 10 = RHS proved 2. a) If cos q 3 = 1 2 p + 1 p , then prove that : cosθ = 1 2 p3 + 1 p3 Solution Here, cos q 3 = 1 2 p + 1 p LHS = cosθ = 4 cos3 q 3 – 3 cos q 3 = 4 . 1 8 p + 1 p 3 – 3 2 p + 1 p = 1 2 p + 1 p p + 1 p 2 – 3 = 1 2 p + 1 p p2 + 2 . p . 1 p + 1 p2 – 3 = 1 2 p + 1 p p2 + 2 + 1 p2 – 3 = 1 2 p + 1 p p2 – 1 + 1 p2 = 1 2 p3 + 1 p3 = RHS proved b) If sin q 2 = 1 2 p + 1 p , then prove that : cosθ = – 1 2 p2 + 1 p2 Solution LHS = cosθ = 1 – 2 sin2 q 2 = 1 – 2 . 1 4 p + 1 p 2 = 1 – 1 2 p2 + 2 . p . 1 p2 + 1 p2 = 1 – 1 2 p2 + 1 p2 + 2 = 1 2 2 – p2 – 1 p2 – 2 = –1 2 p2 + 1 p2
Vedanta Optional Mathematics Teacher's Guide ~ 10 267 = RHS proved 3. Prove the following a) 1 – sinA cosA = 1 – tan A2 1 + tan A2 Solution LHS = 1 – sinA cosA = 1 – 2 tan A/2 1 + tan 2 A/2 1 – tan 2 A/2 1 + tan 2 A/2 = 1 + tan2 A2 – 2 tan A2 1 – tan2 A2 = 1 – tan A2 2 1 + tan A2 1 – tan A2 = 1 – tan A2 1 + tan A2 = RHS proved b) 1 – 2 sin2 p4 – θ2 = sin θ Solution LHS = 1 – 2 sin2 p4 – θ2 = cos2 p4 – θ2 = cos p2 – θ = sinθ = RHS proved c) 1 – tan2 p4 – θ4 1 + tan 2 p4 – θ4 = sin θ2
268 Vedanta Optional Mathematics Teacher's Guide ~ 10 Solution LHS = 1 – tan2 p 4 – θ 4 1 + tan2 p 4 – θ 4 = cos2 p 4 – θ 4 = cos p 2 – θ 2 = sin θ 2 = RHS proved 4. Prove that a) cos4 θ 2 – sin4 θ 2 = cosθ Solution LHS = cos4 θ 2 – sin4 θ 2 = cos2 θ 2 + sin2 θ 2 cos2 θ 2 – sin2 θ 2 = 1 . cosθ = cosθ = RHS proved. b) 2 sinθ – sin2θ 2 sinθ + sin2θ = tan Solution LHS = 2 sinθ – sin2θ 2 sinθ + sin2θ = 2 sinθ – 2 sinθ . cosθ 2 sinθ + 2 sinθ . cosθ = 2 sinθ (1 – cosθ) 2 sinθ (1 + cosθ) = 2 sin2 θ 2 2 cos2 θ 2 = tan2 θ 2 = RHS proved.
Vedanta Optional Mathematics Teacher's Guide ~ 10 269 c) sin2 θ 1 + cos2 θ . cos θ 1 + cos θ = tan θ2 Solution LHS = sin2 θ 1 + cos2 θ . cos θ 1 + cos θ = sin2θ cos θ 2 cos 2 θ . cos θ 1 + cos θ = sin θ 1 + cos θ = 2 sin θ2 . cos θ2 2 cos2 θ2 = tan θ2 = RHS proved d) 1 + sin θ 1 – sin θ = tan2 p4 + θ2 Solution RHS = tan2 p4 + θ2 = tan p4 + θ2 2 = tan p4 + tan θ2 1 – tan p4 . tan θ2 2 = 1 + sin θ2 cos θ2 2 1 – sin θ2 cos θ2 2 = cos θ2 + sin θ2 2 cos θ2 – sin θ2 2 = cos 2 θ2 + sin 2 θ2 + 2 cos θ2 . sin θ2 cos 2 θ2 + sin 2 θ2 – 2 cos θ2 . sin θ2
270 Vedanta Optional Mathematics Teacher's Guide ~ 10 = 1 + sinθ 1 – sinθ = LHS proved e) cos θ 2 – 1 + sinθ sin θ 2 – 1 + sinθ = tan θ 2 Solution LHS = cos θ 2 – 1 + sinθ sin θ 2 – 1 + sinθ = cos θ 2 – sin2 θ 2 + cos2 θ 2 + 2 sin θ 2 . cos θ 2 sin θ 2 – sin2 θ 2 + cos2 θ 2 + 2 sin θ 2 . cos θ 2 = cos θ 2 – sin θ 2 + cos θ 2 2 sin θ 2 – sin θ 2 + cos θ 2 2 = cos θ 2 – sin θ 2 – cos θ 2 sin θ 2 – sin θ 2 – cos θ 2 = –sin θ 2 –cos θ 2 = tan θ 2 = RHS proved. 5. Prove the following a) tan p 4 + θ 2 = secθ + tanθ Solution LHS = tan p 4 + θ 2 = tan p 4 + tan θ 2 1 – tan p 4 . tan θ 2
Vedanta Optional Mathematics Teacher's Guide ~ 10 271 = 1 + tan θ2 1 – tan θ2 = 1 + sin q/2 cos q/2 1 – sin q/2 cos q/2 = cos θ2 + sin θ2 cos θ2 – sin θ2 × cos θ2 + sin θ2 cos θ2 + sin θ2 = cos θ2 + sin θ2 2 cos 2 θ2 – sin 2 θ2 = 1 + sin θ cos θ = 1 cos θ + sin θ cos θ = sec θ + tan θ = RHS proved. b) tan p4 – θ2 = 1 – sin θ 1 + sin θ Solution RHS = 1 – sin θ 1 + sin θ = sin2 θ2 + cos 2 θ2 – 2 sin θ2 . cos θ2 sin2 θ2 + cos 2 θ2 + 2 sin θ2 . cos θ2 = cos θ2 – sin θ2 2 cos θ2 + sin θ2 2 = cos θ2 – sin θ2 cos θ2 + sin θ2 Dividing numerator and denominator by cos θ2 , we get
272 Vedanta Optional Mathematics Teacher's Guide ~ 10 = 1 – tan θ 2 1 + tan θ 2 = tan p 4 – tan θ 2 1 + tan p 4 . tan θ 2 = tan p 4 – θ 2 = LHS proved c) sec p 4 + θ 2 . sec p 4 + θ 2 = 2 secθ Solution LHS = sec p 4 + θ 2 . sec p 4 + θ 2 = 1 cos p 4 – θ 2 – 1 cos p 4 + θ 2 = 1 cos p 4 . cos θ 2 + sin p 4 . sin θ 2 . 1 cos p 4 . cos θ 2 + sin p 4 . sin θ 2 = 1 1 2 . cos θ 2 + 1 2 . sin θ 2 . 1 1 2 . cos θ 2 + 1 2 . sin θ 2 = 2 cos θ 2 + sin θ 2 cos θ 2 – sin θ 2 = 2 cos2 θ 2 – sin2 θ 2 = 2 cosθ = 2 secθ = RHS proved. d) tan p 4 – θ 2 = cosθ 1 + sinθ Solution LHS = tan p 4 – θ 2 = tan p 4 – tan θ 2 1 + tan p 4 . tan θ 2
