Trigonometric Identities

Trigonometric Identities

Trigonometric Identities

A mathematical statement which is always true for every condition or for any values of its variable is known as an identity. The mathematical statement which is only true for particular values of variable is called an equation.


Rules for proving trigonometric identities:

(a) Start with left hand side (L.H.S.) and reduce it to the right hand side (R.H.S.) if left hand side is complicated.

(b) Start with right hand side and reduce it to the left hand side if right hand side is complicated.

(c) If left hand side and right hand side both are equivalent, reduce both of them into lowest term.

(d)If complicated, transpose or apply the method of cross multiplication to change the form of identity. Then, prove that new L.H.S. = new R.H.S.

(e) We apply the operations of addition, subtraction, multiplication and division similar as in case of the simplification of numbers or variables.

(f)  We use almost all the algebraic formula (given below) in case of the trigonometry also.

-      (a + b)2 = a2 + 2ab + b2

-      (a – b)2 = a2 – 2ab + b2

-      a2 – b2 = (a + b)(a – b)

-      (a + b)3 = a3 + 3a2b + 3ab2 + b3

-      (a – b)3 = a3 – 3a2b + 3ab2 – b3

-      a3 + b3 = (a + b)(a2 – ab + b2)

-      a3 – b3 = (a – b)(a2 + ab + b2) etc.

 

Basic Trigonometric Formulae:

1.   Reciprocal Formulae:

sinθ . cosecθ = 1

sinθ = 1/cosecθ

cosecθ = 1/sinθ

cosθ . secθ = 1

cosθ = 1/secθ

secθ = 1/cosθ

tanθ . cotθ = 1

tanθ = 1/cotθ

cotθ = 1/tanθ


2.   Quotient Formulae:

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

tanθ = secθ/cosecθ

cotθ = cosecθ/secθ

 

3.   Pythagorean Formulae

sin2θ + cos2θ = 1

sec2θ – tan2θ = 1

cosec2θ – cot2θ = 1

sin2θ = 1 – cos2θ

sec2θ = 1 + tan2θ

cosec2θ = 1 + cot2θ

cos2θ = 1 – sin2θ

tan2θ = sec2θ – 1

cot2θ = cosec2θ – 1

 

 

Proving Trigonometric Identities:

Worked out Examples:

Example 1: Prove that: (sinA + cosA)2 – (sinA – cosA)2 = 4sinA cosA

Solution: L.H.S. = (sinA + cosA)2 – (sinA – cosA)2

                         = sin2A + 2sinAcosA + cos2A – (sin2A – 2sinAcosA + cos2A)

                         = sin2A + cos2A + 2sinAcosA – (sin2A + cos2A – 2sinAcosA)

                         = 1 + 2sinAcosA – (1 – 2sinAcosA)

                         = 1 + 2sinAcosA – 1 + 2sinAcosA

                         = 4sinAcosA

                         = R.H.S. proved.

 

Example 2: Prove that: sec2θ (1 – cos2θ) =  tan2θ

Solution: L.H.S. = sec2θ (1 – cos2θ)

                          = 1/cos2θ . sin2θ     [ sec2θ = 1/cos2θ and 1 – cos2θ = sin2θ]

                          = sin2θ/cos2θ

                          = tan2θ

                          = R.H.S. proved.

 

Example 3: Prove that: sin2θ . tan2θ = tan2θ – sin2θ

Solution: L.H.S. = sin2θ . tan2θ

                          = sin2θ . (sec2θ – 1)

                          = sin2θ . sec2θ – sin2θ

                          = sin2θ . 1/cos2θ – sin2θ

                          = sin2θ/cos2θ – sin2θ

                          = tan2θ – sin2θ

                          = R.H.S. proved.

 

Example 4: Prove that: tan4A + tan2A = sec4A – sec2A

Solution: L.H.S. = tan4A + tan2A

                          = tan2A (tan2A + 1)

                          = (sec2A – 1) sec2A

                          = sec4A – sec2A

                          = R.H.S. proved.

 

Example 5: Prove that: 1/(tanθ + cotθ) = sinθ .cosθ Solution: L.H.S. = 1/(tanθ + cotθ) = 1/(sinθ/cosθ  + cosθ/sinθ) = 1/((〖sin〗^2 θ + 〖cos〗^2 θ)/(sinθ .  cosθ)) = 1/(1/(sinθ .  cosθ)) = sinθ .cosθ = R.H.S. proved.


