TRIPLE 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)

Identity 1 : sin3A = 3sinA - 4sin³A

Proof : 

sin3A = sin(2A + A)

= sin2AcosA + cos2AsinA

= 2sinAcosAcosA + (1 - 2sin²A)sinA

= 2sinAcos²A + sinA - 2sin³A

= 2sinA(1 - sin²A) + sinA - 2sin³A

= 2sinA - 2sin³A + sinA - 2sin³A

= 3sinA - 4sin³A

Identity 2 : cos3A = 4cos³A - 3cosA

Proof : 

cos3A = cos(2A + A)

= cos2AcosA - sin2AsinA

= (2cos²A - 1)cosA - 2sinAcosAsinA

= 2cos³A - cosA - 2cosAsin²A

= 2cos³A - cosA - 2cosA(1 - cos²A)

= 2cos³A - cosA - 2cosA + 2cos³A

= 4cos³A - 3cosA

Identity 3 : tan3A = (3tanA - tan³A)/(1 - 3tan²A)

Proof : 

tan3A = tan(2A + A)

= (tan2A + tanA)/(1 - tan2AtanA)

Using double angle identity tan2A = 2tanA/(1 - tan²A), 

= (3tanA - tan³A)/(1 - 3tan²A)

Double and Triple Angle Identities

Sine : 

sin2A = 2sinAcosA

sin2A = 2tanA/(1 + tan²A)

sin3A = 3sinA - 4sin³A

Cosine : 

cos2A = cos²A - sin²A

cos2A = 2cos²A - 1

cos2A = 1 - 2sin²A

cos2A = (1 - tan²A)/(1 + tan²A)

cos3A = 4cos³A - 3cosA

Tangent : 

tan2A = 2tanA/(1 - tan²A)

tan3A = (3tanA - tan³A)/(1 - 3tan²A)

Problem 1 :

Show that 

tan 3x tan 2x tan x = tan 3x - tan 2x - tan x 

Solution :

tan 3x = tan (2x + x)

Expanding using the formula of

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

tan 3x = (tan 2x + tan x)/(1 - tan 2x  tan x)

tan 3x (1 - tan 2x  tan x) = tan 2x + tan x

tan 3x - tan 3x tan 2x  tan x = tan 2x + tan x

tan 3x - tan 2x - tan x = tan 3x tan 2x  tan x

Hence it is proved.

Problem 2 :

Prove that

tan 20° tan 40° tan 60° tan 80° = 3

Solution :

L.H.S

= tan 20° tan 40° tan 60° tan 80°

= tan 20° tan 40°.  √3 tan 80°

=  √3 tan20° tan (60° - 20°) tan (60° + 20°)

tan (60° - 20°) = (tan 60 - tan 20)/(1 + tan 60 tan 20)

tan 60 = √3

= (√3 - tan 20)/(1 + √3 tan 20)

tan (60° + 20°) = (tan 60 + tan 20)/(1 - tan 60 tan 20)

= (√3 + tan 20)/(1 - √3 tan 20)

=  √3 tan20° ((√3 - tan 20)/(1 + √3 tan 20)) ((√3 + tan 20)/(1 - √3 tan 20))

=  √3 tan20° (√3² - tan² 20)/(1² - (√3 tan 20)²)

=  √3 tan20° (3 - tan² 20)/(1 - 3 tan² 20)

=  √3 (3tan20° - tan³ 20)/(1 - 3 tan³ 20)

=  √3 tan 3(20)

=  √3 tan 60

=  √3 √3

= 3

R.H.S

Problem 3 :

(cos³θ - cos 3θ) / cos θ + (sin³θ + sin 3θ) / sin θ = 3

Solution :

L.H.S

= (cos³θ - cos 3θ) / cos θ + (sin³θ + sin 3θ) / sin θ

cos 3θ = 4 cos³θ - 3 cos θ

sin 3θ = 3 sin θ - 4 sin³ θ

(cos³θ - cos 3θ) / cos θ = [cos³θ - (4 cos³θ - 3 cos θ)] / cos θ

= [cos³θ - 4 cos³θ + 3 cos θ] / cos θ

= [-3 cos³θ + 3 cos θ] / cos θ

= 3cos θ(-cos²θ + 1)/ cos θ

= 3cos θ(1 - cos²θ)/ cos θ

= 3(1 - cos²θ) ------(1)

(sin³θ + sin 3θ) / sin θ

= [sin³θ + (3 sin θ - 4 sin³ θ)] / sin θ

= [sin³θ + 3 sin θ - 4 sin³ θ] / sin θ

= [-3 sin³ θ + 3 sin θ] / sin θ

= 3 sin θ [-sin² θ + 1] / sin θ

= 3 (1 - sin² θ) ------(2)

(1) + (2)

= 3(1 - cos²θ) + 3(1 - sin²θ)

= 3(sin²θ) + 3(cos²θ)

= 3(sin²θ + cos²θ)

= 3(1)

= 3

Problem 4 :

(sin 3x + sin x) sin x + (cos 3x - cos x) cos x = 0

Solution :

L.H.S

(sin 3x + sin x) sin x + (cos 3x - cos x) cos x

= (3 sin x - 4 sin³ x + sin x) sin x + [(4 cos³ x - 3 cos x) - cos x)] cos x

= (4 sin x - 4 sin³ x) sin x + (4 cos³ x - 4 cos x) cos x

= 4 sin² x - 4 sin⁴ x + 4 cos⁴ x - 4 cos² x

= 4 sin² x - 4 cos² x - 4 sin⁴ x + 4 cos⁴ x

= 4(sin² x - cos² x) - 4(sin⁴ x - cos⁴ x)

= 4(sin² x - cos² x) - 4[(sin² x - cos² x)(sin² x + cos² x)]

= 4(sin² x - cos² x) - 4[1(sin² x - cos² x)]

= 4 sin² x - 4 cos² x - 4sin² x + 4 cos² x

= 0

Hence it is proved.

Problem 5 :

The value of

cos A cos (60° - A)cos (60° + A)

is equal to…..

A) (1/2) cos 3A      B) cos 3A

C) 1/4 cos 3A         D) 4cos3A

Solution :

cos A cos (60° - A) cos (60° + A)

Expanding cos (60° - A) :

cos (A - B) = cos A cos B + sin A  sin B

= cos 60 cos A + sin 60 sin A

= (1/2) cos A + (√3/2) sin A

= (cos A)/2 + (√3 sin A)/2

= (cos A + √3 sin A)/2

Expanding cos (60° + A) :

cos (A + B) = cos A cos B - sin A  sin B

= cos 60 cos A - sin 60 sin A

= (1/2) cos A - (√3/2) sin A

= (cos A)/2 - (√3 sin A)/2

= (cos A - √3 sin A)/2

cos A cos (60° - A)cos (60° + A)

= cos A (cos² A - 3sin²A)/4

= cos A (cos² A - 3(1 - cos²A))/4

= cos A (cos² A - 3 + 3cos²A)/4

= cos A (4cos² A - 3)/4

= (4cos³ A - 3 cos A)/4

= (1/4) cos 3A

So, option C is correct.

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