What is the range of y = (sinx-2) (cosx-2)

What is the range of y = (sinx-2) (cosx-2)

The original formula = sinxcosx-2 (SiNx + cosx) + 4
Let t = SiNx + cosx
T belongs to negative change sign 2 to change sign 2, sinxcosx = (T ^ 2-1) 2, then calculate by yourself!

Value domain, y = 1 / SiNx + 2 / cosx

f(x)＝y＝1/sinx+2/cosx
＝(cosx + 2sinx)/(sinxcosx)
＝2√5 (1/√5 cosx + 2/√5 sinx ) / sin(2x)
2 √ 5 sin (x + a) / sin (2x), where a = arcsin (1 / √ 5) ≈ 0.4636
When x ∈ (- π / 2,0),
There are: sin (- π / 2 + a) = - 2 / √ 5 < 0
sin( 0 + A) ＝ 1/√5 > 0
So there are:
f(－π/2+)＝+∞
f( 0- )＝－∞
Since f (x) is continuous in (- π / 2,0), the range of F (x) is r

Find the range of y = (SiNx + 2) (cosx + 2) Find the value range of y = (SiNx + 2) times (cosx + 2)

y=(sinx+2)(cosx+2)=sinxcosx+2(sinx+cosx)+4=1/2sin2x+2√2sin(x+π/4)+4=-1/2cos(2x+π/2)+2√2sin(x+π/4)+4=-1/2(1-2sin(x+π/4)^2)+2√2sin(x+π/4)+4=sin(x+π/4)^2+2√2sin(x+π/4)+7/2=[sin(x+π/4)+√2]^2+...

What is the range of y = (1-sinx) / (2-cosx) Thankyou!

y=(1-sinx)/(2-cosx),
2y-ycosx=1-sinx,
sinx-ycosx=1-2y,
sin(x-t)=(1-2y)/√（1+y^2),
∴|(1-2y)/√（1+y^2)|

The range of y = (3cosx-1) / (SiNx + 2),

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The range of y = (3cosx + 1) / (SiNx + 2)

y=(3cosx+1)/(sinx+2)
ysinx+2y=3cosx+1
ysinx-3cosx=1-2y
√(y^2+9)sin(x+φ)=1-2y
Then - 1 ≤ (1-2y) / √ (y ^ 2 + 9) ≤ 1
By solving the inequality, y ∈ [- 2 / 3,2] is obtained

Y = SiNx - √ 3cosx, X belongs to [0, π / 2]

y=sinx-√3cosx
=2(1/2sinx-√3/2cosx)
=2sin(x-π/3)
∵x∈【0,π/2】
∴x-π/3∈【-π/3,π/6】
The value range is [- √ 3,1]

Find the function y = 3cosx The range of 2 + SiNx___ .

Set points P (SiNx, cosx), q (- 2, 0),
Then y
3 can be regarded as the slope of the line between the moving point P and the point Q on the unit circle, as shown in the figure on the right
Let qp1 be the equation y = K (x + 2), that is, kx-y + 2K = 0,
Then the distance d from the center of the circle (0, 0) to it is d = | 2K|
k2+1=1，
The solution is K1=-
Three
3 or K2=
Three
3，
So-
Three
3≤y
3≤
Three
3, i.e. - 1 ≤ y ≤ 1,
So the answer is: [- 1,1]

When x belongs to (π / 6,7 π / 6), the minimum value and maximum value of function y = 3-sinx-2cos? X Please explain in detail, thank you

When x belongs to (π / 6,7 π / 6), SiNx belongs to (- 1 / 2,1], so when SiNx = -..., it can be seen that SiNx belongs to (- 1 / 2,1]

Given that the maximum value of the function y = 2cos2 ^ x-sinx + B, X ∈ [3 π / 4,3 π / 2] is 9 / 8, try to find its minimum value

The original formula = - 2 (SiNx) ^ 2 - SiNx + B + 2
Let SiNx = t
Radical 2 / 2 > = t > = - 1
Using the maximum value to find B
Finally, the minimum value is obtained