HALF ANGLE IDENTITIES

Half angle identities are closely related to the double angle identities. We can use half angle identities when we have an angle that is half the size of a special angle.

For example, sin 15° can be computed by writing 

sin15° = sin(30°/2)

Also one can find exact values for some angles using half-angle identities.

Using the following double angle identities, we can derive triple angle identities.

sin2A = 2sinAcosA

cos2A = 2cos²A - 1

cos2A = 1 - 2sin²A

tan2A = 2tanA/(1 - tan²A)

If we take 2A = θ or A = θ/2 in the double angle identities, we get identities in terms of angle θ/2. 

Identity 1 : sin(θ/2) = ± √[(1 - cosθ)/2]

Proof : 

cos2A = 1 - 2sin²A

Taking 2A = θ and A = θ/2,

cosθ = 1 - 2sin²(θ/2)

cosθ + 2sin²(θ/2) = 1

2sin²(θ/2) = 1 - cosθ

sin²(θ/2) = (1 - cosθ)/2

sin(θ/2) = ± √[(1 - cosθ)/2]

Identity 2 : cos(θ/2) = ± √[(1 + cosθ)/2]

Proof : 

cos2A = 2cos²A - 1

Taking 2A = θ and A = θ/2,

cosθ = 2cos²(θ/2) - 1

2cos²(θ/2) = 1 + cosθ

cos²(θ/2) = (1 + cosθ)/2

cos(θ/2) = ± √[(1 + cosθ)/2]

Identity 3 : tan(θ/2) = sinθ/(1 + cosθ) = (1 - cosθ)/sinθ

Proof : 

cos2A = (1 - tan²A)/(1 + tan²A)

Solve for tanA.

cos2A(1 + tan²A) = (1 - tan²A)

cos2A + cos2Atan²A = 1 - tan²A

tan²A + cos2Atan²A = 1 - cos2A

tan²A(1 + cos2A) = 1 - cos2A

tan²A = (1 - cos2A)/(1 + cos2A)

tanA = √[(1 - cos2A)/(1 + cos2A)]

Taking 2A = θ and A = θ/2,

tan(θ/2) = √[(1 - cosθ)/(1 + cosθ)]

In the fraction (1 - cosθ)/(1 + cosθ), multiplying both the numerator and denominator by (1 + cosθ) 

= √[(1² - cos²θ)/(1 + cosθ)²]

= √[(1 - cos²θ)/(1 + cosθ)²]

= √[sin²θ/(1 + cosθ)²]

= sinθ/(1 + cosθ)

And also, in the fraction, (1 - cosθ)/(1 + cosθ), multiplying both the numerator and denominator by (1 - cosθ)

= √[(1 - cosθ)²/(1² - cos²θ)]

= √[(1 - cosθ)²/sin²θ]

= (1 - cosθ)/sinθ

Practice Problems

Problem 1 : 

Find the value of sin(22.5°). 

Solution : 

Write 22.5° in terms of a special angle. 

22.5° = 45°/2

sin(θ/2) = ± √[(1 - cosθ)/2]

Substitute θ = 45°. 

sin(45°/2) = ± √[(1 - cos45°)/2]

sin(22.5°) = ± √[(1 - √2/2)/2]

= ± √[(2 - √2)/4]

= ± √(2 - √2)/2

Since 22.5° lies in the first quadrant sin(22.5°) is positive. 

sin(22.5°) = √(2 - √2)/2

Problem 2 : 

Find the value of cos(15°). 

Solution : 

Write 15° in terms of a special angle. 

15° = 30°/2

cos(θ/2) = ± √[(1 - cosθ)/2]

Substitute θ = 30°. 

cos(30°/2) = ± √[(1 - cos30°)/2]

cos(15°) = ± √[(1 - √3/2)/2]

= ± √[(2 - √3)/4]

= ± √(2 - √3)/2

Since 15° lies in the first quadrant cos(15°) is positive. 

cos(15°) = √(2 - √3)/2

Problem 3 : 

Find the value of tan(157.5°). 

Solution : 

tan(157.5°) = tan(180° - 22.5°)

Since 157.5° lies in the second quadrant, tangent is negative. 

= -tan(22.5°)

= -tan(45°/2)

= -sin(45°)/(1 + cos45°)

= - (√2/2)/(1 - √2/2)

= - (√2/2)/[(2 - √2)/2]

= - √2/(2 - √2)

Problem 4 :

Use half angle identities to find the exact value of

tan 7π/8

Solution : 

Half angle formula for tan θ is 

tan (θ/2) = (1 - cos θ)/sin θ

let θ = 7π/4, θ/2 = 7π/8

tan (7π/8) = (1 - cos 7π/4)/sin 7π/4 ----(1)

Value of sin 7π/4 :

= sin (π + 3π/4)

Lies in third quadrant, then 

sin (π + 3π/4) = -sin (3π/4)

= -sin (π - π/4)

= - sin (π/4)

 = -√2/2

Value of cos 7π/4 :

= cos (π + 3π/4)

Lies in third quadrant, then 

cos (π + 3π/4) = -cos (3π/4)

= -cos (π - π/4)

= cos (π/4)

 = √2/2

By applying these values in (1), we get

tan (7π/8) = (1 - (√2/2)) / (-√2/2)

= (2 - √2)/2 x (-2/√2)

= (-2 + √2)/√2

Multiplying by √2 in both numerator and denominator, we get

= [(√2 - 2)/√2] x = (√2/√2)

= (2 - 2√2)/2

= 1 - √2

so, the value of tan (7π/8) is 1 - √2

Problem 5 :

Use half angel identities to show 

cot² θ/2 = (sec θ + 1) / (sec θ - 1)

Solution : 

L.H.S :

= cot² θ/2

cot² θ/2 = 1/tan² θ/2 

= 1/[(1 - cos θ) / (1 + cos θ)]

= 1 x [(1 + cos θ) / (1 - cos θ)]

= (1 + cos θ) / (1 - cos θ)

= (1 + 1/sec θ)) / (1 - 1/sec θ))

= [(sec θ + 1)/sec θ] [(sec θ - 1)/sec θ]

= [(sec θ + 1) / sec θ] [sec θ / (sec θ - 1)]

= (sec θ + 1) / (sec θ - 1)

R.H.S

Hence it is proved

Problem 6 :

Use half angel identities to show 

sec² θ/2 = 2/(1 + cos θ)

Solution : 

L.H.S

= sec² θ/2

= 1/(cos² θ/2)

= 1/(cos θ/2)²

= 1 / [√(1 + cos θ)/2]²

= 1/[(1 + cos θ)/2]

= 2/(1 + cos θ).

R.H.S

So, it is proved.

Problem 7 :

Using the half angle identities to find the exact value of each expression :

 sin 22 1/2°

Solution :

sin θ/2 = (1 - cos θ)/2

Let θ = 45, then θ/2 = 22 1/2

sin 45/2 = (1 - cos 45)/2

= (1 - (√2/2))/2

= ((2 - √2)/2)/2

= (2 - √2)/4

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