Can you tell me the definition domain of Tan α, sin α, cos α, and the range of values of Tan α, sin α, cos α, and so on

Can you tell me the definition domain of Tan α, sin α, cos α, and the range of values of Tan α, sin α, cos α, and so on

Define domain value domain
And X ≠ K π + π / 2 Tan α∈ R
Sin α α∈ r sin α∈ [- 1,1] or (- 1 ≤ sin α ≤ 1)
Cos α α∈ r cos α∈ [- 1,1] or (- 1 ≤ cos α ≤ 1)

Find the definition and range of the function y = √ [1 - √ 2 cos (π / 2-x)] The answer given in the book is just this [0, √ (1 + √ 2)] and I don't know what's going on. I've read it several times, and that's the answer in the book

Define domain
1 - √2 cos(π/2-x)>=0
1 - √2 sinx>=0
-1

The definition domain of function y = √ [cos (SiNx)] is, and the range of value is

The definition domain should ensure cos (SiNx) > 0. For any x belonging to R, the value domain of SiNx is [- 1,1], which can ensure that the cost t is located in quadrants 1,4, so the domain is defined as R;
The range is [- cos (1), cos (1)]

What is the value range of COS (SiNx) under y = radical? The answer is [cos1,1 under radical]

－π/2<-1≤sinx≤1<π/2, 
cos1≤cos(sinX) ≤1 
√cos1≤√cos(sinX) ≤1 
The value range of function y is [√ cos1,1]

The definition domain of the function f (x) = root cos (SiNx) is

As long as cos (SiNx) ≥ 0
And SiNx ∈ [- 1,1]
That is - 1 radian to 1 radian
In this range, the cosine is always positive
So the definition domain is r

Under the function y = radical, the definition field of sinx-1 is () and the value field is () Explain the process in detail

The definition domain is x = 2K π + 1 / 2 π, because the root must be a non negative number, while the definition domain of SiNx is - 1 to 1, so SiNx can only be equal to 1. According to the periodic image of sinusoidal curve, we can know the condition of definition domain, so the value range is 0

Let f (x) = - ACOS ^ 2x - asinx + 3 / 2A + B (a ≠ 0) have the domain of [- π / 3, π / 2] and the range of value [- 4,5], and find the values of a and B

If SiNx ∈ [- π / 3, π / 2], then SiNx ∈ [- 3 / 3, π / 2], then SiNx ∈ [- √ 3 / 2,1] makes M = SiNx, then G (m) = 2am / / am + 2 / A + B = 2A (m-1 / 4) 2 + 3A / 8 + B (1) when a ＞ 0, G (√ 3 / 2 / 2) = 5, G (1 / 4) = - 4G (1 / 4) = - 4G (- (3 / 3 / 2) = - 4G (1 / 4) = - 4G (3 / 3 / 2) = - 4G (1 / 4) = - 4G (3 / 3 / 2) = - 4G (1 / 4) = - 4G (1 / 4) = - 4G (1 / 4 5 is 2

Given that a > 0, the function f (x) = cos? X-asinx + B, the value range is [- 4,0] (1) try to find the value of a and B (2) find the value of X with the maximum and minimum value of Y

(1) F (x) = (cosx) ^ 2-asinx + B = - (SiNx) ^ 2-asinx + B + 1 = - (SiNx + A / 2) ^ 2 + A ^ 2 / 4 + B + 1 when SiNx = 1, f (x) gets the minimum value - (1 + A / 2) ^ 2 + A ^ 2 / 4 + B + 1 = B-A = - 4, B = A-4 when 02 does not match. (2) f (x) = - (SiNx + 1) ^ 2 when f (x) reaches the minimum value - 4, SiNx = 1, that is, X = 2K π + π / 2

Given that the definition domain of the function f (x) = x2-2ax + 2A + 4 is R and the range is [1, + ∞), then the value of a is______ .

f（x）=x2-2ax+2a+4=（x-a）2-a2+2a+4，
From the value range of F (x) is [1, + ∞), the minimum value of F (x) is 1,
Then - A2 + 2A + 4 = 1, that is, a2-2a-3 = 0,
So the answer is: a = - 1 or a = 3

Given the function f (x) = - A (cosx) ^ 2-asinx + 3A / 2 + B (a ≠ 0), the domain of definition is [- Π / 2, Π / 2], and the range of value is [- 4,5], so we can find the values of a and B

f(x)=-a(1-sin²a)-asinx+3a/2+b
=a/2+b+asin²x-asinx
=a（sinx-1/2）²+a/4+b
X ∈ [- Π / 2, Π / 2] so SiNx ∈ (- 1,1)
So when a > 0, the minimum value is SiNx = 1 / 2
a/4+b=-4
When the maximum SiNx = - 1
9a/4+a/4+b=5
A = 4, B = - 5
So when a < 0 SiNx = 1 / 2, the maximum value is taken
At this time, a / 4 + B = 5
The minimum value is obtained when the maximum value SiNx = - 1
9A / 4 + A / 4 + B = - 4
A = - 4 B = 6 is obtained