Given Tan θ = root 2, find the value of ((2cos square θ / 2) - (sin θ - 1)) / root 2Sin (θ + π / 4)? Given Tan θ = root 2, find the value of ((2cos square θ / 2) - (sin θ - 1)) / Radix 2Sin (θ + π / 4)

Given Tan θ = root 2, find the value of ((2cos square θ / 2) - (sin θ - 1)) / root 2Sin (θ + π / 4)? Given Tan θ = root 2, find the value of ((2cos square θ / 2) - (sin θ - 1)) / Radix 2Sin (θ + π / 4)

According to: 2cos (θ / 2) - 1 = cos θ radical 2Sin (θ + π / 4 = Radix 2 (Radix 2 / 2 * sin θ + Radix 2 / 2 * cos θ) = cos θ + sin θ, therefore: ((2cos square θ / 2) - (sin θ - 1)) / Radix 2Sin (θ + π / 4) = (COS θ - sin θ) / (sin θ + cos θ) = (1-tan θ) / (Tan θ + 1)

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

Cot (sin θ) * Tan (COS θ) > 0 try to point out the quadrant where θ is located Because - 10 Zero

tan(cosθ)>0
If the tangent value is positive, the corresponding angle is indeed in the first three quadrants
But the angle cos θ here is also the cosine value of angle θ,
The cosine values range from [- 1,1] as the number of radians of the angle,
From - 1 radian to 1 radian,
[- 1,0) in the fourth quadrant, (0,1] in the first quadrant, ②
① At the same time
Only 0

According to the following conditions, which quadrant angle θ is determined as (1) sin θ < 0 and Tan θ > 0 (2) cos θ × cot θ < 0

SIN α + | +
--------
- | -
tan a - | +
--------The third quadrant
+ | -
cos a - | +
-------
- | +
Cot A is the same as Tan a in the third or fourth quadrant

If Tan α + cot α = - 25 / 12, and in the second quadrant, sin α + cos α is equal to

one-fifth
Replace Tana with sina / cosa and COTA with cosa / Sina

It is known that α is the fourth quadrant angle and f (α) = sin (π - α) * cos (2 π - α) * Tan (- α + 3 π / 2) / cos (- α - π), (1) simplify f (a) (2) If cos (A-3 π / 2) = 1 / 5, find the value of F (a). (3) if a = 1860 °, find the value of F (a)

If cos (A-3 π / 2) = 1 / 5, then sin α = - 1 / 5, namely cos α = 2 √ 6 / 5 or cos α = - 2 √ 6 / 5, that is, f (a) = - cos α = cos α = 1 / 5, that is, cos α = 2 √ 6 / 5 or cos α = - 2 √ 6 / 5 or cos α = - 2 √ 6 / 5, that is, f (a) = - cos α = cos α = 2 √ 6 / 5 or cos α = - 2 √ 6 / 5, namely, f (a) = - cos α = cos α = 2 √ 6 / 5 or cos α = - 2 √ 6 / 5, that is, f (a) = - cos α = - cos α = - cos α = 2 √ 6, 6/ 5 or =

If Tan θ = 2, then the value of 1 + sin θ cos θ is A,7/3 B,7/5 C,5/4 D5/3 If (sin θ + cos θ) / (sin θ - cos θ) = 2, then the value of sin θ cos θ is A. 3 / 4 B, plus or minus 3 / 10 C, 3 / 10 d, negative 3 / 10 Both questions are multiple choice questions, but please give the process as much as possible,

1. Because Tan θ = 2, sin2 θ = 2 * 2 / (1 + 2 ^ 2) = 4 / 5 (universal formula), so 1 + sin θ cos θ = 1 + (sin2 θ) / 2 = 1 + 2 / 5 = 7 / 5, so choose B2. Because (sin θ + cos θ) / (sin θ - cos θ) = 2, so sin θ + cos θ = 2Sin θ - 2cos θ and sin θ = 3cos θ, then Tan θ = sin θ / cos θ = 3

If sin α = 4 / 5, α is the second quadrant angle, then cos (- α)=

If sin α = 4 / 5, α is the second quadrant angle
Then cos α = - √ [1 - (4 / 5) 2] = - 3 / 5
So cos (- α) = cos α = - 3 / 5

If sin α + cos α = 1 / 5 α is the second quadrant angle sin α - cos α=

Because sin α + cos α = 1 / 5, so (sin α + cos α) ^ 2 = 1 + 2Sin α cos α = 1 / 25, so 2Sin α cos α = - 24 / 25
Because (sin α - cos α) ^ 2 = 1-2sin α cos α = 1 - (- 24 / 25) = 49 / 25
Because α is the second quadrant angle, sin α is positive, cos α is negative, so sin α - cos α is positive
Therefore, the square root of (sin α - cos α) ^ 2 shows that sin α - cos α = 7 / / 5

Cos (π / 6 - α) = 1 / 3, if α is the fourth quadrant angle, find sin (5 π / 6 + α)

Replace with a
cos(π/6-a)=1/3
cos[π-(π/6-a)]=-1/3
cos(5π/6+a)=-1/3
2K π - π / 2, 2K π + π / 3 < 5 π / 6 + a < 2K π + 5 π / 6
So sin (5 π / 6 + a) > 0
sin²(5π/6+a)+cos²(5π/6+a)=1
So sin (5 π / 6 + a) = 2 √ 2 / 3