2(sin^6 A+cos^6 A)-3(sin^4 A+cos^4 A)+1=0
Asked by | 29th Aug, 2012, 09:17: PM
SIN6 A + COS 6A IS IN FORM OF A3 + B3
SIN6A+ COS 6 A = ( SIN 2 A + COS 2A ) 3 - 3 SIN2A * COS2A ( SIN 2A + COS 2A )
SIN4A + COS 4A = ( SIN 2A + COS 2A)2 - 2 ( SIN 2A * COS 2A)
=> 1 2 - 2 ( SIN 2A * COS 2A)
Now, LHS = 2(sin6A + cos6A) - 3(sin4 + cos4) +1 =
2 (13 - 3 SIN2A * COS2A ) - 3 ( 1 2 - 2 SIN 2A * COS 2A) + 1
= 2 - 6 SIN 2A * COS 2A - 3 + 6 SIN 2A * COS 2A + 1
= 0
Answered by | 29th Aug, 2012, 11:36: PM
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