It is known that f (x) = log2 (xcosa + Sina), a belongs to (π / 2,3 π / 4) 1.f（cosA/[1+sinA])=0 2. When x belongs to [- 1,1], f (x) is always greater than or equal to - 1, and the value range of sin (a + [π / 4]) can be obtained 3) If f (1-4sina)

It is known that f (x) = log2 (xcosa + Sina), a belongs to (π / 2,3 π / 4) 1.f（cosA/[1+sinA])=0 2. When x belongs to [- 1,1], f (x) is always greater than or equal to - 1, and the value range of sin (a + [π / 4]) can be obtained 3) If f (1-4sina)

The denominator of cosa / 1 + sina is multiplied by 1-sina at the same time
Cosa / 1 + Sina = [(1-sina) * cosa] / [(1 + Sina) * (1-sina)]
=[（1-sinA）*cosA]/(cosA)^2
=（1-sinA）/cosA
So f (COSA / 1 + Sina) = f [(1-sina) / cosa] = log2 [(1-sina) / cosa * cosa + Sina] = Log1 = 0
2. If x belongs to [- 1,1], f (x) is always greater than or equal to - 1, then the value range of sin (a + [π / 4]) can be obtained
If x ∈ [- 1,1], f (x) ≥ - 1, then:
First, xcosa + Sina > 0
===> √(1+x^)[sinA(1/√(1+x^))+cosA(x/√(1+x^))]＞0
===> √(1+x^)sin(A+ф)＞0
===>A + ф ∈ (0, π), where: Tan ф = x ∈ [- 1,1], so: ф ∈ [- π / 4, π / 4] (as for 2K π, don't consider it first, because it has no effect on the value)
===> A∈(-π/4,5π/4)
So: a + (π / 4) ∈ (0,3 π / 2)
Then, sin [a + (π / 4)] ∈ (0,1] （1）
Secondly, f (x) ≥ - 1
===> log2(xcosA+sinA)≥-1
===> xcosA+sinA≥1/2
===> √(1+x^)[sinA(1/√(1+x^))+cosA(x/√(1+x^))]≥1/2
Let x = 1, then:
===> √2[sinA(√2/2)+cosA(√2/2)]≥1/2
===> √2sin[A+(π/4)]≥1/2
===> sin[A+(π/4)]≥√2/4……………………………… （2）
LIANLI (1) (2) obtained:
√2/4≤sin[A+(π/4)]≤1
3. F (1-4sina) = log2 (1-4sina) cosa + Sina), is this a uppercase or lowercase?

A belongs to (π, 3 / 2 π), and the inclination of xsina + ycosa + 1 = 0 is equal to

Slope = - Sina / cosa = - Tana
Let the inclination angle be P
Then tap = - Tana = Tan (- a)
-3π/2

The positional relationship between the straight line xsina + ycosa = 1 + cosa and the circle X & # 178; + (Y-1) &# 178; = 4 is

The distance from the center of the circle to the straight line = | cosa-1-cosa | / √ [(Sina) ^ 2 + (COSA) ^ 2] = 1,
And the radius of the circle is 2,
Due to 1

What is the positional relationship between the line xsina + ycosa = R and the circle x ^ 2 + y ^ 2 = R?
Tangent B. intersect C. depart D. change with a

Distance from center of circle to straight line = | 0 + 0-r | / √ (Sin & sup2; a + cos & sup2; a) = R / 1 = R
The radius is √ R
Obviously r > 0
therefore
01,√R