Vedanta Optional Mathematics Teacher's Guide ~ 10 273 = 1 – sin q/2 cos q/2 1 + sin q/2 cos q/2 = cos θ2 – sin θ2 cos θ2 + sin θ2 = cos θ2 – sin θ2 cos θ2 + sin θ2 × cos θ2 + sin θ2 cos θ2 + sin θ2 = cos 2 θ2 – sin 2 θ2 sin θ2 + cos θ2 2 = cos θ sin 2 θ2 + cos 2 θ2 + 2 sin θ2 . cos θ2 = cos θ 1 + sin θ = RHS proved. e) cot θ2 + p4 – tan θ2 – p4 = 2 cosθ 1 + sin θ Solution LHS = cot θ2 + p4 – tan θ2 – p4 = cot θ2 . cot p4 – 1 cot θ2 + cot p4 – tan θ2 – tan p4 1 + tan θ2 . tan p4 = cos q/2 sin q/2 – 1 cos q/2 sin q/2 + 1 – sin q/2 cos q/2 – 1 1 + sin q/2 cos q/2 = cos θ2 – sin θ2 cos θ2 + sin θ2 – sin θ2 – cos θ2 cos θ2 + sin θ2 = cos θ2 – sin θ2 cos θ2 + sin θ2 + cos θ2 – sin θ2 cos θ2 + sin θ2
274 Vedanta Optional Mathematics Teacher's Guide ~ 10 = 2 cos θ 2 – sin θ 2 cos θ 2 + sin θ 2 = 2 cos θ 2 – sin θ 2 cos θ 2 + sin θ 2 × cos θ 2 + sin θ 2 cos θ 2 + sin θ 2 = 2 cos2 θ 2 – sin2 θ 2 cos θ 2 + sin θ 2 2 = 2 cosθ cos2 θ 2 + sin2 θ 2 + 2 sin θ 2 . cos θ 2 = 2 cosθ 1 + sinθ = RHS proved. f) tan p 4 + θ 2 + tan p 4 – θ 2 = 2 secθ Solution LHS = tan p 4 + θ 2 + tan p 4 – θ 2 = tan p 4 + tan θ 2 1 – tan p 4 . tan θ 2 + tan p 4 – tan θ 2 1 + tan p 4 . tan θ 2 = 1 + sin q/2 cos q/2 1 – sin q/2 cos q/2 + 1 – sin q/2 cos q/2 1 + sin q/2 cos q/2 = cos θ 2 + sin θ 2 cos θ 2 – sin θ 2 + cos θ 2 – sin θ 2 cos θ 2 + sin θ 2 = cos θ 2 + sin θ 2 2 + cos θ 2 – sin θ 2 2 cos θ 2 – sin θ 2 cos θ 2 + sin θ 2 = 1 + sinθ + 1 – sinθ cos2 θ 2 – sin2 θ 2 = 2 cosθ = 2 secθ = RHS proved. 6. Prove that
Vedanta Optional Mathematics Teacher's Guide ~ 10 275 a) (cosa – cosb) 2 + (sina – sinb) 2 = 4 sin2 a – b 2 Solution LHS = (cosa – cosb) 2 + (sina – sinb) 2 = cos2 a – 2 cosa . cosb + cos2 b + sin2 a – 2 sina . sinb + cos2 b = 2(cos2 a + sin2 a) + (cos2 b + sin2 b) – 2(cosa . cosb + sina . sinb) = 1 + 1 – 2 cos(a – b) = 2 – 2 cos(a – b) = 2[1 – cos(a – b)] = 2 . 2 sin2 a – b 2 = 4 sin2 a – b 2 = RHS proved b) (sina + sinb) 2 + (cosa + cosb) 2 = 4 cos2 a – b 2 Solution LHS = (sina + sinb) 2 + (cosa + cosb) 2 = sin2 a + sin2 b + 2 sina . sinb + cos2 a + cos2 b – 2 cosa . cosb = (sin2 a + cos2 a) + (sin2 b + sin2 a) + 2(sina . sinb + cosa . cosb) = 1 + 1 + 2 cos(a – b) = 2 + 2 cos(a – b) = 2[1 + cos(a – b)] = 2 . 2 cos2 a – b 2 = 4 cos2 a – b 2 = RHS proved 7. Prove the following a) cos 2p 15 . cos 4p 15 . cos 8p 15 . cos 16p 15 = 1 16 Solution LHS = cos 2p 15 . cos 4p 15 . cos 8p 15 . cos 16p 15 = 1 2 sin 2p 15 2 sin 2p 15 . cos 2p 15 . cos 4p 15 . cos 8p 15 . cos 16p 15
276 Vedanta Optional Mathematics Teacher's Guide ~ 10 = 1 2 sin 2p 15 sin 4p 15 . cos 4p 15 . cos 8p 15 . cos 16p 15 = 1 4 sin 2p 15 sin 8p 15 . cos 8p 15 . cos 16p 15 = 1 8 sin 2p 15 sin 16p 15 . cos 16p 15 = 1 16 sin 2p 15 sin 32p 15 = 1 16 sin 2p 15 . sin 2p + 2p 15 = 1 16 sin 2p 15 . sin 2p 15 = 1 16 = RHS proved. 8. Proved that: 1 + cos p 8 1 + cos 3p 8 1 + cos 5p 8 1 + cos 7p 8 = 1 8 Solution LHS = 1 + cos p 8 1 + cos 3p 8 1 + cos 5p 8 1 + cos 7p 8 = 1 + cos p 8 1 + cos 3p 8 1 + cos p – 3p 8 1 + cos p – p 8 = 1 + cos p 8 1 + cos 3p 8 1 – cos 3p 8 1 – cos p 8 = 1 – cos2 p 8 1 – cos2 3p 8 = sin2 p 8 . sin2 3p 8 = sin2 p 8 . sin2 p 2 – p 8 = sin2 p 8 . cos2 p 8
Vedanta Optional Mathematics Teacher's Guide ~ 10 277 = 14 2 sin p8 . cos p8 2 = 14 sin2 p8 2 = 14 sin p4 2 = 14 1 2 2 = 18 = RHS proved. 8. Prove that: tan 7 1°2 = 6 – 3 + 2 – 2 Solution LHS = tan 7 1°2 = tan 15° 2 = sin 15° 2 cos 15° × 2 sin 15° 2 2 sin 15° 2 = 2 sin 2 15° 2 sin 15° = 1 – cos 15° sin 15° = 1 – cos(45° – 30°) sin(45° – 30°) = 1 – cos 45° . cos 30° – sin 45° . sin 30° sin 45° . cos 30° – cos 45° . sin 30° = 1 – 12 . 3 2 – 12 . 12 12 . 3 2 – 12 . 12 = 2 2 – 3 – 1 3 – 1 × 3 + 1 3 + 1 = 2 6 – 3 – 3 + 2 2 – 3 – 1 3 – 1
