![Trigonometric Ratios of Compound Angles Trigonometric Ratios of Compound Angles](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhtv2p2flxrSR7UVaKS3625DTnRT2bWjOYAQxHUd1u9YOAWS6UhpbLGuXDcPnXEfWH7T5QI6aVaSRXm6IsXnQAEOwQC9yscW6x3U3KDx2JPs7Ywz0LVXxbhD7QpZyGrzQm-0i7F7cvj595x/s16000/Trigonometric+Ratios+of+Compound+Angles.png)
Trigonometric Ratios of Compound Angles
The sum or difference of two or more
angles is said to be a compound angle.
If A and B are two angles then A + B
or A - B are known as the compound angles.
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We express the trigonometrical ratios of compound angle A + B and A - B in terms of trigonometrical ratios of the angles A and B.
The trigonometrical ratios of the angles
A + B and A - B are known as the addition
and subtraction formula
respectively.
Trigonometric Ratios of Compound Angle A + B (Addition Formula)
Let a revolving line start from OX and
trace out an angle XOY = A and revolve further through an angle YOZ = B.
Therefore, ∠XOZ = A + B.
![Trigonometric Ratios of Compound Angle A + B (Addition Formula) Figure: Trigonometric Ratios of Compound Angle A + B (Addition Formula) Figure:](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3uqDzqSDAMUS7drHeEqxeJbKJ6co2je4BP29f5IRtZKAt0WRXj4h0d-7n3mENK8CE3TpasxrazSNB9AM5tstWiY-FcG3CKYnjS3_pAvClIpKcjv8l0MeYEvnJnwoEvYjai4lL9BaFI2kw/s16000/trigonometric+ratios+of+A+%252B+B+figure.png)
Let, P be any point on OZ. Draw PM
perpendicular to OX and PN perpendicular to OY. From N draw NQ perpendicular to
OX and NR perpendicular to PM.
Here,
∠RPN = 90° - ∠PNR [∵ NR⊥MP]
= ∠RNO [∵ PN⊥OY]
= ∠NOQ [∵ RN ∥ OX]
= A
Again,
RMQN is a rectangle. So, MR = QN and RN = MQ.
Now
from right angle ΔOMP,
![Trigonometric Ratios of Compound Angle A + B (Addition Formula) sin(A + B) = "MP" /"OP" = "MR + RP" /"OP" = "QN + RP" /"OP" = "QN" /"OP" +"RP" /"OP" = "QN" /"ON" . "ON" /"OP" + "RP" /"NP" . "NP" /"OP" = sinA cosB + cosA sinB cos(A + B) = "OM" /"OP" = "OQ – MQ" /"OP" = "OQ – RN" /"OP" = "OQ" /"OP" – "RN" /"OP" = "OQ" /"ON" . "ON" /"OP" – "RN" /"NP" . "NP" /"OP" = cosA cosB – sinA sinB Hence, the sine formula and cos formula of compound angle (A + B) are: sin(A + B) = sinA cosB + cosA sinB cos(A + B) = cosA cosB - sinA sinB](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9z5XK6pun9uvbTRy3tPd7mBsAtMHF4brbfa0DOp9w-tUJbaRvALzbDwU-UybIkM1WvZFmiEFUKlybgewb_bIdIlQO-p7vo74vk_lOFzdRdWOaXnvmhxb33YYfDhDca0MCak3WXqOs99YY/s16000/trigonometric+ratios+of+A+%252B+B.png)
