Given that α is the angle of the second quadrant, then α 2 is the angle in which quadrant () A. The first and second quadrant angles B. Second and third quadrant angle C. The first and third quadrant angles D. The third and fourth quadrants

∵ α is the angle of the second quadrant, ᙽ 2K π + π
2＜α＜2kπ+π，k∈z，
∴kπ+π
4＜α
2＜kπ+π
2, K ∈ Z, so α
2 is the first and third quadrant angle,
Therefore, C

Given that angle a is the second quadrant angle, the quadrant where angle A / 2 is located?

Because the angle a is the second quadrant angle, there is 360n + 90 < a < 360n + 180
180n+45 < a/2

Given that a is the angle of the second quadrant, what quadrant is 2 of a?

A is the angle of the second quadrant, so 2K π + π / 2

It is known that sin a = m (the absolute value of M is less than or equal to 1), and the value of Tan a of the sphere

It can be concluded that: cosa = ± √ (1-sin? A) = ± √ (1-m? 2)
So there are:
tana=sina/cosa
=±m/√(1-m²)
=±m√(1-m²)/(1-m²)

Cos (π / 6 - α) = m, the absolute value of M is less than or equal to 1, find cos (5 π / 6 + α) + sin (2 π / 3 - α)

cos(5π/6+α)=cos[π-(π/6-α)]=-cos(π/6-α)=-m
sin(2π/3-α)=cos[π/2-(2π/3-α)]=cos(-π/6+α)=cos(π/6-α)=m
So the original formula = 0

If the absolute value of cos α = 1 / 5,5 π / 2 < α < 3 π, then the value of sin α / 2 is?

5π/2＜α＜3π ∴5π/4＜α/2＜3π/2 ∴sina/2

If the absolute value of COS (a) is 1 / 5 5 / 2 π < a < 3 π, then the value of sin (A / 2) is

Because 5 π / 2 < a < 3 π, a is the angle of the second quadrant
So 5 π / 4 < A / 2 < 3 π / 2 A / 2 is the angle of the third quadrant
Because | cos (a) | = 1 / 5
So cos (a) = - 1 / 5 = 1 - 2 (sin a) squared
Sina = - radical ([1 - (- 1 / 5)] / 2) = - (Radix 15) / 5

Comparison of absolute values of sin and COS

0-45.. sin is less than cos
45-90 sin is greater than cos (cycle 360 or half cycle 180 can be added)

Let α be the angle of the second quadrant, and │ cos α / 2 │ = - cos α / 2, then what quadrant is α / 2?

Since α is the angle of the second quadrant, α ∈ (2k π + π / 2,2k π + π)
So α / 2 ∈ (K π + π / 4, K π + π / 2). (K ∈ z)
So α / 2 can only fall in the first and third quadrants
Moreover, cos α / 2 < 0
The cosine of the first quadrant is positive, so α / 2 can only be the angle of the third quadrant

Let α be the second quadrant 2|=-cosα Then α, 2 Angle 2 belongs to () A. First quadrant B. Second quadrant C. The third quadrant D. Fourth quadrant

∵ α is the second quadrant angle,
∴90°+k•360°＜α＜180°+k•360°，k
∴45°+k•180°＜α
2＜90°+k•180° k∈Z
∴α
2 in the first or third quadrant,
∵|cosα
2|=-cosα
2，
∴cosα
2＜0
∴α
Angle 2 is in the third quadrant
Therefore, C

Given that α is the angle of the second quadrant, then α 2 is the angle in which quadrant () A. The first and second quadrant angles B. Second and third quadrant angle C. The first and third quadrant angles D. The third and fourth quadrants