278 Vedanta Optional Mathematics Teacher's Guide ~ 10 = 2 6 – 2 3 + 2 2 – 4 2 = 6 – 3 + 2 – 2 = RHS proved Questions for practice 1. Prove that: tan A 2 =± 1 – cosA 1 + cosA 2. If cos 330° = 3 2 , prove that : sin 165° = 1 2 2 – 3 3. Prove that: 1 + cosθ + sinθ 1 – cosθ + sinθ = cot θ 2 4. Prove that: secθ + tanθ = tan p 4 + θ 2 5. If sin θ 3 = 1 2 a + 1 a , prove that: sinθ = 1 2 a3 + 1 a3 6. Prove that: tan p 4 + θ 2 = 1 + sinθ 1 – sinθ = secθ + tanθ 7. Prove that: sin2A 1 + cos2A × cosA 1 + cosA = tan A 2 8. Prove that: cot 71° 2 = 2 + 3 + 4 + 6
Vedanta Optional Mathematics Teacher's Guide ~ 10 279 Transformation of Trigonometric Formula of sine and cosine Estimated Periods: 5 1. Objectives S,N. Level Objectives (i) Knowledge(k) To tell the formulae of transformation of trigonometric ratios of sine and cosine. (ii) Understanding(U) To explain to derive the formulae of trannsformation of trigonometric ratios. (ii) Application(A) To solve the problems of transformation of trigonometric formulae. (iv) Higher Ability (HA) To solve harder problems of transformation of trigonometric formulae. 2. Required teaching materials Formula chart of trigonometric ratio of compound angles and transformation formula of trigonometric ratios. Teaching strategies – Review the formulae of trigonometric ratios of compound angles. – List the trigonometric ratios of compound angles as given below: sin(A + B) = sinA . cosB + cosA . sinB ... ... ... (i) sin(A – B) = sinA . cosB – cosA . sinB ... ... ... (ii) cos(A + B) = cosA . cosB – sinA . sinB ... ... ... (iii) cos(A – B) = cosA . cosB + sinA . sinB ... ... ... (iv) – Adding and subtracting above identities we get the required formulae. Example adding (i) and (ii), we get 2 sinA . cosB = sin(A + B) + sin(A – B) Subtracting (ii) from (i), we get 2 cosA . sinB = sin(A + B) – sin(A – B) Similarly, explain to get the following results 2 cosA . cosB = cos(A – B) + cos(A – B) 2 sinA . sinB = sin(A – B) – cos(A + B) – Again, discuss how to derive the following formulae sinC + sinD = 2 sin C + D 2 . cos C – D 2 sinC – sinD = 2 cos C + D 2 . sin C – D 2 cosC + cosD = 2 cos C + D 2 . cos C – D 2 cosC – cosD = 2 sin C + D 2 . sin C – D 2
280 Vedanta Optional Mathematics Teacher's Guide ~ 10 List of formula: 1. 2 sinA . cosB = sin(A + B) + sin(A – B) 2. 2 cosA . sinB = sin(A + B) – sin(A – B) 3. 2 cosA . cosB = cos(A + B) + cos(A – B) 4. 2 sinA . sinB = cos(A – B) – cos(A + B) 5. sinC + sinD = 2 sin C + D 2 . cos C – D 2 6. sinC – sinD = 2 cos C + D 2 . sin C – D 2 7. cosD – cosC = 2 sin C + D 2 . sin C – D 2 8. If a b = c d then a+b a–b = c+d c–d is called componendo and dividendo. Some solved problems 1. Prove the following: a) cos75° + cos15° = 3 2 Solution LHS = cos75° + cos15° = 2 cos 75° + 15° 2 cos 75° – 15° 2 = 2 cos45° . cos30° = 2 . 1 2 . 3 2 = 3 2 = RHS proved. b) sin75° – sin15° = 1 2 Solution LHS = sin75° – sin15° = 2 cos 75° + 15° 2 . sin 75° – 15° 2 = 2 cos45° . sin30°
Vedanta Optional Mathematics Teacher's Guide ~ 10 281 = 2 . 1 2 . 1 2 = 1 2 = RHS proved. c) cos52° + cos68° + cos172° = 0 Solution LHS = cos52° + cos68° + cos172° = 2 cos 52° + 68° 2 . cos 52° – 68° 2 + cos(180° – 8°) = 2 cos60° . cos(–8°) – cos8° = 2 . 1 2 . cos8° – cos8° = cos8° – cos8° = 0 = RHS proved. 2. Prove that: a) cos5A + sin3A sin5A – sin3A = cotA Solution LHS = cos5A + sin3A sin5A – sin3A = 2 cos 5A + 3A 2 . cos 5A – 3A 2 2 cos 5A + 3A 2 . sin 5A – 3A 2 = cos4A . cosA cos4A . sinA cos(–θ) = cosθ = cotA = RHS proved. b) cos40° – cos60° sin60° – sin40° = tan50° Solution LHS = cos40° – cos60° sin60° – sin40° = 2 sin 40° + 60° 2 . sin 40° – 60° 2 2 cos 40° + 60° 2 . sin 40° – 60° 2 = sin50° . sin10° cos50° . sin10° = tan50° = RHS proved.
