For any real x find the minimum and maximum value for the expression 3sin^2x +3sinxcosx + 7cos^2x.

Asked by acv27joy | 10th Dec, 2018, 01:10: PM

Expert Answer:

3sin2x +3sinxcosx + 7cos2x
Use cos2x = 1 - 2sin2x = 2cos2 x - 1
sin2x = 2sinxcosx
begin mathsize 16px style 3 sin squared straight x plus 3 sinxcosx plus 7 cos squared straight x
equals 3 open parentheses fraction numerator 1 minus cos 2 straight x over denominator 2 end fraction close parentheses plus 3 over 2 sin 2 straight x plus 7 open parentheses fraction numerator 1 plus cos 2 straight x over denominator 2 end fraction close parentheses
equals fraction numerator 3 minus 3 cos 2 straight x plus 3 sin 2 straight x plus 7 plus 7 cos 2 straight x over denominator 2 end fraction
equals fraction numerator 10 plus 4 cos 2 straight x plus 3 sin 2 straight x over denominator 2 end fraction
equals 5 plus 1 half open parentheses 4 cos 2 straight x plus 3 sin 2 straight x close parentheses
Using space the space condition
minus square root of straight a squared plus straight b squared end root less or equal than asinx plus bcosx less or equal than square root of straight a squared plus straight b squared end root
you space can space solve space it space further space to space find space minimum space and space maximum space value space of space 3 sin squared straight x plus 3 sinxcosx plus 7 cos squared straight x end style 

Answered by Sneha shidid | 12th Dec, 2018, 09:49: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.