If (absolute value of cos θ) / cos θ + sin θ / (absolute value of sin θ) = 0, try to judge the sign of sin (COS θ) * cos (sin θ) RT. When the final edge of θ is in the second quadrant, the sign is "+"

If (absolute value of cos θ) / cos θ + sin θ / (absolute value of sin θ) = 0, try to judge the sign of sin (COS θ) * cos (sin θ) RT. When the final edge of θ is in the second quadrant, the sign is "+"

I'm sorry for wrong reading, sorry ╮╮╮╮╮╮╮╭╭╮╮╭╭╭╭╮╮╮╮╮╮╮╮╭╭╮╮╮╭╭╭╭╭╭╭╭╭╭╭╭╭╭

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 And because │ cos α / 2 │ = - cos α / 2, cos α / 2 can be obtained

Only the absolute value of a negative number is its opposite number. Therefore, according to cos α / 2 = - cos α / 2, cos α / 2 is obtained

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

If angle a belongs to the second quadrant and | cosa / 2 | = - cos (A / 2), then angle A / 2 belongs to the second quadrant?

Angle a belongs to the second quadrant, 2K π + π / 2

If a is the second quadrant angle and cosa / 2 = - 1-cos square under the root [(л - a) / 2], then what quadrant angle is a / 2

third quadrant
If a is the second quadrant angle, then a ∈ (90 ℃＋ K × 360 °, 180 ° + K × 360 °),
Then a / 2 ∈ (45 ° C + K × 180 ° C, 90 ° C + K × 180 ° C) is a / 2 ∈,
cosa/2=-√{1-cos²[(π-a)/2]}, => cosa/2≤0,
=>A / 2 ∈ (225 ° C + K × 360 ° o, 270 ° C + K × 360 ° o), that is, a / 2 is in the third quadrant

If a belongs to the second quadrant and COS A / 2 = - cos A / 2, then a / 2 angle belongs to the second quadrant A. The first quadrant angle B, the second quadrant angle c, the third quadrant angle D, and the fourth quadrant angle

A should be an angle greater than 360 degrees, such as 480 degrees, in the second quadrant, then a / 2 is in the third quadrant

If θ is the second quadrant angle, and cos θ / 2 = - cos θ / 2, then what quadrant does θ / 2 belong to

If θ is the second quadrant angle, then 2K π + π / 2

It is proved that 1 + sec α + Tan α / 1 + sec α - Tan α = 1 + sin α / cos α

It is proved that: 1 + sec α + Tan α / 1 + sec α - Tan α = (1 + 1 / cos α + sin α / cos α) / (1 + 1 / cos α - sin α / cos α) = (COS α + 1 + sin α) / (COS α + 1-sin α) = (COS α + 1 + sin α) = (COS α + 1 + sin α) / [(COS α + 1) σ - Sin 2] = (2cos α + 2 + 2Sin α + 2

To prove the identity: (Tan a + sec A-1) / (Tan a - sec a + 1) = (1 + sin a) / cos a

Let 2 β = α left = (1 + sec α + Tan α) / (1 + sec α - Tan α) = (COS α + sin α + 1) / (COS α - sin α + 1) = (COS α + 1 + sin α) / (COS α + 1-sin α) = (Cos2 β + 1 + sin2 β) / (Cos2 β + 1-sin2 β) = [2cos ^ 2 (β) + 2Sin β cos β] / [2cos ^ 2 (...)

It is proved that Tan θ sin θ = sec θ - cos θ)

tana=sina/cosa
seca=1/cosa
Left = Sina * Sina / cosa
Right = (1-cosa * COSA) / cosa = Sina * Sina / cosa
So left and right are equal