CONDITIONAL IDENTITIES SOLVED PROBLEMS

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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