Trigonometric Ratios of Compound Angles

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Trigonometric Ratios of Compound Angles

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:

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,


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

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" )


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"

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:

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,


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

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"


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" )

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" )


Worked Out Examples


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" )


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"
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.


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.


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.


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.


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.

Do you have any questions regarding the trigonometric ratios of compound angles?

 

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About author: Min Rajbanshi

I am a school math teacher and a blogger since 2012. I do blogging on a variety of topics including math. Over the years, I've explored and written about a wide range of topics, with a particular emphasis on mathematics. My passion for education extends beyond the classroom, as I enjoy sharing knowledge and insights on various subjects through my blog.

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