282 Vedanta Optional Mathematics Teacher's Guide ~ 10 c) cos80° + cos20° sin80° – sin20° = 3 Solution LHS = cos80° + cos20° sin80° – sin20° = 2 cos 80° + 20° 2 . cos 80° – 20° 2 2 cos 80° + 20° 2 . sin 80° – 20° 2 = cos30° sin30° = 3 2 1 2 = 3 2 × 2 = 3 = RHS proved. d) cos8° + sin8° cos8° – sin8° = tan53° Solution LHS = cos8° + sin8° cos8° – sin8° = cos8° + sin(90° – 82°) cos8° – sin(90° – 82°) = cos8° + cos82° cos8° – cos82° = 2 cos 8° + 82° 2 . cos 8° – 82° 2 2 sin 8° + 82° 2 . sin 82° –8° 2 = cos37° sin37° = cot37° = cot(90° – 53°) = tan53° = RHS proved. e) cos10° – sin10° cos10° + sin10° = cot55°
Vedanta Optional Mathematics Teacher's Guide ~ 10 283 Solution LHS = cos10° – sin10° cos10° + sin10° = cos10° – sin(90° – 80°) cos10° + sin(90° – 80°) = cos10° – cos80° cos10° + cos80° = 2 sin 10° + 80° 2 . sin 80° – 10° 2 2 cos 10° + 80° 2 . cos 10° – 80° 2 = sin45° . sin35° cos45° . cos35° = tan35° = tan(90° – 55°) = cot55° = RHS proved. f) cos(40° + A) + cos(40° – A) sin(40° + A) – sin(40° – A) = cotA Solution LHS = cos(40° + A) + cos(40° – A) sin(40° + A) – sin(40° – A) = 2 cos 40° + A + 40° – A 2 . cos 40° + A – 40° + A 2 2 cos 40° + A + 40° – A 2 . sin 40° + A – 40° + A 2 = cosA sinA = cotA = RHS proved. 3. Prove the following. a) sinA . sin2A + sin3A . sin6A sinA . cos2A + sin3A . cos6A = tan5A Solution LHS = sinA . sin2A + sin3A . sin6A sinA . cos2A + sin3A . cos6A = 2 sinA . sin2A + 2 sin3A . sin6A 2 sinA . cos2A + 2 sin3A . cos6A = cos(A – 2A) – cos(A + 2A) + cos(3A – 6A) – cos(3A + 6A) sin(A + 2A) + sin(A – 2A) + sin(3A + 6A) + sin(3A – 6A)
284 Vedanta Optional Mathematics Teacher's Guide ~ 10 = cosA – cos3A + cos3A – cos9A sin3A – sinA + sin9A – sin3A = cosA – cos9A sin9A – sinA = 2 sin A + 9A 2 . sin A + 9A 2 2 cos A + 9A 2 . sin A + 9A 2 = tan5A = RHS proved. b) cos2A . cos3A – cos2A . cos7A sin4A . sin3A – sin2A . sin5A = sin7A + sin3A sinA Solution LHS = cos2A . cos3A – cos2A . cos7A sin4A . sin3A – sin2A . sin5A = 2 cos2A . cos3A – 2 cos2A . cos7A 2 sin2A . sin3A – 2 sin2A . sin5A = cos(2A + 3A) + cos(2A – 3A) – cos(2A + 7A) – cos(2A – 7A) cos(4A – 3A) – cos(4A + 3A) – cos(2A – A) + cos(2A + 5A) = cos5A + cosA – cos9A – cos5A cosA – cos7A – cos3A + cos7A = cosA – cos9A cosA – cos3A = 2 sin A + 9A 2 . sin 9A – A 2 2 sin A + 3A 2 . sin 3A – A 2 = sin5A . sin4A sin2A . sinA = sin5A . 2 sin2A . cos2A sin2A . sinA = 2 sin5A . cos2A sinA RHS = sin7A + sin3A sinA = 2 sin 7A + 3A 2 . cos 7A – 3A 2 sinA = 2 sin5A . cos2A sinA LHS = RHS proved
Vedanta Optional Mathematics Teacher's Guide ~ 10 285 c) cos7A + cos3A – cos5A – cosA sin7A – sin3A – sin5A + sinA = cot2A Solution LHS = cos7A + cos3A – cos5A – cosA sin7A – sin3A – sin5A + sinA = (cos7A – cosA) + (cos3A – cos5A) (sin7A + sinA) – (sin3A + sin5A) = 2 sin 7A + A 2 . sin 7A – A 2 + 2 sin 3A + 5A 2 . sin 5A – 3A 2 2 sin 7A + A 2 . cos 7A – A 2 – 2 sin 3A + 5A 2 . cos 3A – 5A 2 = 2 sin4A [sin3A + sinA] 2 sin4A [cos3A – cosA] = sinA – sin3A cos3A – cosA = 2 cos A + 3A 2 . sin A – 3A 2 2 sin 3A + A 2 . sin A – 3A 2 = cos2A sin2A = cot2A = RHS proved. d) sinA + sin3A + sin5A + sin7A cosA + cos3A + cos5A + cos7A = tan4A Solution LHS = sinA + sin3A + sin5A + sin7A cosA + cos3A + cos5A + cos7A = (sinA + sin7A) + (sin3A + sin5A) (cosA + cos7A) + (cos3A + cos5A) = 2 sin A + 7A 2 . cos A – 7A 2 + 2 sin 3A + 5A 2 . cos 3A – 5A 2 2 cos A + 7A 2 . cos A – 7A 2 + 2 cos 3A + 5A 2 . cos 3A – 5A 2 = 2 sin4A [cos3A + cosA] 2 cos4A [cos3A + cosA] = tan4A = RHS proved. e) sin5A – sin7A – sin4A + sin8A cos4A – cos5A – cos8A + cos7A = cot6A Solution LHS = sin5A – sin7A – sin4A + sin8A cos4A – cos5A – cos8A + cos7A
