Find the value range of the function y = sin α / │ sin α + │ cos α │ / cos α + Tan α / │ Tan α

Find the value range of the function y = sin α / │ sin α + │ cos α │ / cos α + Tan α / │ Tan α

Alpha is in the first quadrant
sinα>0,cosα>0,tanα>0
y=1+1+1=3
Alpha is in the second quadrant
sinα>0,cosα<0,tanα<0
y=1-1-1=-1
Alpha is in the third quadrant
sinα<0,cosα<0,tanα>0
y=-1-1+1=-1
Alpha is in the fourth quadrant
sinα<0,cosα>0,tanα<0
y=-1+1-1=-1
The range is {- 1,3}

The value range of the function y = sin θ + cos θ + 2 is Use of knowledge points, and the use of what formula!

Give a root sign 2, y = sin θ + cos θ + 2 = √ 2 [(sin θ) / √ 2 + (COS θ) / √ 2] + 2
For this thing: (sin θ) / √ 2 + (COS θ) / √ 2
We get √ 2 (sin θ) / 2 + √ 2 (COS θ) / 2
You know sin45 = cos45 = √ 2 / 2
Using the formula: sin (a + b) = sinacosb + cosasinb
So the above is: sin (θ + 45)
So the original formula is sin (θ + 45) + 2, and then we get the answer: [1,3]
You see, I haven't answered the question for such a long time, just passing by to see you this question is too simple

Find the definition and range of the function y = √ [1 - √ 2 cos (π / 2-x)]

1－√2cos(π/2－x)≥0
√2sinx≤1
SiNx ≤√ 2 / 2, using the image of function y = SiNx
2kπ－5π/4≤x≤2kπ＋π/4,
Thus, the domain of its definition is [2K π - 5 π / 4,2k π + π / 4], where k is an integer
Value range [0, under root sign (1 plus root 2)]

Given the function y = cos (2x - π / 3), if the definition domain is [π / 6, π / 2], find the function range if the value field is [root (3) / 2,1]

The definition domain is [π / 6, π / 2]
Then the range of 2x - π / 3 is
Zero

The definition domain of COS (SiNx) is () and the value field is ()

cos(sinx)>=0
So 2K π - π / 2

Given that a > 0, the definition domain of the function f (x) = cos2x asinx + B is [0,2 π], and the range of value is [- 4,0]. Try to find the values of a and B

f(x)＝(1−sin2x)−asinx+b＝−(sinx+a
2)2+a2
4+b+1．
Let t = SiNx, and t ∈ [- 1,1] be obtained by X ∈ [0,2 π], then y = f (x) = − (T + a)
2)2+a2
4+b+1，
From a > 0, the axis of symmetry t = − a
2＜0，
① When − a
2 ≤− 1, that is, when a ≥ 2, the function gets the minimum value when t = 1, and the function gets the maximum value when t = - 1
0−a+b＝−4
a+b＝0 ，
A = 2, B = - 2;
② When − a
2 < 0, i.e. 0 < a < 2, t = − a
When t = 1, the function gets the minimum value
A2
4+b+1＝0
b−a＝−4 ，
A = - 2 or a = - 6 (omitted)
∴a=2，b=-2．

If A.B.C is the length of three sides of the triangle ABC, then the absolute value of a plus B minus C plus the root sign a minus B minus C is the square of

a. B. C is the length of three sides of the triangle ABC
a+b>c
a|a+b-c|+√(a-b-c)^2
=a+b-c+b+c-a
=2b

Given that the sides of the triangle ABC are ABC and ABC satisfies the absolute value of the root a minus three plus B-4 + the square of C-5 is equal to 0, try to find the shape of the triangle ABC?

∵ the absolute value of radical (A-3) + B-4 + (C-5) 2 = 0
Radical (A-3) ≥ 0
The absolute value of B-4 ≥ 0
（c-5）²≥0
∴a=3,b=4,c=5
∵3²+4²=5²
ν Δ ABC is a right triangle

It is known that the three sides of the triangle ABC are a, B, C, and the absolute value of a square-6a + 9 + (B-4) + C minus 5 = 0 Online wait, solve, good-natured people to help ah! Answer the question right, I will give a reward I'm asking you a question. Thank you! Given that the three sides of the triangle ABC are a, B and C, if a + B = 10, ab = 18 and C = 8 are satisfied, then the triangle is_____ triangle

The absolute value of a square-6a + 9 + (B-4 under the root sign) + C minus 5 = 0
（a-3)²+√(b-4)+│c-5│=0
a-3=0,b-4=0,c-5=0
a=3,b=4,c=5
∴c²=a²+b²
⊿ ABC is a right triangle

If there are three sides of A.B.C triangle ABC, the absolute value of the square of the root sign (a-b-c) is + (- a-b) +c) The cube root of the cube is equal to?

a. The three sides of B.C triangle ABC
b+c>a
a-b-cb
a+c-b>0
Square of root sign (a-b-c) absolute value of - (a-b + C) + (- a-b)
+c) Cube root of the cube of
=-(a-b-c)-(a-b+c)+(-a-b+c)
=-a+b+c-a+b-c-a-b+c
=b+c-3a