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Further Trigonometric Identities Exercise 13.1 5. (f) tan 5x = tan 2x, 0 < x < 360 sin 5 x sin 2 x = cos 5 x cos 2 x sin 5x cos 2x = cos 5x sin 2x sin 5x cos 2x cos 5x sin 2x = 0 sin (5x 2x) = 0 sin 3x = 0 0 < 3x < 1080 Basic angle, = 0 3x = 0, 180, 360, 540, 720, 900, 1080 (reject 0 & 1080) x = 60, 120, 180, 240, 300 6. (b) 7. (e) Prove: (g) Prove: tan A + tan B = (h) Prove: tan 2 x tan 1 1 = , 0 < x < 2 1 tan 2 x tan 1 2 1 tan (2x 1) = 1 < 2x 1 < 4 1 2 Basic angle, = 0.4636 2x 1 = 0.4636, + 0.4636, 2 + 0.4636, 3 + 0.4636 2x = 1.4636, 4.6052, 7.7468, 10.888 x = 0.732, 2.30, 3.87, 5.44 sin( A B) sin( A B) = cot A cos( A B) cos( A B) [sin A cos B cos A sin B] [sin A cos B cos A sin B] L.H.S. = [cos A cos B sin A sin B] [cos A cos B sin A sin B] 2 cos A sin B = 2 sin A sin B = cot A = R.H.S. sin( A B ) cos A cos B sin A cos B cos A sin B R.H.S. = cos A cos B sin A cos B cos A sin B = + cos A cos B cos A cos B = tan A + tan B = L.H.S. sin( A B) tan A tan B = cos( A B) 1 tan A tan B L.H.S. = [sin A cos B + cos A sin B] [cos A cos B + sin A sin B] sin A cos B cos A sin B cos A cos B sin A sin B = cos A cos B cos A cos B = [tan A + tan B] [1 + tan A tan B] tan A tan B = = R.H.S. 1 tan A tan B Ex 13: Further Trigonometric Identities Prepared by Mr Ang KS 1 9. (a) (b) (c) 12. 4 3 , sin B = , and A & B in the same quadrant (3rd quadrant) y 3 5 tan A tan 45 tan (A + 45) = 3 A 1 tan A tan 45 4 4 = [ + 1] [1 ] 4 5 3 3 = 7 cos (A B) = cos A cos B + sin A sin B y 3 4 4 3 = ( )( ) + ( )( ) 5 5 5 5 4 B 24 = 3 25 5 sin (B A) = sin B cos A cos B sin A 3 3 4 4 = ( )( ) ( )( ) 5 5 5 5 7 = 25 Given: tan A = x x Show: sin (A + B) sin (A B) = sin2 A sin2 B L.H.S. = [sin A cos B + cos A sin B] [sin A cos B cos A sin B] = (sin A cos B)2 (cos A sin B)2 = sin2 A (1 sin2 B) (1 sin2 A)sin2 B = sin2 A sin2 A sin2 B sin2 B + sin2 Asin2 B = sin2 A sin2 B = R.H.S. (a) sin2 3x = sin2 x, 0 x 180 sin2 3x sin2 x = 0 sin (3x + x) sin (3x x) = 0 sin 4x sin 2x = 0 0 2x 360 or 0 4x 720 sin 4x = 0 or sin 2x = 0 Basic angle, = 0 Basic angle, = 0 4x = 0, 180, 360, 540, 720 2x = 0, 180, 360 x = 0, 45, 90, 135, 180 x = 0, 90, 180 x = 0, 45, 90, 135, 180 (b) Show: sin 15 sin 75 = 1 4 L.H.S. = sin 75 sin 15 = sin (45 + 30) sin (45 30) = sin2 45 sin2 30 1 1 2 =( ) ( )2 2 2 1 1 1 = = = R.H.S. 2 4 4 End of