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Show that the min value of 4x 2 +4x |sin@| - cos 2 @ is -1
hi, p = 4x2+4x |sin@| - cos2@ is -1 = 4x2+4x |sin@| -1 + sin2@ ( sin2@ = |sin@| ^2 ) =4x2+4x |sin@| + |sin@| ^2 - 1 = ( 2x + |sin@| )^2 -1 give min . value when the squre part will be equal to zero. hence min. value = -1
hi,
p = 4x2+4x |sin@| - cos2@ is -1
= 4x2+4x |sin@| -1 + sin2@ ( sin2@ = |sin@| ^2 )
=4x2+4x |sin@| + |sin@| ^2 - 1
= ( 2x + |sin@| )^2 -1
give min . value when the squre part will be equal to zero.
hence min. value = -1
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