Tangent Formula of Compound Angle (A + B)
![Tangent Formula of Compound Angle (A + B) tan(A + B) = "sin(A + B)" /"cos(A + B)" = "sinA cosB + cosA sinB" /("cosA cosB " -" sinA sinB" ) = ("sinA cosB + cosA sinB" /"cosA cosB" )/(("cosA cosB " -" sinA sinB" )/"cosA cosB" ) = ("sinA cosB" /"cosA cosB" + "cosA sinB" /"cosA cosB" )/("cosA cosB" /"cosA cosB" - "sinA sinB" /"cosA cosB" ) = "tanA + tanB" /"1 - tanA tanB" Hence, the tangent formula of compound angle (A + B) is, tan(A + B) = "tanA + tanB" /("1 " -" tanA tanB" )](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizFSGQk-qll1VFX0BO2ggYCLg6qcMBR96QK5jLjsqhLImm-TB7GM83uAOwGgaFYbqqdfx_jEDUNao9QFukjfDHbBaR-F1zle0kuzj4Wl6YdXHzty6m5MeN3J7oUDs4joW-XoxhHtAcpSxM/s16000/Tangent+Formula+of+Compound+Angle+%2528A+%252B+B%2529.png)
Cotangent Formula of Compound Angle (A + B)
![Cotangent Formula of Compound Angle (A + B) cot(A + B) = "cos(A + B)" /"sin(A + B)" = ("cosA cosB" - "sinA sinB" )/("sinA cosB" + "cosA sinB" ) = (("cosA cosB" - "sinA sinB" )/"sinA sinB" )/(("sinA cosB" + "cosA sinB" )/"sinA sinB" ) = ("cosA cosB" /"sinA sinB" - "sinA sinB" /"sinA sinB" )/("sinA cosB" /"sinA sinB" + "cosA sinB" /"sinA sinB" ) = ("cotA cotB" - "1" )/"cotB + cotA" Hence, the cotangent formula of compound angle (A + B) is, cot(A + B) = ("cotA cotB" - "1" )/"cotB + cotA"](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjm1jbYaSXOKw-oUJ6xRgHiarKkHBd6k3ZeTq4HLW2YKD8KhmNI1N7-gKhn3l0X5KlsUgfcFYd3Ri6b4hKT2GOUnEIGQP45NAUDiRf8jsS33UVinniDJRDny3HbzaiZvx-X4Q5TSUVzp4gy/s16000/Cotangent+Formula+of+Compound+Angle+%2528A+%252B+B%2529.png)
Trigonometric Ratios of Compound Angle A – B (Subtraction Formula)
Let a revolving line start from OX and
trace out an angle XOY = A and then revolve back through an angle YOZ = B.
Therefore, ∠XOZ = A - B.
![Trigonometric Ratios of Compound Angle A – B (Subtraction Formula) Figure: Trigonometric Ratios of Compound Angle A – B (Subtraction Formula) Figure:](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhUonr4XSeq_1Kq7If6LiBd2CM8tJtSbIM2PJ_uJpiVKtduHHu79fxLEBoGLPAsflwNXsoV-q3QelAXghJWzlTu53PuaMlIru8uVjcSpbYG-wBJki-gM-GMj2yHymgisevczn4GeDwP7xYC/s16000/trigonometric+ratios+of+A+-+B+figure.png)
Let, P be any point on the line OZ. Draw
PM perpendicular to OX and PN perpendicular to OY. From N draw NQ perpendicular
to OX and NR perpendicular to MP produced.
Here,
∠RPN = 90° - ∠PNR [∵ PR⊥NR]
= ∠RNY [∵ PN⊥OY]
= ∠XOY [∵ OX ∥ NR]
= A
Again,
QMRN is a rectangle. So, QN = MR and QM = NR.
