DOUBLE ANGLE IDENTITIES

Double angle identities are a special case of the sum identities. That is, when the two angles are equal, the sum identities are reduced to double angle identities. They are useful in solving trigonometric equations and also in the verification of trigonometric identities.

Further double angle identities can be used to derive the reduction identities (power reducing identities). Also double angle identities are used to find maximum or minimum values of trigonometric expressions.

Identity 1 : sin2A = 2sinAcosA

Proof : 

We know that

sin(α + β) = sinα cosβ + cosα sinβ

Taking α = β = A, we have 

sin(A + A) = sinA cosA + cosA sinA

sin2A = 2sinA cosA

Identity 2 : cos2A = cos²A - sin²A

Proof : 

We know that

cos(α + β) = cosα cosβ - sinα sinβ

Taking α = β = A, we have 

cos(A + A) = cosA cosA + sinA sinA

cos2A = cos²A - sin²A

From cos2A = cos²A - sin²A, 

cos2A = cos²A - (1 - cos²A)

= cos²A - 1 + cos²A

cos2A = 2cos²A - 1

And also, 

cos2A = (1 - sin²A) - sin²A

= 1 - sin²A - sin²A

cos2A = 1 - 2sin²A

Identity 3 : tan2A = 2tanA/(1 - tan²A)

Proof : 

We know that

tan(α + β) = (tanα + tanβ)/(1 - tanα tanβ)

Taking α = β = A, we have 

tan(A + A) = (tanA + tanA)/(1 - tanA tanA)

tan2A = 2tanA/(1 - tan²A)

Identity 4 : sin2A = 2tanA/(1 + tan²A)

Proof : 

We know that

sin2A = 2sinA cosA

= (2sinA cosA)/1

= (2sinA cosA) / (sin²A + cos²A)

Divide both numerator and denominator by cos²A.

= [(2sinA cosA)/cos²A] / [(sin²A + cos²A)/cos²A]

= [2sinA/cosA] / [sin²A/cos²A + cos²A/cos²A]

= 2tanA/(tan²A + 1)

sin2A = 2tanA/(1 + tan²A)

Identity 5 : cos2A = (1 - tan²A)/(1 + tan²A)

Proof : 

We know that

cos2A = cos²A - sin²A

= (cos²A - sin²A)/1

= (cos²A - sin²A)/(cos²A + sin²A)

Divide both numerator and denominator by cos²A.

= [(cos²A - sin²A)/cos²A] / [(sin²A + cos²A)/cos²A]

cos2A = (1 - tan²A)/(1 + tan²A)

Practice Problems

Problem 1 : 

Find the value of sin2θ, when sinθ = 12/13, θ lies in the first quadrant. 

Solution : 

Using a right angle, we can find that cosθ = 5/13. 

sin2θ = 2sinθcosθ

= 2(12/13)(5/13)

= 120/169

Problem 2 : 

Find the value of cos2A, when tanA = 16/63, A lies in the first quadrant. 

Solution : 

Using a right angle, we can find that sinA = 16/65. 

cos2A = 1 - 2sin²A

= 1 - 2(16/65)²

= 1 - 2(256/4225)

= 1 - 512/4225)

= (4225 - 512)/4225

= 3713/4225

Problem 3 : 

Prove that :

cos⁴A - sin⁴A = cos2A

Solution : 

cos⁴A - sin⁴A : 

= (cos²A)² - (sin²A)²

= (cos²A + sin²A)(cos²A - sin²A)

= 1(cos2A)

= cos2A

Problem 4 : 

Prove that : 

(sinA + cosA)² - (sinA - cosA)² = 2sin2A

Solution : 

(sinA + cosA)² - (sinA - cosA)² :

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

= sin²A + cos²A + 2sinAcosA - sin²A - cos²A + 2sinAcosA

= 4sinAcosA

= 2 ⋅ 2sinAcosA

= 2sin2A

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