262sinx(sinx+cosx)
= 12. (2sin2x + 2sinxcosx )
=12. (1-cos2x + sin2x )
now -√2 ≤ sin a - cos a ≤ +√2
hence 1-√22 ≤ sinx(sinx+cosx)≤1+√22
find the maximum and minimum value of sinx(sinx+cosx).
i got the minimum value as [1-√2]/2 and the maximum value as [1+√2]/2.
want to confirm it.
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2 Answers
262
3thanks aditya.
i did it like this,
i multiplied and divided it by 2 to get,
or, 2(sinx)[sin(x+45)]√2
or, [cos45 - cos(2x+54)]√2
or, [1 - √2 cos(2x+54)]2
now the given function will be minimum at the maximum value of cos(2x+45) that is 1 and maximum at the minimum value of cos(2x+45) that is -1.
so,
1-√22≤ sinx(sinx+cosx)≤1+√22