286 Vedanta Optional Mathematics Teacher's Guide ~ 10 = (sin5A – sin7A) – (sin4A – sin8A) (cos4A – cos8A) – (cos5A – cos7A) = 2 cos 5A + 7A 2 . sin 5A – 7A 2 – 2 cos 4A + 8A 2 . sin 4A – 8A 2 2 sin 4A + 8A 2 . sin 8A – 4A 2 – 2 sin 5A + 7A 2 . sin 7A – 5A 2 = cos6A [–sinA + sin2A] sin6A [sin2A – sinA] = cot6A = RHS proved. f) sin(p + 2)θ – sinpθ cospθ – cos(p + 2)θ = cot(p + 1)θ Solution LHS = sin(p + 2)θ – sinpθ cospθ – cos(p + 2)θ = 2 cos p + 2 + p 2 θ . sin p + 2 – p 2 θ 2 sin p + p + 2 2 θ . sin p + 2 – p 2 θ = cos(p + 1)θ . sinθ sin(p + 1)θ . sinθ = cot(p + 1)θ = RHS proved. g) (sin4A + sin2A) . (cos4A – cos8A) (sin7A + sin5A) . (cosA – cos5A) = 1 Solution LHS = (sin4A + sin2A) . (cos4A – cos8A) (sin7A + sin5A) . (cosA – cos5A) = 2 sin 4A + 2A 2 . cos 4A – 2A 2 . 2 sin 4A + 8A 2 . sin 8A – 4A 2 2 sin 7A + 5A 2 . cos 7A – 5A 2 . 2 sin A + 5A 2 . sin 5A – A 2 = sin3A . cosA . sin6A . sin2A sin6A . cosA . sin3A . sin2A = 1 = RHS proved. 4. Prove that: a) sin20° . sin40° . sin60° . sin80° = 3 16
Vedanta Optional Mathematics Teacher's Guide ~ 10 287 Solution LHS = sin20° . sin40° . sin60° . sin80° = 3 2 . 12 (2 sin20° . sin40°) . sin80° = 34 cos(20° – 40°) – cos(20° + 40°) . sin80° = 34 {cos20° – cos60°} . sin80° = 34 cos20° . sin80° – 34 . 12 . sin80° = 38 sin(20° + 80°) – sin(20° – 80°) – 38 sin80° = 38 sin100° + 38 sin60° – 38 sin80° = 38 sin(180° – 80°) – 38 sin80° + 38 . 3 2 = 3 16 + 38 sin80° – 38 sin80° = 3 16 = RHS proved. b) sin10° . sin30° . sin50° . sin70° = 1 16 Solution LHS = sin10° . sin30° . sin50° . sin70° = sin10° . 12 . sin50° . sin70° = 14 (2 sin10° . sin50°) sin70° = 14 {cos(10° – 50°) – cos(10° + 50°)} . sin70° = 14 {cos40° – cos60°} . sin70° = 14 cos40° . sin70° – 14 . 12 . sin70° = 18 (2 cos40° . sin70°) – 18 sin70° = 18 {sin110° – sin(–30°)} – 18 sin70° = 18 sin110° + 18 . 12 – 18 sin70°
288 Vedanta Optional Mathematics Teacher's Guide ~ 10 = 1 8 sin(180° – 70°) + 1 16 – 1 8 sin70° = 1 8 sin70° – 1 8 sin70° + 1 16 = 1 16 = RHS proved. d) cos40° . cos100° . cos160° = 1 8 Solution LHS = cos40° . cos100° . cos160° = 1 2 [2 cos40° . cos100°] . cos160° = 1 2 [cos40° + 100°) + cos(40° – 100°)] . cos160° = 1 2 [cos140° + cos60°] . cos160° = 1 2 cos140° . cos160° + 1 2 . 1 2 . cos160° = 1 4 (cos300° + cos20°) + 1 4 . cos160° = 1 4 . 1 2 + 1 4 cos20° – 1 4 cos20° = 1 8 = RHS proved. e) tan20° . tan40° . tan80° = 3 Solution LHS = tan20° . tan40° . tan80° = sin20° cos20° . sin40° cos40° . sin80° cos80° Numerator = sin20° . sin40° . sin80° Simplify it to get, 3 8 Again, denominator = cos20°, cos40°, cos80° Simplify it to get, 1 8 Then, LHT = 3 8 1 8 = 3 = RHS proved.
Vedanta Optional Mathematics Teacher's Guide ~ 10 289 5. Prove that: a) sin(45° + θ) . sin(45° – θ) = 1 2 cos2θ Solution LHS = sin(45° + θ) . sin(45° – θ) = 1 2 {2 sin(45° + θ) . sin(45° – θ)} = 1 2 {cos(45° + θ – 45° + θ) – cos(45° + θ + 45° – θ)} = 1 2 {cos2θ – cos90°} = 1 2 cos2θ = RHS proved. b) cos(45° + θ) . cos(45° – θ) = 1 2 cos2θ Solution LHS = cos(45° + θ) . cos(45° – θ) = 1 2 {cos(45° + θ + 45° – θ) + cos(45° + θ – 45° + θ)} = 1 2 [cos90° + cos2θ] = 1 2 cos2θ = RHS proved. c) cosθ . cos(60° – θ) . cos(60° + θ) = 1 4 cos3θ Solution LHS = cosθ . cos(60° – θ) . cos(60° + θ) = 1 2 cosθ {2 cos(60° – θ) . cos(60° + θ)} = 1 2 cosθ {cos(60° – θ + 60° + θ) + cos(60° – θ – 60° – θ)} = 1 2 cosθ {cos120° + cos2θ} = 1 2 cosθ –1 2 + 1 2 cosθ . cos2θ = – 1 4 cosθ + 1 4 {cos3θ . cosθ} =– 1 4 cosθ + 1 4 cos3θ + 1 4 cosθ = 1 4 cos3θ = RHS proved.