Ex 13.1 Ex 13: Further Trigonometric Identities Prepared by Mr Ang KS 2 Exercise 13.2 2. (e) cos 2y 1 = 3sin y, (1 2sin2 y) 1 = 3sin y 2sin2 y + 3sin y = 0 sin y(2sin y + 3) = 0 0 < y < 360 sin y = 0 sin y = Basic angle, = 0 y = 0, 180, 360 y = 180 3. or 3 (No solution as 1 sin y 1) 2 (reject 0 & 360, since 0 < y < 360) (f) (sin x cos x)2 = 2, 0 < x < 360 sin2 x + cos2 x 2sin x cos x = 2 1 sin 2x = 2 sin 2x = 1 0 < 2x < 720 Basic angle, = 90 2x = 180 + 90, 360 90, 540 + 90, 720 90 x = 135, 315 (b) 5sin x cos2 x + 2cos x = 0, 0 x 2, or 0 2x 4 cos x(5sin x cos x + 2) = 0 1 4 cos x = 0 or 5( sin 2x) + 2 = 0, i.e. sin 2x = 2 5 Basic angle, = Basic angle, = 0.9273 2 x = , 2 2x = + 0.9273, 2 0.9273, 3 + 0.9273, 4 0.9273 2 2 3 x = , , 2.03, 2.68, 5.18, 5.82 2 2 (e) 2 tan 2 tan x = 1 tan2 x, 0 x 2, so 0 2x 4 2 tan x tan 2( )=1 1 tan 2 x 1 tan 2x = = 0.4577 tan 2 Basic angle, = 0.4292 2x = 0.4292, 2 0.4292, 3 0.4292, 4 0.4292 x = 1.36, 2.93, 4.50, 6.07 Ex 13: Further Trigonometric Identities Prepared by Mr Ang KS 3 (f) tan 2x = 3tan x, 2 tan x = 3tan x 1 tan 2 x 0 x 2 2tan x = 3tan x(1 tan2 x) 2tan x 3tan x(1 tan2 x) = 0 tan x [2 3(1 tan2 x)] = 0 tan x(3tan2 x 1) = 0 tan x = 0 tan x = or 1 3 Basic angle, = 0 Basic angle, = 6 , , + , 2 6 6 6 6 5 7 11 x = 0, , 2, , , , 6 6 6 6 x = 0, , 2 6. x= Given: cot = k and is acute. 1 cot = k tan = k (a) (b) (d) sin 2 = 2sin cos = 2( k 2 1 1 1 2k = 2 k 1 sec 2 = 1 cos 2 k )( ) k k2 1 k2 1 cot ( + 45) = 1 tan ( + 45) (c) tan tan 45 =1 = 1 (1 2sin2 ) 1 tan tan 45 1 1 1 =1 k = 1 [1 2( 2 )] 1 k 1 1 k k2 1 2 1 k k 1 =1[ ] =1 k k k2 1 k2 1 1 k k =1[ ] = 2 k k 1 k 1 k 1 k 1 =1 = k 1 k 1 tan tan 2 2 tan tan 2 = then, tan 3 = 2 1 tan tan 2 1 tan 1 1 1 1 2k 2k = 2( ) [1 2 ] =( + 2 ) (1 2 ) k k k k k 1 k 1 k (k 2 1) 2k k 2 1 2 k 2 1 2k 2 = = k k2 k (k 2 1) k (k 2 1) = k2 2 2 k k 1 = k (k 2 1) 3k 3 1 k 3 3k k (k 2 1) = 2k 2 k 1 = 3k 3 1 k (k 2 3) Ex 13: Further Trigonometric Identities Prepared by Mr Ang KS 4 7. (e) (f) 9. sin A 1 = tan A 1 cos A 2 A A A A 2 sin cos 2 sin cos 2 2 2 2 L.H.S. = = A A 1 [2 cos 2 1] 2 cos 2 2 2 A sin 1 2 = = tan A A 2 cos 2 Prove: = R.H.S. 1 tan 2 A = sec 2A 1 tan 2 A sin 2 A sin 2 A L.H.S. = [1 + ] [1 ] cos 2 A cos 2 A cos 2 A sin 2 A cos 2 A sin 2 A = cos 2 A cos 2 A cos 2 A 1 = cos 2 A cos 2 A 1 = = sec 2A = R.H.S. cos 2 A Prove: Given: a = cos x + sin x and b = cos x sin x (a) Prove: cos 2x = ab L.H.S. = cos2 x sin2 x = (cos x + sin x)(cos x sin x) = ab = R.H.S. (b) Prove: sin 2x = a2 1 R.H.S. = (cos x + sin x)2 1 = cos2 x + sin2 x + 2sin x cos x 1 = 1 + sin 2x 1 = L.H.S. (c) Prove: a2 + b2 = 2 R.H.S. = (cos x + sin x)2 + (cos x sin x)2 = (cos2 x + sin2 x + 2sin x cos x) + (cos2 x + sin2 x 2sin x cos x) =1+1 =2 = L.H.S. End of Ex 13.2 Ex 13: Further Trigonometric Identities Prepared by Mr Ang KS 5 Exercise 13.3 3. (d) 3 7cos + 24sin , 0 < < 360 Let 24sin 7cos = Rsin ( ) Then R = 24 2 (7) 2 = 25 and 7 24 so = 16.26 tan = Since 25 25sin ( 16.26) 25 3 25 3 7cos + 24sin 25 + 3 Max. value = 28, when sin ( 16.26) = 1 = 90 + 16.26 = 106.3 Min. value = 22, when sin ( 16.26) = 1 = 270 + 16.26 = 286.3 4. (d) sin x + 2cos x = 2 , Let sin x + 2cos x = Rsin (x + ) 0 < x < 360 Then R = 12 2 2 = 5 and sin x + 2cos x = 2 5 sin (x + 63.43) = 2 2 =2 1 so = 63.43 tan = 2 63.43 < x + 63.43 < 423.23 5 Basic angle, = 39.23 x + 63.43 = 180 39.23, 360 + 39.23 x = 77.3, 335.8 sin (x + 63.43) = (f) cos x + e sin x = 2, 0 < x < 360 Let cos x + e sin x = Rcos (x ) Then R = 2 e 2 = 4.154 cos x + e sin x = 2 4.154cos (x 40.87) = 2 2 cos (x 40.87) = 4.154 Basic angle, = 61.22 x 40.87 = 61.22, 360 61.22 x = 102.1, 339.7 Ex 13: Further Trigonometric Identities Prepared by Mr Ang KS and tan = e so = 40.87 40.87 < x 40.87 < 400.87 6 5. (b) 2cot x = 3 + 2cosec x, 2 cos x 2 =3+ sin x sin x 3 sin x 2 2 cos x = sin x sin x 2cos x = 3sin x + 2 2cos x 3sin x = 2 Let 2cos x 3sin x = Rcos (x + ) Then R = (sin x 0) 2 2 (3) 2 = 13 2cos x 3sin x = 2 13 cos (x + 56.31) = 2 2 cos (x + 56.31) = 13 Basic angle, = 56.31 x + 56.31 = 56.31, 360 56.31 x = 247.4 8. 0 < x < 360 and tan = so = 56.31 56.31 < x + 56.31 < 416.31 Given: ORQ = , so RQX = 90 and PQX = PX In PQX, sin = , so PX = 2sin 2 YR In QRY, cos = , so YR = 8cos 8 So OR = OY + YR = PX + YR = (2sin + 8cos ) cm (a) Let 2sin + 8cos = Rsin ( + ) Then R = (b) 2 2 82 = 68 3 2 and Q 2 cm P X 90 8 cm O Y R 8 =4 2 so = 75.96 tan = Since 68 68 sin ( + 75.96) 68 68 OR 68 For max. OR, sin ( + 75.96) = 1 = 90 75.96 = 14.0 68 sin ( + 75.96) = 6 OR = 6 6 sin ( + 75.96) = Note that: 68 Basic angle, = 46.69 so + 75.96 = 180 46.69 = 57.4 x > 0 x + 75.96 > 75.96 End of Ex 13.3 Ex 13: Further Trigonometric Identities Prepared by Mr Ang KS 7 Exercise 13.4 2. (f) sin 11 5 sin 12 12 = 2cos [( 11 5 11 5 + ) 2] sin [( ) 2] 12 12 12 12 2 sin 3 4 1 1 = 2( )( ) 2 2 = 2cos 3. 