Now
from right angle ΔOMP,
![Trigonometric Ratios of Compound Angle A – B (Subtraction Formula) sin(A – B) = "MP" /"OP" = "MR – PR" /"OP" = "QN – RP" /"OP" = "QN" /"OP" – "PR" /"OP" = "QN" /"ON" . "ON" /"OP" – "PR" /"NP" . "NP" /"OP" = sinA cosB – cosA sinB cos(A – B) = "OM" /"OP" = "OQ + QM" /"OP" = "OQ + NR" /"OP" = "OQ" /"OP" + "NR" /"OP" = "OQ" /"ON" . "ON" /"OP" + "NR" /"NP" . "NP" /"OP" = cosA cosB + sinA sinB Hence, the sine formula and cos formula of compound angle (A - B) are: sin(A – B) = sinA cosB – cosA sinB cos(A – B) = cosA cosB + sinA sinB](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgqL_0YwGAr3jMr65AoLCmVO77yMRlE3aEe9c5DLMhMGfafsvbxE_l4lHbr4QvVlxPrDZK2fEw-KHBEcL9qjHSIi9SXp8UAIHFVpVno33-BUUlMH5yIiBKJ-IaA50ux0uPs3I26fNh9V0_Z/s16000/trigonometric+ratios+of+A+-+B.png)
Tangent Formula of Compound Angle (A – B)
![Tangent Formula of Compound Angle (A – B) tan(A – B) = ("sin(A " -" B)" )/("cos(A " -" B)" ) = ("sinA cosB " -" cosA sinB" )/"cosA cosB + sinA sinB" = (("sinA cosB " -" cosA sinB" )/"cosA cosB" )/("cosA cosB + sinA sinB" /"cosA cosB" ) = ("sinA cosB" /"cosA cosB" - "cosA sinB" /"cosA cosB" )/("cosA cosB" /"cosA cosB" + "sinA sinB" /"cosA cosB" ) = ("tanA " -" tanB" )/"1 + tanA tanB" Hence, the tangent formula of compound angle (A – B) is, tan(A – B) = ("tanA " -" tanB" )/"1 + tanA tanB"](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgswa82EaIR9FOvPpkBZNsZynv9_C5DK9HGq1tZcy1fnE74HGXey5fGe35ae70vcIkL0-R4MOCYLFnpMPKWojC71nEey6av3x869MFyhPu8NpmJK7Hm16_MKSQCMzm0UJQkRElFnC0nYPxK/s16000/Tangent+Formula+of+Compound+Angle+%2528A+-+B%2529.png)
Cotangent Formula of Compound Angle (A - B)
![Cotangent Formula of Compound Angle (A - B) cot(A – B) = ("cos(A" - "B)" )/("sin(A" - "B)" ) = "cosA cosB + sinA sinB" /("sinA cosB" - "cosA sinB" ) = ("cosA cosB + sinA sinB" /"sinA sinB" )/(("sinA cosB" - "cosA sinB" )/"sinA sinB" ) = ("cosA cosB" /"sinA sinB" + "sinA sinB" /"sinA sinB" )/("sinA cosB" /"sinA sinB" - "cosA sinB" /"sinA sinB" ) = "cotA cotB + 1" /("cotB" - "cotA" ) Hence, the cotangent formula of compound angle (A - B) is, cot(A - B) = "cotA cotB + 1" /("cotB" - "cotA" )](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgAxYqmQUyO8g2RuCd1ARoWVMz3QiOquMgfSwFYz2K3FHBtnqOjLlhvccZQOdXrG-ZB4ZutK_O8G-yQWdLvp6pHT4Fg-KqafnQo5hlvNbfhfaKYZeWrnS_wcOFAZyI7Yn7NJveaBdoqUnaH/s16000/Cotangent+Formula+of+Compound+Angle+%2528A+-+B%2529.png)
List of Trigonometric Formula for Compound Angles A + B and A – B