290 Vedanta Optional Mathematics Teacher's Guide ~ 10 6. Prove that: cos(36° – θ) . cos(36° + θ) + cos(54° + θ) . cos(54° – θ) = cos2θ Solution LHS = cos(36° – θ) . cos(36° + θ) + cos(54° + θ) . cos(54° – θ) = 1 2 {cos(36° – θ + 36° + θ) + cos(36° – θ – 36° – θ)} + 1 2 {cos(54° + θ + 54° – θ) + cos(54° + θ – 54° + θ)} = 1 2 {cos72° + cos2θ + cos108° + cos2θ} = 1 2 . 2 cos2θ + 1 2 {cos72° + cos(180° – 72°)} = cos2θ + 1 2 {cos72° – cos72°} = cos2θ + 1 2 . 0 = cos2θ = RHS proved. 7. Prove that: a) tan p 4 + θ + tan p 4 – θ = 2 sec2θ Solution LHS = tan p 4 + θ + tan p 4 – θ = tan p 4 + tanθ 1 – tan p 4 . tanθ – tan p 4 – tanθ 1 + tan p 4 . tanθ = 1 + sin q cos q 1 – sin q cos q + 1 – sin q cos q 1 + sin q cos q = cosθ + sinθ cosθ – sinθ + cosθ – sinθ cosθ + sinθ = (cosθ + sinθ) 2 + (cosθ – sinθ) 2 cos2 θ – sin2 θ = cos2 θ + sin2 θ + 2 sinθ. cosθ + cos2 θ + sin2 θ – 2 sinθ cosθ cos2θ = 1 + 1 cos2θ = 2 1 cos2θ = 2 sec2θ = RHS proved.
Vedanta Optional Mathematics Teacher's Guide ~ 10 291 b) tan p 4 + θ – tan p 4 – θ = 2 tan2θ Solution LHS = tan p 4 + θ – tan p 4 – θ = tan p 4 + tanθ 1 – tan p 4 . tanθ – tan p 4 – tanθ 1 + tan p 4 . tanθ = 1 + sin q cos q 1 – sin q cos q – 1 – sin q cos q 1 + sin q cos q = cosθ + sinθ cosθ – sinθ – cosθ – sinθ cosθ + sinθ = (cosθ + sinθ) 2 – (cosθ – sinθ) 2 cos2 θ – sin2 θ = cos2 θ + sin2 θ + 2 sinθ. cosθ – cos2 θ – sin2 θ + 2 sinθ cosθ cos2θ = sin2θ + cos2θ cos2θ = 2 sin2θ cos2θ = 2 tan2θ = RHS proved. 8. Prove the following. a) (cosA + cosB)2 + (sinA + sinB)2 = 4cos2 A – B 2 Solution LHS = (cosA + cosB)2 + (sinA + sinB)2 = 2 cos A + B 2 . cos A – B 2 2 + 2 sin A + B 2 . cos A – B 2 2 = 4 cos2 A – B 2 cos2 A + B 2 + sin2 A + B 2 = 4 cos2 A – B 2 = RHS proved. b) (CosB – cosA)2 + (sinA – sinB)2 = 4 sin2 A – B 2 Solution LHS = (CosB – cosA)2 + (sinA – sinB)2 = 2 sin A + B 2 . sin A – B 2 2 + 2 cos A + B 2 . sin A – B 2 2
292 Vedanta Optional Mathematics Teacher's Guide ~ 10 = 4 sin2 A + B 2 sin2 A – B 2 + 4 cos2 A + B 2 . sin2 A – B 2 = 4 sin2 A – B 2 sin2 A + B 2 + cos2 A + B 2 = 4 sin2 A – B 2 . 1 = 4 sin2 A – B 2 = RHS proved. c) sin2 π 8 + A 2 – sin2 π 8 – A 2 = 1 2 sinA Solution LHS = sin2 π 8 + A 2 – sin2 π 8 – A 2 = 1 2 1 – cos2 π 8 + A 2 – 1 2 1 – cos2 π 8 – A 2 = 1 2 – 1 2 cos π 4 + A – 1 2 + 1 2 cos π 4 – A = 1 2 cos π 4 – A – cos π 4 – A = 1 2 2 sin π 4 – A + π 4 + A 2 . sin π 4 + A – π 4 + A 2 = sin π 4 . sinA = 1 2 sinA = RHS proved. d) sin2 A + sin2 (A + 120°) + sin2 (A – 120°) = 3 2 Solution LHS = sin2 A + sin2 (A + 120°) + sin2 (A – 120°) = sin2 A + 1 2 {1 – cos2 (A + 120°) + 1 – cos2 (A – 120°)} = 1 2 + 1 2 + sin2 A – 1 2 {cos(2A + 240°) + cos(2A – 240°)} = 1 + sin2 A – 1 2 . 2 cos 2A + 240° + 2A – 240° 2 cos 2A + 240° – 2A – 240° 2 = 1 + sin2 A – cos2A . cos240° = 1 + sin2 A – cos2A cos240° = 1 + sin2 A – (1 – 2 sin2 A) . –1 2 = 1 + sin2 A + 1 2 – sin2 A = 1 + 1 2 = 3 2 = RHS proved.
