# CONDITIONAL TRIGONOMETRIC IDENTITIES PROBLEMS

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.

Abbreviations used :

L.H.S -----> Left hand side

R.H.S -----> Right hand side

Problem 1 :

If A + B + C = 180°, prove that

sin2A + sin2B + sin2C  =  4sinAsinBsin C

Solution :

L.H.S :

=  sin2A + sin2B + sin2C

Use the formula of (sin C + sin D).

=  2sin[(2A + 2B)/2]cos[(2A - 2B)/2] + sin2C

=  2sin(A + B)cos(A - B) + sin2C -----(1)

A + B + C  =  180°

A + B  =  180 - C

sin(A + B)  =  sin(180 - C)  =  sinC

(1)-----> =  2sinCcos(A - B) + 2sinCcosC

=  2sinC[cos(A - B) + cosC]

=  2sinC{cos(A - B) + cos[180 - (A + B)]}

=  2sinC{cos(A - B) + cos(A + B)} -----(2)

cosC - cosD  =  -2sin(C + D)/2sin(C - D)/2

Here C  =  A - B and D =  A  + B

C  + D  =  A - B + A + B  =  2A

C  - D  =  A - B - A - B  =  -2B

(2)----->  =  2sinC[-2sin(2A/2)sin(-2B/2)]

=  2sinC[2sinAsinB]

=  4sinAsinBsinC

=  R.H.S

Problem 2 :

If A + B + C = 180°, prove that

cosA + cosB - cosC  =  -1 + 4cos(A/2)cos(B/2)sin(C/2)

Solution :

L.H.S :

=  cosA + cosB - cosC

=  2cos(A + B)/2cos(A - B)/2 - cosC

=  2cos(180 - C)/2cos(A - B)/2 - cosC

=  2cos[90 - (C/2)]cos(A - B)/2 - cosC

=  2sin(C/2)cos(A - B)/2 - [1 - 2sin2(C/2)]

=  2sin(C/2)cos(A - B)/2 - 1 + 2sin2(C/2)

=  -1 + 2sin(C/2)cos(A - B)/2 + 2sin2(C/2)

=  -1 + 2sin(C/2){cos(A - B)/2 + sin[180 - (A + B)]/2}

=  -1 + 2sin(C/2){cos(A - B)/2 + sin[90 - (A + B)/2]}

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

Use the formula of (cosC + cosD).

Here C  =  (A - B) / 2 and D  =  (A + B)/2

C + D  =  2A/2  =  A

C - D  =  -2B/2  =  -B

(1)----->  =  -1 + 2sin(C/2)[2cos(A/2)cos(-B/2)]

=  -1 + 4sin(C/2)[cos(A/2)cos(B/2)]

=  -1 + 4cos(A/2)cos(B/2)sin(C/2)

=  R.H.S

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