![List of Trigonometric Formula for Compound Angles A + B and A – B 1. sin(A + B) = sinA cosB + cosA sinB 2. cos(A + B) = cosA cosB – sinA sinB 3. tan(A + B) = "tanA + tanB" /("1 " -" tanA tanB" ) 4. cot(A + B) = ("cotA cotB" - "1" )/"cotB + cotA" 5. sin(A – B) = sinA cosB – cosA sinB 6. cos(A – B) = cosA cosB + sinA sinB 7. tan(A – B) = ("tanA " -" tanB" )/"1 + tanA tanB" 8. cot(A - B) = "cotA cotB + 1" /("cotB" - "cotA" )](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj0sJyFsAeWy-QqDquUvj2J0KNKPiJd7Zxt0SHOPYgK4_L5-beeOiW_ajl54zU3IazIbRLG0ZhqTrEDQGuiq_JyfL2cE_PMizk-p6t8Nmtlpsk8b7dE2mp0VnHy4Kz0qWPWhKPv2mU1aKpy/s16000/List+of+Trigonometric+Formula+for+Compound+Angles.png)
Worked Out Examples
![Example 1: Example 1: Find the value of: (a) sin75° (b) tan15° Solution: sin75° = sin(45° + 30°) = sin45° cos30° + cos45° sin30° = "1" /√("2" ) . √("3" )/"2" + "1" /√("2" ) . "1" /"2" = √("3" )/("2" √("2" )) + "1" /("2" √("2" )) = (√("3" ) + "1" )/("2" √("2" )) tan15° = tan(45° - 30°) = ("tan45°" - "tan30°" )/"1 + tan45° tan30°" = ("1" - "1" /√("3" ))/("1" + "1" ."1" /√("3" ) ) = (√("3" ) - "1" )/(√("3" ) + "1" ) = (√("3" ) - "1" )/(√("3" ) + "1" ) × (√("3" ) - "1" )/(√("3" ) - "1" ) = (√("3" ) - "1" )^"2" /((√("3" ))^"2" - "1" ^"2" ) = ((√("3" ))^"2" – "2" .√("3" )."1" + "1" ^"2" )/("3 " - "1" ) = ("4" – "2" √("3" ) )/"2" = "2" – √("3" )](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiV4-rbHDhG-HVQxFeUm3cFMzXdZEYecrj-H7OvqFyvDgyaQe2CyvOhGU94TiZ_HzQTMffVGP8CXGfbg9TxHxWperH6Y9FgdPXYaMKd8Kz1Obv79jiqpG3OBjyqIsA0umgmw48U9G1dy40M/s16000/example+1.png)
![Example 2: Example 2: If sinA = "3" /"5" and cosB = "5" /"13" , find the value of sin(A + B) Solution: Here, sinA = "3" /"5" and cosB = 5/13 ∴ cosA = √("1" – 〖"sin" 〗^"2" "A" ) = √("1" – "9" /"25" ) = √("16" /"25" ) = "4" /"5" ∴ sinB = √("1" – 〖"cos" 〗^"2" "B" ) = √("1" – "25" /"169" ) = √("144" /"169" ) = "12" /"13" Now, sin(A + B) = sinA cosB + cosA sinB = "3" /"5" . "5" /"13" + "4" /"5" . "12" /"13" = "15" /"65" + "48" /"65" = "63" /"65"](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhoZpQHZ7RwNyXbR-LToA2eoP2hBPbDAw_RyoTH2qKGtngLqRX8LDW-xao2jySRpfbgQL8CTr2uIzcKgxQ_tigfMYXvsB3bI2q8B7KdOwe6wi4IN6bdnsSlnnlsgts0tmmvAhskLhJGvFkz/s16000/example+2.png)
![Example 3: Example 3: If sinA = "1" /√("10" ) and sinB = "1" /√("5" ) , show that A + B = "Ï€" /"4" Solution: Here, sinA = "1" /√("10" ) and sinB = "1" /√("5" ) ∴ cosA = √("1" – 〖"sin" 〗^"2" "A" ) = √("1" – "1" /"10" ) = √("9" /"10" ) = "3" /√("10" ) ∴ cosB = = √("1" – 〖"sin" 〗^"2" "B" ) = √("1" – "1" /"5" ) = √("4" /"5" ) = "2" /√("5" ) Now, sin(A + B) = sinA cosB + cosA sinB = "1" /√("10" ) . "2" /√("5" ) + "3" /√("10" ) . "1" /√("5" ) = "2" /√("50" ) + "3" /√("50" ) = "5" /√("50" ) = "1" /√("2" ) = sin45° = sin "Ï€" /"4" ∴ A + B = "Ï€" /"4" . Proved.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEglVuysrSMiCt0Hfg9lhyphenhyphen3mLPyOVwtrspfT8rQyMXxUZh1KHkmosZiUMgYyv57kxt0eaPyGaKy4la7nyIQTw-kS-jsernZIFnEI-7rVCbdmhOsHx8NbC5ZPsHurjpWD21VAf9KjMEgFvu55/s16000/example+3.png)