Vedanta Optional Mathematics Teacher's Guide ~ 10 293 9. a) If sinα + sinb = 1 4 and cosα + cosb = 1 2 , prove that: tan a + b 2 = 1 2 Solution Here, sinα + sinb = 1 4 or, 2 sin a + b 2 . cos a – b 2 = 1 4 ... ... ... (i) and cosα + cosb = 1 2 or, 2 cos a + b 2 . cos a – b 2 = 1 2 ... ... ... (ii) Dividing (i) by (ii), we get, tan(α + b) = 1 2 proved. b) If sinx = k siny, then prove that: tan x – y 2 = k – 1 k + 1 tan x + y 2 Solution Here, k = sinx siny RHS = k – 1 k + 1 tan x + y 2 = sinx siny – 1 sinx siny + 1 tan x + y 2 = sinx – siny siny × siny sinx + siny tan x + y 2 = 2 cos x + y 2 . sin x – y 2 2 sin x + y 2 . cos x – y 2 tan x + y 2 = cot x + y 2 . tan x + y 2 . tan x – y 2 = tan x – y 2 = RHS proved. c) If sin(A + B) = k sin(A – B), then prove that: (k – 1) tanA = (k + 1) tanB. Solution Here, = sin(A + B) sin(A – B) = k 1 By using componendo and dividendo sin(A + B) + sin(A – B) sin(A + B) – sin(A – B) = k + 1 k – 1
294 Vedanta Optional Mathematics Teacher's Guide ~ 10 or, 2 sinA . cosB 2 cosA . sinB = k + 1 k – 1 or, tanA tanB = k + 1 k – 1 (k – 1) tanA = (k + 1) tan B proved. 10. Prove that: sin2 x – sin2 y sinx . cosx – siny . cosy = tan(x y) Solution LHS = sin2 x – sin2 y sinx . cosx – siny . cosy = 2 sin2 x – sin2 y 2 sinx . cosx – 2 siny . cosy = 1 – cos2x – 1 + cos2y sin2x – sin2y = cos2y – cos2x sin2x – sin2y = 2 sin 2y + 2x 2 . sin 2x – 2y 2 2 cos 2x + 2y 2 . sin 2x – 2y 2 = tan(x y) = RHS proved. 11. Prove that cos 2π 7 + cos 4π 7 + cos 6π 7 = – 1 2 Solution LHS = cos 2π 7 + cos 4π 7 + cos 6π 7 = 1 2 sin p 7 2 cos 2π 7 . sin p 7 + 2 cos 4π 7 . sin p 7 + 2 cos 6π 7 . sin p 7 = 1 2 sin p 7 sin 2π 7 + p 7 – sin 2π 7 – p 7 + sin 4π 7 + p 7 – sin 4π 7 – p 7 + sin 6π 7 + p 7 – sin 6π 7 – p 7 = 1 2 sin p 7 sin 3π 7 – sin p 7 + sin 5π 7 – sin 3π 7 + sin π – sin 5π 7 = 1 2 sin p 7 –sin p 7 + 0 = –1 2 = RHS proved.
Vedanta Optional Mathematics Teacher's Guide ~ 10 295 11. Prove that: x = y cot a + b 2 , if x cosα + y sinα = x cosβ + y sinβ. Solution Here, x cosα + y sinα = x cosβ + y sinβ or, x(cosα – cosb) = y(sinβ – sina) or, x . 2 sin a + b 2 sin b – a 2 = y 2 cos b + a 2 sin b – a 2 x = y cot a + b 2 proved. Questions for practice Prove the following 1. cos15° . sin75° = 2 + 3 4 2. sin5x – sin3x cos5x + cos3x = tan4x 3. cos20° – sin20° cos20° + sin25° = tan25° 4. cosθ – cos2θ + cos3θ sinθ – sin2θ + sin3θ = cot2θ 5. sin8θ . cosθ – sin6θ . cos3θ cos2θ . cosθ – sin3θ . sin4θ = tan2θ 6. cosec π 4 – θ . cosec π 4 + θ = 2 cosθ 7. sin10° . sin50° . sin60° . sin70° = 3 16 8. cos80° . cos140° . cos160° = 1 8 9. cosθ . cos(60° – θ) . cos(60° + θ) = 1 4 cos3θ 10. cos20° . cos40° . cos80° . cos240° = – 1 16 11. cos π 7 . cos 2π 7 . cos 4π 7 = – 1 8 12. If sin2x + sin2y = 1 3 and cos2x + cos2y = 1 2 , then show that: tan(x + y) = 2 3 . Conditional Trigonometric Identities
296 Vedanta Optional Mathematics Teacher's Guide ~ 10 Estimated Periods: 6 1. Objectives S,N. Level Objectives (i) Knowledge(k) To state the conditional identities related to angles of a triangle. (ii) Understanding(U) To explain to derive the conditional trigonometric identities. (ii) Application(A) To solve the problems related to conditional identities in trigonometry. (iv) Higher Ability (HA) To solve the order problems related to conditional identities. 2. Teaching Materials List of formula of conditional trigonometric identity in a chart paper. 3. Teaching Learning Stategies: – Review the formulae of transformation of trigonometric ratios. – Discuss components of a triangle 3 sides and 3 angles. – If A, B and C are angles of a triangle. Then A + B + C = π or A + B = π – C Taking sin, cosine and tangent ratios on both side, find different identities. Example sin(A + B) = sin(π – C) = sinC cos(A + B) = cos (π – C) = –cosC tan(A + B) = tan (π – C) = –tanC – Similarly, from A 2 + B 2 = π 2 – C 2 find different conditional identities. – Again from 2A + 2B = 2π – 2C find different conditional identities. – Solve some question from exercise of the text book and give guidance to the students to solve the problems in the same exercise. List of formula: 1. 2 sinA . cosB = sin(A + B) + sin(A – B) 2. 2 cosA . sinB = sin(A + B) – sin(A – B) 3. 2 sinA . sinB = cos(A – B) – cos(A + B) 4. 2 cosA . cosB = cos(A + B) + cos(A – B) 5. sinC + sinD = 2 sin C + D 2 . cos C – D 2 6. sinC – sinD = 2 cos C + D 2 . sin C + D 2
