Given that the point P (a + 3, - 2-2a) is on the bisector of the second and fourth quadrants, find the value of a to the power of 2007-a

Given that the point P (a + 3, - 2-2a) is on the bisector of the second and fourth quadrants, find the value of a to the power of 2007-a

The point P (a + 3, - 2-2a) is on the bisector of the second and fourth quadrants
a+3＝-(-2-2a)
A=1
The 2007 power of a - a = 0

If the point a (- 3 + A, 2A + 9) is on the bisector of the second quadrant, then the value of a is___ .

∵ point a (- 3 + A, 2A + 9) is on the bisector of the second quadrant,
∴（-3+a）+（2a+9）=0，
A = - 2
So the answer is: - 2

Given the quadrant of angle a, find the quadrant of 2A A / 2 A / 3 Given the quadrant of angle a 2A A/2 A/3 Quadrant How to do? Not at all. Thank you!

But there is a simple way to X / N type. Take a / 2 as an example to draw a rectangular coordinate system. Then divide 1 and 3 quadrants 2 and 4 into two equal parts by drawing diagonals, and then start from the first quadrant

If α is the second quadrant angle, try to determine the position of the final edge of 2 α, α / 2, α / 3 respectively And explain the reasons 4kπ+π

Beta Quadrant
So 2K π + π / 2

Given cosa = - 1 / 3, a is the second quadrant angle, sin (a + b) = 1, find the value of COS (2a + b)

Because sin (a + b) = 1, cos (a + b) = 0
Because cosa = - 1 / 3, a is the second quadrant angle, so Sina = 2 √ 2 / 3
cos（2a+b)=cos（a+b)cosa-sin(a+b)sina
=-2√2/3

If cos α = 2 3, α is the fourth quadrant angle, then sin (α - 2 π) + sin (- α - 3 π) cos (α - 3 π)=______ .

According to the meaning of the question, ∵ cos α = 2
3, α is the fourth quadrant angle,
∴sinα＝−
Five
Three
∵sin（α-2π）+sin（-α-3π）cos（α-3π）=sinα+sin（-α+π）cos（α+π）=sinα-sinαcosα
∴sin（α-2π）+sin（-α-3π）cos（α-3π）=−
Five
3−(−
Five
3)×2
3＝−
Five
Nine
So the answer is: --
Five
Nine

Suppose that the angle of the second quadrant is implied, Sina = 3 / 5, and find the value of sin (37 π / 6-2a),

In this paper, we propose a new method to solve this problem

Let α be the angle of the second quadrant, sin α = 3 5, find sin (37 π) 6 − 2 α)

Because sin (37 π)
6−2α)＝sin(π
6−2α)，
sinα＝3
5⇒cosα＝−4
5 (α is Ⅱ)
sin2α＝−24
25cos2α＝1−2sin2α＝7
25 ------ (6 points)
So sin (π)
6−2α)＝7+24
Three
50 -------- (13 points)

Let a be the second quadrant angle, Sina = 3 / 5, and find the value of sin (37 π / 6-2a) Please write a description of each step,

So cosa = - 4 / 5 sin2a = 2sinacosa = - 24 / 25 cos2a = 1 - 2 (Sina) ^ 2 = 7 / 25 sin (37 π / 6-2a) = sin [6 π + (π / 6-2a)] = sin (π / 6-2a) = sin (π / 6-2a) = sin (π / 6) cos2a - cos (π / 6) sin2a = (7 + 24 √ 3) / 50

Let a be the second quadrant angle, Sina = 3 / 5, and find sin (30)_ 2A)

A is the second quadrant angle, Sina = 3 / 5,
So: cosa = - 4 / 5
Then: sin2a = 2sinacosa = - 24 / 25
cos2a=cos^2a-sin^2a=7/25
sin(30-2a)=sin30cos2a-cos30sin2a
=7/50+24√3/50
=(7+24√3)/50