三角函數最大值問題

12.y=sin^2x+sinxcosx+cos2x
y=(1/2)(1-cos2x)+(1/2)sin2x+cos2x
y=(1/2)-(1/2)cos2x+(1/2)sin2x+cos2x
y=(1/2)+(1/2)sin2x+(1/2)cos2x
y=(1/2)(1+sin2x+cos2x)
y=(1/2)[1+sqrt(2)cos(2x-pi/4)]
極大值=(1/2)[1+sqrt(2)(1)]
=[1+sqrt(2)]/2
x值:2x-pi/4=0
x=pi/8

13.y=sinxcosx+sinx+cosx+2
=sin2x/2+sqrt(2)cos(x-pi/4)+2
考慮sin2x/2及cos(x-pi/4)
當x=pi/4時，雙方同時達最大值
最大值=1/2+sqrt(2)(1)+2
=5/2+sqrt(2)

As follows:

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圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/03-4.gif