Example 6: Prove that: (1 -〖 sin〗^4 A)/(〖cos〗^4 A)=2〖sec〗^2 A-1 Solution: L.H.S. = (1 -〖 sin〗^4 A)/(〖cos〗^4 A) = (1^2  – 〖(〖 sin〗^2 A)〗^2)/〖(〖cos〗^2 A)〗^2 = ((1 +〖 sin〗^2 A)(1 - 〖 sin〗^2 A))/(〖cos〗^2 A .〖cos〗^2 A) = ((1 + 〖 sin〗^2 A) 〖 .cos〗^2 A)/(〖cos〗^2 A .〖cos〗^2 A) = ((1 + 〖 sin〗^2 A))/(〖cos〗^2 A) = 1/(〖cos〗^2 A)+ (〖 sin〗^2 A)/(〖cos〗^2 A) = 〖sec〗^2 A+〖tan〗^2 A = 〖sec〗^2 A+〖sec〗^2 A-1 = 〖2sec〗^2 A-1 = R.H.S. proved.


Example 7: Prove that: (〖cos〗^2 A)/(1-tanA)+(〖sin〗^2 A)/(1-cotA)=1+cosA sinA Solution: L.H.S. = (〖cos〗^2 A)/(1-tanA)+(〖sin〗^2 A)/(1-cotA) = (〖cos〗^2 A)/(1- sinA/cosA)+(〖sin〗^2 A)/(1- cosA/sinA) = (〖cos〗^2 A)/((cosA- sinA)/cosA)+(〖sin〗^2 A)/((sinA- cosA)/sinA) = (〖cos〗^3 A)/(cosA- sinA)  -  (〖sin〗^3 A)/(cosA- sinA) = (〖cos〗^3 A - 〖sin〗^3 A )/(cosA- sinA) = (〖(cosA - sinA)(cos〗^2 A + cosA sinA + 〖sin〗^2 A) )/(cosA - sinA) = 〖cos〗^2 A+〖cosA sinA+sin〗^2 A = 〖cos〗^2 A 〖+ sin〗^2 A+cosA sinA = 1+cosA sinA = R.H.S. proved.


Example 8: Prove that: (tanA + secA - 1)/(tanA – secA + 1)=(1 + sinA)/cosA Solution: L.H.S. = (tanA + secA - 1)/(tanA – secA + 1) = ((tanA + secA) – 〖(sec〗^2 A - 〖tan〗^2 A))/(tanA – secA + 1)       [∵ 1 = sec2A – tan2A] = ((secA + tanA)  – (secA + tanA)(secA - tanA))/(tanA – secA + 1) = ((secA + tanA)[1 – (secA - tanA)])/(tanA – secA + 1) = ((secA + tanA)(1 – secA + tanA])/(1 – secA + tanA) = secA + tanA = 1/cosA+ sinA/cosA = (1 + sinA)/cosA = R.H.S. proved.


Example 9: Prove that: 1/(cosecA + cotA)-  1/sinA=1/sinA-  1/(cosecA - cotA) Solution: L.H.S. = 1/(cosecA + cotA)-  1/sinA = ( co〖sec〗^2 A - 〖cot〗^2 A)/(cosecA + cotA)- cosecA       [∵ 1 = cosec2A – cot2A] = ( (cosecA + cotA)(cosecA - cotA))/(cosecA + cotA)- cosecA = cosecA-cotA-cosecA = - cotA.    R.H.S. = 1/sinA  -1/(cosecA - cotA) = cosecA-( co〖sec〗^2 A - 〖cot〗^2 A)/(cosecA - cotA)       [∵ 1 = cosec2A – cot2A] = cosecA-( (cosecA + cotA)(cosecA - cotA))/(cosecA - cotA) = cosecA-(cosecA+cotA) = cosecA-cosecA-cotA = - cotA.  Hence, L.H.S. = R.H.S. proved.


Example 10: Prove that: (3-4〖sin〗^2 A)(〖sec〗^2 A-4〖tan〗^2 A)=(3- 〖tan〗^2 A)(1-4〖sin〗^2 A) Solution: L.H.S. = (3-4〖sin〗^2 A)(〖sec〗^2 A-4〖tan〗^2 A) = (3-4〖sin〗^2 A)(1/(〖cos〗^2 A)- (4〖sin〗^2 A)/(〖cos〗^2 A)) = (3-4〖sin〗^2 A)((1 - 4〖sin〗^2 A)/(〖cos〗^2 A)) = ((3 - 4〖sin〗^2 A)/(〖cos〗^2 A))(1-4〖sin〗^2 A) = ((3 )/(〖cos〗^2 A)  - (4〖sin〗^2 A)/(〖cos〗^2 A))(1-4〖sin〗^2 A) = (3〖sec〗^2 A-4〖tan〗^2 A)(1-4〖sin〗^2 A) = [3〖(1+tan〗^2 A)-4〖tan〗^2 A](1-4〖sin〗^2 A) = (3+3〖tan〗^2 A-4〖tan〗^2 A)(1-4〖sin〗^2 A) = (3-〖tan〗^2 A)(1-4〖sin〗^2 A) = R.H.S. proved.

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