The maximum and minimum values of y = SiNx + cosx 1 / Radix (1 + | sin 2x |)

The maximum and minimum values of y = SiNx + cosx 1 / Radix (1 + | sin 2x |)

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Find the maximum value of the function f (x) = sin2x + SiNx + cosx

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

Given that f (cosx) = cos2x, find f (SiNx)

f（cosx）=cos2x=2cosx*cosx-1
So f (SiNx) = 2sinx * SiNx - 1
=-cos2x

If f (SiNx) = cos2x, then f (1 / 2)=___ ,f(cosx)=___

Solution 1: F (SiNx) = cos2x = 1-2 (SiNx) ^ 2
So f (x) = 1-2x ^ 2
f(1/2)=1/2
f(cosx)=1-2(cosx)^2=-cos2x
Solution 2: F (1 / 2) = f (sin30 ') = cos60' = 1 / 2
f(cosx)=f(sin(x+90'))=cos(2x+180')=-cos2x

Given that f (SiNx) = 5 + cos2x, it is proved that f (cosx) = 5-cos2x

f（sinx）＝5＋cos2x =5+1-2sin^2x=4-2sin^2x
f(x)=4-2x^2
f（cosx）=4-2cos^2x=5－cos2x

(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

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From SiNx / 2 + cosx / 2 = 1 / 2, we get: (SiNx / 2 + cosx / 2) ^ 2 = 1 / 4, and then (SiNx / 2) ^ 2 + (cosx / 2) ^ 2 + 2sinx / 2cosx / 2 = 1 / 4, 1 + SiNx = 1 / 4, SiNx = - 3 / 4cos (2x) = cos 2 (x) - Sin 2 (x) a, cos2x = - 1 / 8 can be calculated

Known - Π / 2

∵ known - Π / 2  SiNx + cosx) ∵ is known to be - Π / 2  (SiNx + cosx) ∵ s = 1 / 25; cosx ﹥ 0 > SiNx
sinxcosx=-12/25;
(cosx-sinx)²=1-2sinxcosx=49/25;
∴cosx-sinx=7/5;
∴cos2x=cos²x-sin²x=(cosx+sinx)(cosx-sinx)=(7/5)(1/5)=7/25;

If (1 + SiNx + cosx) / (1 + SiNx cosx) = 1 / 2, then the value of TaNx / 2 is equal to?

(1+sinx+cosx)/(1+sinx-cosx)
=[2sin(x/2)cos(x/2)+2cos(x/2)cos(x/2)]/2sin(x/2)cos(x/2)+2sin(x/2)sin(x/2)]
=tan(x/2)=1/2

Range y= (sinx-1) / (cosx-2)

From - 1 ≤ SiNx ≤ 1, - 2 ≤ sinx-1 ≤ 0 is obtained
From - 1 ≤ cosx ≤ 1, - 3 ≤ cosx-2 ≤ - 1 is obtained
So the range of y = (sinx-1) / (cosx-2) is [0,2]