Vedanta Optional Mathematics Teacher's Guide ~ 10 297 7. cosC + cosD = 2 cos C + D 2 . cos C – D 2 8. cosC – cosD = 2 sin C + D 2 . sin D – C 2 = –2 sin C + D 2 . sin C – D 2 9. sin(A + B) = sin(π – C) = sinC sin A 2 + B 2 = sin π 2 – C 2 = cos C 2 cos(A + B) = cos(π – C) = –cosC cos A 2 + B 2 = cos π 2 – C 2 = sin C 2 tan(A + B) = tan(π – C) = –tanC tan A 2 + B 2 = tan π 2 – C 2 = tan C 2 sin(2A + 2B) = sin(2π – 2C) = –sin2C cos(2A + 2B) = cos(2π – 2C) = cos2C tan(2A + 2B) = tan(2π – 2C) = –tan2C Some solved problems 1. If A + B + C = 180°, show that: sin(B + 2C) + sin(C + 2A) + sin(A + 2B) = sin(A – C) + sin(B – A) + sin(C – B) Solution Here, A + B + C = 180° or, A + B = 180° – C sin(A + B) = sin(180° – C) = sinC LHS = sin(B + 2C) + sin(C + 2A) + sin(A + 2B) (∴180° = π) = sin(B + C + C) + sin(C + A + A) + sin(A + B + B) = sin(π – A + C) + sin(π – B + A) + sin(π – C + B) = sin{π – (A – C)} + sin{π – (B – A)} + sin{π – (C – B)} = sin(A – C) + sin(B – A) + sin(C – B) = RHS proved. 2. If A + B + C = π, prove that: a) cot A 2 . cot B 2 . cot C 2 = cot A 2 + cot B 2 + cot C 2
298 Vedanta Optional Mathematics Teacher's Guide ~ 10 b) tan A 2 . tan B 2 + tan B 2 . tan C 2 + tan C 2 . tan A 2 = 1. Solution a) cot A 2 . cot B 2 . cot C 2 = cot A 2 + cot B 2 + cot C 2 Here, A + B + C = πc or, A 2 + B 2 = π 2 – C 2 cot A 2 + B 2 = cot π 2 – C 2 or, cot A 2 . cot B 2 – 1 cot B 2 + cot A 2 = tan C 2 or, cot A 2 . cot B 2 – 1 cot B 2 + cot A 2 = 1 cot C 2 cot A 2 . cot B 2 . cot C 2 = cot A 2 + cot B 2 + cot C 2 proved. b) tan A 2 . tan B 2 + tan B 2 . tan C 2 + tan C 2 . tan A 2 = 1. Solution Here, A + B + C = πc or, A 2 + B 2 = π 2 – C 2 tan A 2 + B 2 = tan π 2 – C 2 or, tan A 2 + tan B 2 1 – tan A 2 . tan B 2 = cot C 2 or, tan A 2 + tan B 2 1 – tan A 2 . tan B 2 = 1 tan C 2 tan A 2 . tan B 2 + tan B 2 . tan C 2 + tan C 2 . tan A 2 = 1 proved. 3. If A + B + C = π, prove that: a) cos2A + cos2B – cos2C = 1 – 4 sinA . sinB . cosC Solution
Vedanta Optional Mathematics Teacher's Guide ~ 10 299 Here, A + B + C = π or, A + B = π – C cos(A + B) = cos(π – C) = –cosC LHS = cos2A + cos2B – cos2C = 2 cos 2A + 2B 2 . cos 2A – 2B 2 –(2 cos2 C – 1) = 2 cos(A + B) . cos(A – B) – 2 cos2 C + 1 = 2(–cosC) . cos(A – B) – 2 cos2 C + 1 = 1 – 2 cosC [cos(A – B) + cosC] = 1 – 2 cosC [cos(A – B – cos(A + B)] = 1 – 2 cosC . 2 sinA . sinB = 1 – 4 sinA . sinB . cosC = RHS proved. 4. If A + B + C = pc , prove the following: a. SinA + sinB – sinC = 4 sinA 2 . sinB 2 . cos C 2 Solution Here, A + B + C = pc A 2 + B 2 = pc 2 – C 2 ∴ sin A 2 – B 2 = sin pc 2 – C 2 = cos C 2 and cos A 2 – B 2 = cos pc 2 – C 2 = sinC 2 Here, LHS = sinA + sinB – sinC = 2sin A + B 2 . cos A – B 2 – 2sinC 2 . cos C 2 = 2cosC 2 . cos A – B 2 . 2sinC 2 . cos C 2 = 2cosC 2 cos A – B 2 – sinC 2 = 2cosC 2 cos A – B 2 – cos A + B 2 = 2cosC 2 . 2sinA 2 . sinB 2 = 4sinA 2 . sinB 2 . cos C 2
300 Vedanta Optional Mathematics Teacher's Guide ~ 10 = RHS proved. b. –SinA + sinB + sinC = 4cosA 2 . sinB 2 . sinC Solution Here, A + B + C = pc A 2 + B 2 = pc 2 – C 2 ∴ sin A 2 + B 2 = sin pc 2 – C 2 = cos C 2 and cos A 2 + B 2 = sinC 2 LHS = sinB + sinC – sinA = 2sin B + C 2 . cos B – C 2 – 2sinA 2 . cos A 2 = 2cosA 2 cos B – C 2 – sinA 2 = 2cosA 2 cos B – C 2 – cos B + C 2 = 2cosA 2 2sinB 2 . sinC 2 = 4cosA 2 . sinB 2 . sinC 2 = RHS proved. 5. If A + B + C = pc , prove that a. cos2A + cos2B – cos2C = 1 – 4sinA . sinB . cosC Solution Here, A + B + C = p A + B = p – C ∴ cos(A + B) = cos(p – C) = –cosC LHS = cos2A + cos2B – cos2C = 2cos 2A + 2B 2 . cos 2A – 2B 2 – (2cos2 C – 1) = 2cos(A + B) . cos(A – B) – 2cos2 C + 1 = –2cosC . cos(A – B) – 2cos2 C + 1 = 1 – 2cosC[cos(A – B) + cosC] = 1 – 2cosC[cos(A – B) – cos(A + B)] = 1 – 2cosC . 2sinA . sinB = 1 – 4sinA . sinB . cosC = RHS proved.