Trigonometric Ratios of Compound Angles

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     [  NRMP]

               = RNO   [  PNOY]

               = 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     [  PRNR]

               = RNY    [  PNOY]

               = 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 = "π" /"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 "π" /"4"  ∴  A + B = "π" /"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.

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