Let α be the second quadrant angle and | cos α 2|＝−cosα 2, then α 2 is No.2______ Quadrant angle

Let α be the second quadrant angle and | cos α 2|＝−cosα 2, then α 2 is No.2______ Quadrant angle

∵ α is the second quadrant angle, ᙽ α
2 is the first or third quadrant angle,
∵|cosα
2|＝−cosα
2，∴cosα
2 < 0, i.e. α
2 is the third quadrant angle
So the answer is: 3

Sin θ * cos θ is known

The absolute value of cos θ = cos θ. So cos θ 0
tanθ

If θ is the third quadrant angle and satisfies the absolute value of cos θ / 2 = - cos θ / 2, then θ / 2 is A first quadrant Second quadrant B C fourth quadrant D two and four quadrants

|cosθ/2|=-cosθ/2
explain
cosθ/2

Let α be the second quadrant angle and cos α 2=- 1−cos2(π−α 2) Then α 2 is () A. First quadrant angle B. Second quadrant angle C. Third quadrant angle D. Fourth quadrant angle

Alpha is the second quadrant angle, alpha
2 is the first or third quadrant angle
∵cosα
2=−
1−sin2α
2=-|cosα
2|，
∴α
2 is the third quadrant angle
Therefore, C

When α is the second quadrant angle, is the absolute value of sin α / sin α minus the absolute value of cos α / cos α?

When α is the second quadrant angle, sin α > 0, cos α

(1+sinα)/cosα=[1+tan （α/2）]/[1-tan（α/2）] (1+sinα)/cosα=[1+tan（α/2）]/[1-tan（α/2）]

The formula is: Sina = (2tana / 2) / [1 + (Tana / 2) ^ 2] cosa = [1 - (Tana / 2) ^ 2] / [1 + (Tana / 2) ^ 2] / [1 + (Tana / 2) ^ 2] left = {[1 + (Tana / 2) ^ 2 + 2tana / 2] / [1 + (Tana / 2) ^ 2] / [1 + (Tana / 2) ^ 2]} / {[1 - (Tana / 2) ^ 2] / [1 + (Tana / 2) ^ 2]} = {[1 + (Tana / 2) ^ 2 + 2tana / 2} / {1 - (Tana / 2) ^ 2 ^ [1 - (Tana / 2) ^ 2] is [1 (Tana / 2) ^ 2] and [1 - (1 + (Tana / 2)

If Tan (α - π / 4) = 1 / 2 and α belongs to (0, π / 2), then sin α + cos α=

tan(a-π/4)=(tana-1)/(1+tana)=1/2
1+tana=2tana-2
tana=3
Because 0

Senior one mathematics: prove (1-cos α + sin α) / (1 + cos α + sin α) = Tan α / 2

（1－cosα+sinα）/（1+cosα+sinα）=[1-(1-2sin^2(a/2))+sina]/[1+2cos^2(a/2)-1+sina]=[2sin^2(a/2) + 2sin(a/2)cos(a/2)]/[2cos^2(a/2)+2sin(a/2)cos(a/2)]=sin(a/2)/cos(a/2)=tana/2

Given Tan (π - a) = m ^ 2, the absolute value of COS (π - a) = - cosa, find the value of COS (π + a) Calculation process!

The absolute value of COS (π - a) = - cosa, a ∈ Ⅱ, Ⅲ
tan(π-a)=-tana=m^2 ,sina=-cosam^2
1-cos^2a=cos^2am^4,cos^2a=1/(1+m^4)
cos(π+a)=-cosa=-√[1/(1+m^4)]

Given Tan (π - a) = a square, | cos (π - a) | = - cosa, find the value of 1 / cos (π + a) Be quick

In conclusion, a is the angle of the fourth quadrant, then π + A is the angle of the second quadrant, so sin a = - A ^ 2 cosa