CONDITIONAL  TRIGONOMETRIC IDENTITIES

Trigonometric identities are true for all admissible values of the angle involved. There are some trigonometric identities which satisfy the given additional conditions. Such identities are called conditional trigonometric identities.

Example 1 :

If A + B + C = π/2, prove that 

sin2A + sin2B + sin2C = 4cosAcosBcosC

Solution :

sin2A + sin2B + sin2C :

= 2sin(A + B)cos(A - B) + sin2C

= 2sin(90 - C)cos(A - B) + 2sinCcosC

= 2cosCcos(A - B) + 2sinCcosC

= 2cosC[cos(A - B) + sinC]

= 2cosC[cos(A - B) + sin(90 - (A + B)]

= 2cosC[cos(A - B) + cos(A + B)]

= 2cosC[2cosAcos(-B)]

= 2cosC[2cosAcosB]

= 4cosAcosBcosC

Example 2 :

If A + B + C = π/2, prove that 

cos2A + cos2B + cos2C = 1 + 4sinAsinBcosC

Solution :

cos2A + cos2B + cos2C :

Use the identity of (cosC + cosD) for cos2A + cos2B.

= 2cos(A + B)cos(A - B) + cos2C

= 2cos(90 - C)cos(A - B) + 1 - 2sin2C

= 2sinCcos(A - B) + 1 - 2sin2C

= 1 + 2sinC[cos(A - B) - sinC]

= 1 + 2sinC[cos(A - B) - sin(90 - (A + B)]

= 1 + 2sinC[cos(A - B) - cos(A + B)]

= 1 + 2sinC[-2sinAsin(-B)]

= 1 + 2sinC[2sinAsinB]

= 1 + 4sinAsinBsinC

Example 3 :

If triangle ABC is a right triangle and ∠A = π/2, then prove that

(i) cos2B + cos2C = 1

(ii) sin2B + sin2C = 1

(iii) cosB - cosC = -1 + 2√2cos(B/2)sin(C/2)

Solution :

(i) cos2 B + cos2 C = 1 :

In the right triangle ABC above, 

cosθ = Adjacent side/Hypotenuse

cosB = AB/BC

cosC = AC/BC

cos2B + cos2C  =  (AB/BC)2 + (AC/BC)2 

  =  (AB2 + AC2)/BC2

  =  BC2/BC2

  =  1

(ii) sin2 B + sin2 C = 1 :

In the right triangle ABC above, 

sinθ = Opposite side/Hypotenuse

sinB = AC/BC

sinC = AB/BC

 sin2B + sin2C = (AC/BC)2 + (AB/BC)2

= AC2/BC2 + (AB2/BC2)

= (AC2 + AB2)/BC2

= BC2/BC2

= 1

(iii) cosB − cosC = -1 + 2 √2 cos B/2 sin C/2 :

cosB - cosC = 2cos2B - 1 - cosC

= -1 + 2cos2B - cosC ----(1)

In the triangle ABC above, 

A + B + C = π

It is given that A = π/2. Then, 

B + C = π/2

C = π/2 - B

Substitute C = π/2 - B in (1).

= -1 + 2cos2(B/2) - cos(π/2 - B)

= -1 + 2cos2(B/2) - sinB

= -1 + 2cos2(B/2) - 2sin(B/2)cos(B/2)

= -1 + 2cos(B/2)(cos(B/2) - sin(B/2))

B = π/2 - C ----> B/2 = π/4 - C/2

= -1 + 2cos(B/2)[cos(π/4 -  C/2) - sin(π/4 - C/2)]

= -1 + 2cos(B/2)[cos(π/4 -  C/2) - cos(π/2 - (π/4 - C/2))]

= -1 + 2cos(B/2)[cos(π/4 -  C/2) - cos(π/4 + C/2)]

= -1 + 2cos(B/2)[-2sin(π/4)sin(-C/2)]

= -1 + 2cos(B/2)[2(1/√2)sin(C/2)]

= -1 + 2√2cos(B/2)sin(C/2)

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