CONDITIONAL IDENTITIES SOLVED 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.

Problem 1 :

If A + B + C = 180°, prove that

 sin2A + sin2B + sin2C = 2 + 2cosAcosBcosC

Solution :

sin2A + sin2B + sin2C :

= (1 - cos2A)/2 + (1 - cos2B)/2 + (1 - cos2B)/2

= 1/2 - (cos2A)/2 + 1/2 - (cos2B)/2 + 1/2 - (cos2B)/2

= 3/2 - (1/2)[cos2A + cos2B + cos2C]

= 3/2 - (1/2)[2cos(A + B)cos(A - B) + 2cos2C - 1]

= 3/2 - (1/2)[2cos(180 - C)cos(A - B) + 2cos2C - 1]

= 3/2 - (1/2)[-2cosCcos(A - B) + 2cos2C - 1]

= 3/2 - [-cosCcos(A - B) + cos2C] + 1/2

= 3/2 - [-cosCcos(A - B) + cos2C] + 1/2

= 3/2 + 1/2 + cosC[cos(A - B) - cosC]

= 2 + cosC[cos(A - B) - cos(180 - (A + B)]

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

Now let us use the formula for cosC - cosD.

= 2 + cosC[2cosAcosB]

= 2 + 2cosAcosBcosC

Problem 2 :

If A + B + C = 180°, prove that

sin2A + sin2B - sin2C = 2sinAsinBcosC

Solution :

 sin2A + sin2B - sin2C :

= (1 - cos2A)/2 + (1 - cos2B)/2 - (1 - cos2C)/2

= 1/2 - (1/2)[cos2A + cos2B - cos2C]

= 1/2 - (1/2)[2cos(A + B)cos(A - B) - cos2C]

= 1/2 - (1/2)[2cos(180 - C)cos(A - B) - cos2C]

  = 1/2 - (1/2)[-2cosCcos(A - B) - (2cos2C - 1)]

  = 1/2 + cosCcos(A - B) + cos2C - 1/2

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

= cosC[cos(A - B) + cos(180 - (A + B)]

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

= cosC[2sinAsinB]

= 2sinAsinBcosC 

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