4. = 2 2 (e) cos 3x 2sin 2x = cos x, 0 x 180 cos 3x cos x 2sin 2x = 0 3x x 3x x 2sin sin 2sin 2x = 0 2 2 2sin 2x sin x 2sin 2x = 0 2sin 2x (sin x + 1) = 0 sin 2x = 0 or sin x = 1 Basic angle, = 0 Basic angle, = 90 2x = 0, 180, 360 x = 180 + 90, 360 90 x = 0, 90, 180 = 270 (rejected since 0 x 180) x = 0, 90, 180 (f) cos 3x cos x = sin 3x sin x, 0 x 180 3x x 3x x 3x x 3x x 2sin sin = 2cos sin 2 2 2 2 2sin 2x sin x = 2cos 2x sin x 2cos 2x sin x + 2sin 2x sinx = 0 2sin x(cos 2x + sin 2x) = 0 sin x = 0 or sin 2x = cos 2x tan 2x = 1 Basic angle, = 0 Basic angle, = 45 x = 0, 180 2x = 180 45, 360 45 x = 0, 67.5, 157.5, 180 (d) sin 4x + sin 2x + cos 4x = cos 2x, 0x (sin 4x + sin 2x) + (cos 4x + cos 2x) = 0 4x 2x 4x 2x 4x 2x 4x 2x 2sin cos + 2cos cos =0 2 2 2 2 2sin 3x cos x + 2cos 3x cos x = 0 2cos x(sin 3x + cos 3x) = 0 cos x = 0 or sin 3x = cos 3x tan 3x = 1 Basic angle, = Basic angle, = 2 4 x= 3x = , 2 , 3 2 4 4 4 7 11 x = , , , 2 4 12 12 Ex 13: Further Trigonometric Identities Prepared by Mr Ang KS 8 5. 6. cos 6 cos 2 = tan 4 tan 2 cos 6 cos 2 6 2 6 2 6 2 6 2 L.H.S. = [2sin sin ] [2cos cos ] 2 2 2 2 2 sin 4 sin 2 = 2 cos 4 cos 2 = tan 4 tan 2 = R.H.S. (c) Prove: (d) Prove: (f) Prove: cos2 A cos2 B = sin (B A) sin (B + A) L.H.S. = cos2 A cos2 B = (cos A + cos B)(cos A cos B) A B A B A B A B = (2cos cos )(2sin sin ) 2 2 2 2 A B A B A B A B = (2sin cos )(2sin cos ) 2 2 2 2 A B A B = sin 2( ) sin 2( ) 2 2 = sin [(B A)] sin (A + B) = sin (B A) sin (B + A) = R.H.S. sin 6 cos 4 sin 2 = cot 4 cos 6 sin 4 cos 2 sin 6 sin 2 cos 4 L.H.S. = cos 6 cos 2 sin 4 6 2 6 2 6 2 6 2 = [2cos sin + cos 4] [2sin sin sin 4] 2 2 2 2 = [2cos 4 sin 2 + cos 4] [2sin 4 sin 2 sin 4] cos 4 (2 sin 2 1) = sin 4 (2 sin 2 1) = cot 4 = R.H.S. 1 4 x 3x 4 x 3x [2sin cos ] 2 2 2 7x x = sin cos 2 2 7 1 1 a = 1, b = = 3 and c = . 2 2 2 1 1 5x 4x 5x 4x (sin 5x sin 4x) = [2cos sin ] 2 2 2 2 9x x = cos sin 2 2 1 (sin 4x + sin 3x) 2 = Ex 13: Further Trigonometric Identities Prepared by Mr Ang KS 9 7x x 9x x cos + cos sin = sin 4x cos x 2 2 2 2 1 1 L.H.S. = (sin 4x + sin 3x) + (sin 5x sin 4x) 2 2 1 = (sin 5x + sin 3x) 2 1 5 x 3x 5 x 3x = [2sin cos ] 2 2 2 = sin 4x cos x = R.H.S. Show: sin End of Ex 13.4 Ex 13: Further Trigonometric Identities Prepared by Mr Ang KS 10