![Example 4: Example 4: Prove that: ("cos14°" - "sin14°" )/"cos14° + sin14°" = "cot59°" Solution: RHS = cot59° = cot(45° + 14°) = ("cot45° cot14°" -" 1" )/"cot45° + cot14°" = ("cot14°" -" 1" )/"1 + cot14°" = ("cos14°" /"sin14°" " " -" 1" )/("1 + " "cos14°" /"sin14°" ) = (("cos14° " -" sin14°" )/"sin14°" )/("cos14° + sin14°" /"sin14°" ) = ("cos14° " -" sin14°" )/"cos14° + sin14°" = LHS. Proved.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj9JHaImdjS0L_AwuRj6rUPMbkSZh6caQE7nDzcY-qLAe_OV5TBUrwfGg7nqXGM_7v99WUPmMWc31lIGHundpcuQXD5FUgbmQmbSIUQOHXPf1t7q_FdmTT4AX-J9tu6_QAeYOYCEggxrjtk/s16000/example+4.png)
![Example 5: Example 5: Prove that: sin35° + cos35° = √("2" ) cos10° Solution: Here, RHS = √("2" ) cos10° = √("2" ) cos(45° - 35°) = √("2" ) (cos45° cos35° + sin45° sin35°) = √("2" ) ("1" /√("2" ) ."cos35°" + "1" /√("2" ) ."sin35°" ) = cos35° + sin35° = LHS. Proved.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi6eGWE7rlwNHgZE5JXCLNTJTc9vbJrrE1lQ9aj-3Fxd-hQu3cimCfBOWvW-43_faWmuj_rqVERT11kJRM6rVBOXFQS5bnGjgTItzKCDV3dHYqxHphLLzAonrGqwgqdf6iHbwKoU_HhlBnP/s16000/example+5.png)
![Example 6: Example 6: Prove that: tan10° + tan35° + tan10° tan35° = 1 Solution: Here, 10° + 35° = 45° So, tan(10° + 35°) = tan45° or, "tan10° + tan35°" /("1 " –" tan10° tan35°" ) = 1 or, tan10° + tan35° = 1 – tan10° tan35° or, tan10° + tan35° + tan10° tan35° = 1 Hence proved.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi-5cY2Ae9RpOmzb5O-m5q4M4hRaEwMvfUlb1gj3foXbo8clQu3OYtWIAPnLLwMJkXbhUcGDW0ohuYCMpNYi-4a7XzrOxwOA7-KuWH8JRVxYyH5f02Lwez3AVhG0w-oP4v2zIAYJpN-YbZO/s16000/example+6.png)
![Example 7: Example 7: If A, B and C are the angles of a triangle then prove that: tanA + tanB + tanC = tanA tanB tanC. Solution: Here, A, B and C are angles of a triangle, So, A + B + C = 180° or, A + B = 180° - C or, tan(A + B) = tan(180° - C) or, "tanA + tanB" /("1 " –" tanA tanB" ) = - tanC or, tanA + tanB = - tanC + tanA tanB tanC or, tanA + tanB + tanC = tanA tanB tanC Hence proved.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0yljnqAocp6g0V0lUfP82e0t8CsL1XFHEIfbchCy8zgTxDk9xL42B8DbCS9rjqXaPmR6oFdOCDvf4CsLPvhWeHEd9WLwXGezrsQahamX3IkeGnw48gYLjJoHoZZNp4WEX0aDYvwDJ96GD/s16000/example+7.png)
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