It is known that α is the angle of the second quadrant, sin α = 3 5, β is the angle of the first quadrant, cos β = 5 13. Find the value of Tan (2 α - β)

It is known that α is the angle of the second quadrant, sin α = 3 5, β is the angle of the first quadrant, cos β = 5 13. Find the value of Tan (2 α - β)

∵ α is the second quadrant angle, sin α = 3
5，∴cosα=-4
5，tanα=-3
4，tan2α=-24
7，
And ∵ β is the first quadrant angle, cos β = 5
13，∴sinβ=12
13，tanβ=12
5，
∴tan（2α-β）=tan2α−tanβ
1+tan2α•tanβ＝−24
7−12
Five
1−24
7×12
5＝204
Two hundred and fifty-three

If a is the angle of the second quadrant, compare the size of a in sin, a in COS and a in tan

If K is an even number, then π / 4 + 2n π < A / 2 < π / 2 + 2n π is in the first quadrant, Tan A / 2 > cos A / 2. When k is odd, let k = 2n, then π / 4 + 2n π < A / 2 < π / 2 + 2n π

Let β be the second quadrant angle and compare the values of sin θ / 2, cos θ / 2 and Tan θ / 2

θ is the second quadrant angle, π / 4cos, θ / 2
Tan θ / 2 > 1
Sin θ / 2 > 2 / 2 root sign 2
cosθ/2

In this paper, we compare the values of sin θ / 2, cos θ / 2 and Tan θ / 2

tan>sin>cos
I think so~

Given that α is the angle of the second quadrant and COS (α - π / 2) = 1 / 5, the value of (sin (π + α) cos (π - α) Tan (- 3 π / 2 - α)) / (Tan (π / 2 + α) cos (3 π / 2 + α)) is obtained

cos（a-π/2）=1/5
sina=1/5
α is the angle of the second quadrant
cosa=-2√6/5
（sin（π+a）cos（π-a）tan（-3π/2-a））/（tan（π/2+a）cos（3π/2+a））
=-sina(-cosa)tana/((-tana)sina)
=-cosa
=2√6/5

It is known that α is the second quadrant angle, f (α) = sin (π − α) Tan (− α − π) sin(π+α)cos(2π−α)tan(−α)． (I) simplify f (α); (II) if cos (α - 3 π) 2）=-1 3. Find the value of F (α)

（Ⅰ）f（α）=sinα(−tanα)
−sinαcosα(−tanα)=-1
cosα；
（Ⅱ）∵cos（α-3π
2）=cos（3π
2-α）=-sinα=-1
3，
∴sinα=1
3，
∵ α is the second quadrant angle,
∴cosα=-
1−sin2α=-2
Two
3，
Then f (α) = - 1
−2
Two
3=3
Two
4．

Given that θ is the second quadrant angle and sin θ + cos θ = 1 / 5, then Tan θ =?

Because SiNx + cosx = 1 / 5, and (SiNx) ^ 2 + (cosx) ^ 2 = 1
That is (SiNx + cosx) ^ 2-2sinxcosx = 1, so 1 / 25-2sinxcosx = 1
So 2sinxcosx = - 24 / 25
Because (SiNx) ^ 2 + (cosx) ^ 2 = (SiNx cosx) ^ 2 + 2sinxcosx = 1
So (SiNx cosx) ^ 2-24 / 25 = 1, so (SiNx cosx) ^ 2 = 49 / 25
Because of 00, and 2sinxcosx = - 24 / 25 < 0, cosx < 0
So SiNx cosx > 0, so SiNx cosx = 7 / 5
Add SiNx + cosx = 1 / 5 to get: SiNx = 4 / 5, so cosx = - 3 / 5
So TaNx = - 4 / 3

The sign of cosina (cosina) is known

A is the angle of the third quadrant
∴ -1＜cosa＜0 , -1＜sina＜0
/ / cosa and Sina are in the fourth quadrant,
∴ sin（cosa）＜0,cos（sina）＞0
∴ sin（cosa）·cos（sina）＜0
Is negative

Given that α is the third quadrant angle, what is the sign of sin (COS α) × cos (COS α)?

∵ α is the angle of the third quadrant,
∴-1

Tan (COS θ) · cot (sin θ) > 0, try to point out the quadrant where θ is located

Because - 10
Zero