Find the range of y = sinx-1 / cosx + 2

Find the range of y = sinx-1 / cosx + 2

Find the range of y = (sinx-1) / (cosx + 2)
Using the boundedness of trigonometric function (combined with auxiliary angle formula)
ycosx+2y=sinx-1,sinx- ycosx=1+2 y,
√(y²+1)sin(x+α) =1+2 y,
sin(x+α) =(1+2 y)/√(y²+1)
∵|sin(x+α)|≤1
∴|(1+2 y)/√(y²+1)| ≤1
-4/3≤y≤0.
The value range of the function is [- 4 / 3,0]

Y = root 3cosx SiNx range

Do you mean that 3cosx SiNx are all in the root sign, or are only 3 in the root sign? If only 3 is in the root sign, it is very simple, according to these three formulas, as follows: trigonometric function of sum and difference of two corners: cos (α + β) = cos α. Cos β - sin α. Sin β cos (α - β) = cos α. Cos β + sin α. Sin β sin (α

When x ∈ [Π / 6,7 Π / 6], find the maximum and minimum of the function y = 3-sinx-2cos X

y=3-sinx-2(1-sin^2x)
=1-sinx+2sin^2x
=2(sinx-1/4)^2+7/8
Because x ∈ [Π / 6,7 Π / 6], SiNx ∈ [- 1 / 2,1]
When SiNx = 1 / 4, the minimum value of Y is 7 / 8
When SiNx = - 1 / 2 or 1, y max = 2

Given that x ∈ [3 / 4 π, 2 / 3 π], the maximum value of the function y = 2cos ^ 2x SiNx + B is 9 / 8, find its minimum value Good integral, big deal!

(cosx) ^ 2 = 1 - (SiNx) ^ 2 so y = - 2 (SiNx) ^ 2-sinx + 2 + b let a = sinxx ∈ [3 / 4 π, 3 / 2 π] because SiNx is a subtraction function in [1 / 2 π, 3 / 2 π], so x = 3 / 2 π, a min = - 1 x = 3 / 4 π, a max = √ 2 / 2Y = - 2A ^ 2-A + 2 + B = - 2 (a + 1 / 4) ^ 2 + 17 / 8 + B with the opening downward, the axis of symmetry a = - 1 / 4-1

Find the maximum and minimum of the function y = 2cos ^ 2x SiNx, X ∈ [0, π]

Y = 2cos? X-sinx = 2-2sin? X-sinx = - 2 (sin? X + 1 / 2 * SiNx + 1 / 16) + 17 / 8 = - 2 (SiNx + 1 / 4) 2 + 17 / 8 because x ∈ [0, π], SiNx ∈ [0,1] then when SiNx = 0, the maximum value of the function is 2; when SiNx = 1, the minimum value of the function is - 1

The known function y = (SiNx + cosx) square + 2cos square x to find the minimum positive period! Decreasing interval! Maximum minimum value!

y=sin²x+cos²x+2sinxcosx+2(1+cos2x)/2
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
So the maximum value = √ 2 + 2
Minimum = √ 2 / 2
The decrease is 2K π + π / 2 < 2x + π / 4 < 2K π + 3 π / 2
K π + π / 8 so the decreasing interval (K π + π / 8, K π + 5 π / 8)

Given the function y = (SiNx + cosx) + 2cosx, find the decreasing interval of the function? Find the maximum and minimum value?

The original formula = SiNx + cosx + 2sinxcosx + 2cosx = 1 + sin2x + cos2x + 1 = 2 + Radix 2Sin (2x + π / 4) you should know the rest
For adoption

The minimum value of the square of the sum of (cosx) = (2)

The square of (SiNx + cosx) ^ 2 + 2cos ^ x
=2+sin2x+cos2x
=2+√2sin(2x+π/4)
So: maximum = 2 + √ 2,
Minimum = 2 - √ 2
The decline interval is as follows:
2kπ+π/2≤2x+π/4≤2kπ+3π/2
The results are as follows: 1
kπ+π/8≤x≤kπ+5π/8
So its decreasing interval [K π + π / 8, K π + 5 π / 8], K ∈ Z
Maximum = 2 + √ 2,
Minimum = 2 - √ 2

Given the function f (x) = cosx ^ 4 + 2sinxcosx-sinx ^ 4, find the minimum positive period of F (x)

f(x)=(cos²x+sin²x)(cos²x-sin²x)+2sinxcosx
=1×cos2x+sin2x
=√2sin(2x+π/4)
So t = 2 π / 2 = π

The known function f (x) = a (cosx SiNx) - 2sinxcosx (x ∈ [0, π / 2] (1) Let t = cosx SiNx, find the range of T (2) When f (x) min = - 5 / 4, find the value of real number a

In this problem: (1) x goes from 0 to π / 2, and cosx decreases and SiNx increases in the interval. Therefore, cosx is the smallest and SiNx is the largest at π / 2, SiNx is the smallest at 0, and cosx is the largest. Then the range of T = cosx SiNx is [cos π / 2-sin π / 2, cos0-sin0], that is [- 1,1]. (2) the conclusion in (1) is very important because 2sinxc