If sin θ = 3 / 5 and cos θ = - 4 / 5, then the quadrant of 2 θ is

If sin θ = 3 / 5 and cos θ = - 4 / 5, then the quadrant of 2 θ is

sinθ/2=3/5>1/2
2kπ+π/6

If α is the second quadrant angle, then sin (α - 2 π) / √ [1-cos ^ 2 (- α)] + 2 √ [1-cos ^ 2 (π / 2 + α)] / sin (α - 3 π / 2)

Three

It is known that sin θ = (1 - α) / (1 + α), cos θ = (3 α - 1) / (1 + α). If θ is the second quadrant angle, find Tan θ

sin²+cos²=1
So a 2 - 2A + 1 + 9A 2 - 6A + 1 = a 2 + 2A + 1
9a²-10a+1=0
a=1/9,a=1
Beta Quadrant
(1-a)(1+a)>0
(3a-1)/(1+a)

Let θ be the second quadrant angle respectively, then the point P (sin θ, cos θ) is in the second quadrant______ Quadrant

∵ θ is the second quadrant angle,
∴sinθ＞0，cosθ＜0，
The point P (sin θ, cos θ) is in the fourth quadrant
Answer: four

If α is the second quadrant angle, then the point P (sin α, cos α) is the point in the second quadrant

If α is the second quadrant angle
Then sin α > 0, cos α

If α is the angle of the third quadrant, then what quadrant is the point P (sin (α / 2), cos (α / 2)), in which quadrant?

Since α is the angle of the third quadrant, let α∈ (Π + 2K Π, 3 Π / 2 + 2K Π)
So α / 2 ∈ (Π / 2 + 2 Π, 3 Π / 4 + 2 Π), α / 2 ∈ (Π / 2 + 2 Π),
α / 2 may be the angle of the second quadrant or the angle of the fourth quadrant
When α / 2 is the angle of the second quadrant, sin (α / 2) > 0, cos (α / 2)

Given that P: α is the second quadrant angle, Q: sin α > cos α, then what condition is p for Q? On sufficient conditions and necessary conditions

If α is the second quadrant angle, then sin α > 0 > cos α, that is, P can deduce that Q holds, so p is a sufficient condition for Q;
If sin α > cos α, then the angle α is the first, second or third quadrant angle, or its final edge is on the negative half axis of X axis or the positive half axis of Y axis. Obviously, Q cannot deduce P, that is, P is not a necessary condition of Q
To sum up, P is a sufficient and unnecessary condition of Q

If sin α - cos α = - 7 / 5, what quadrant is the angle α

sinα+cosα=-7/5
(sinα+cosα)^2=49/25
1+2sina*cosa=49/25
sinacosa=12/25>0
So Sina and cosa are the same,
Sin α + cos α = - 7 / 5,
So Sina and cosa are negative
So a is the third quadrant angle

When sin α, cos α > 0, judge the quadrant where α is I have to make up for it

Alpha is in the first quadrant
If sin > 0, cos

Known a (2,0), B (0,2), C (COS θ, sin θ) The module of vector OA + OC = radical 7, and θ∈ (0, π), find the angle between vector OB and OC

|（2,0）+（cosα,sinα）|=√7
|（2+cosα,sinα）|=√7
The results are as follows: 1
cosα=1/2
sinα=√3/2
Then insert cos θ = (OB * OC) / (| ob | * | OC |)
The results show that cos θ = √ 3 / 2
The included angle is 60 degrees