Find the maximum value of the function y = sin2x + SiNx cosx

Find the maximum value of the function y = sin2x + SiNx cosx

y=sin2x+sinx-cosx=2sinxcosx-1+sinx-cosx+1=-（sinx-cosx）²+sinx-cosx+1
=-{(radical 2Sin (x - π / 4) - 1 / 2} 2 + 5 / 4
When sin (x - π / 4) = root 2 / 4, the maximum value of Y is 5 / 4
The maximum value of the function y = sin2x + SiNx cosx is 5 / 4

1 + SiNx cosx sin2x / 1 + SiNx cosx is simplified and its maximum value is obtained

(1+sinx-cosx-sin2x)/(1+sinx-cosx)
=(1+sinx-cosx-sin2x)/(1+sinx-cosx);
=(sin^2x+cos^2x+sinx-cosx-2sinxcosx)/(1+sinx-cosx);
=(sin^2x-2sinxcosx+cos^2x+sinx-cosx)/(1+sinx-cosx);
=[(sinx-cosx)^2+(sinx-cosx)]/(1+sinx-cosx);
=(sinx-cosx)*(sinx-cosx+1)/(sinx-cosx+1)
=sinx-cosx
=√2sin(x-π/4)
-√2<=√2sin(x-π/4)<=√2
Therefore, the maximum value of (1 + SiNx cosx-sin2x) / (1 + SiNx cosx) is: √ 2

How to find the maximum value of sin2x + SiNx + cosx

Let f (x) = sin2x + SiNx + cosxf (x) = sin2x + SiNx + cosx = 2sinxcosx + SiNx + cosx = (1 + 2sinxcosx) + SiNx + cosx-1 = (SiNx + cosx) 2 + (SiNx + cosx) - 1 = t? + T-1 let t = SiNx + cosx = y (T), then the maximum value of F (x) is the parabola y (T) = t

Y = the sum of the minimum and maximum of | SiNx | cosx-1 is______

(y+1)^2= (sinx)^2*(cosx)^2
=(1-cosx^2)*(cosx^2)
On the right is the parabola formula about cosx, the maximum value can be found, then we know the range of Y + 1, and then we know the range of Y
Or according to SiNx > 0 and SiNx

(2004 · Anhui) if f (SiNx) = 2-cos2x, then f (cosx) is equal to () A. 2-sin2x B. 2+sin2x C. 2-cos2x D. 2+cos2x

∵f（sinx）=2-（1-2sin2x）=1+2sin2x，
∴f（x）=1+2x2，（-1≤x≤1）
∴f（cosx）=1+2cos2x=2+cos2x．
Therefore, D is selected

It is known that f (SiNx) = cos2x, f (cosx) =?

f(sinx)=cos2x=cos²x-sin²x=1-2sin²x
f(cosx)=1-2cos²x

If f (SiNx) = 3-cos2x, then f (cosx)=______ .

f（cosx）=f[sin(π
2−x)]
=3-cos（π-2x）
=3+cos2x．
So the answer is: 3 + cos2x

Let f '(cosx) = cos2x, find f' (SiNx)

f'(cosx)=cos(2x)=2cos^2x-1
f'(x)=2x^2-1
f'(sinx)=2sin^2x-1=-cos(2x)

(2004 · Anhui) if f (SiNx) = 2-cos2x, then f (cosx) is equal to () A. 2-sin2x B. 2+sin2x C. 2-cos2x D. 2+cos2x

∵f（sinx）=2-（1-2sin2x）=1+2sin2x，
∴f（x）=1+2x2，（-1≤x≤1）
∴f（cosx）=1+2cos2x=2+cos2x．
Therefore, D is selected

If f (SiNx) = 3-cos2x, then f (cosx)=______ .

f（cosx）=f[sin(π
2−x)]
=3-cos（π-2x）
=3+cos2x．
So the answer is: 3 + cos2x