1−2sin(π+2)cos(π+2)=______ .

1−2sin(π+2)cos(π+2)=______ .

1−2sin(π+2)cos(π+2)=
1−2sin2cos2=
(sin2−cos2) 2
∵sin2＞cos2
The original formula = sin2-cos2
So the answer is: sin2-cos2

What is 1-2sin (180 + 2) cos (180-2) equal to under the radical The easier to understand, the better

The original formula = 1-2 [- sin (2)] [- cos (2)] = 1-2sin (2) cos (2) = the square of sin (2) + 2Sin (2) cos (2) + cos (2) = [sin (2) - cos (2)] squared

If cos α + 2Sin α = - radical 5, then Tan α equals () I want to know the specific process

cos^2 a+sin^2 a=1
Cosa + 2sina = - radical 5
Cosa = - root 5 / 5, Sina = - 2 root 5 / 5
So Tana = 2

Cos (10 ° + α) = 2 / 2 root 3 2Sin (α + 15 °) = root 3 Find the degree of acute angle α

Cos (10 ° + α) = root of 2, 3 α = 20 °
2Sin (α + 15 °) = radical 3 α = 45 °

Cos (a) + 2Sin (a) = - radical 5 to find Tan (a)

Cos (a) + 2Sin (a) = - the square of root 5 gives cosa ^ 2 + 4sina ^ 2 + 4sinacosa = 5 (1-sina ^ 2) + 4 (1-cosa ^ 2) + 4sinacosa = 5sina ^ 2-4sinacosa + 4cos ^ a = 0. Suppose cosa is not equal to 0, divide both sides by (tana-2) ^ 2 = 0 of cosa ^ 2

If cos a-2sin a = radical 5, find the value of Tan a

(COS a-2sin a) ^ 2 = 5, the left divided by sin ^ 2 A + cos ^ 2 A is divided by 1. The numerator and denominator are divided by cos ^ 2 A

2Sin + cos = - radical 5 Tan

There is also a known condition: the square of SiNx + the square of cosx = 1. The two formulas form a binary quadratic system, TaNx = SiNx / cosx

Given that the final edge of the angle α crosses the point (- 1, root 3), what is cos α equal to?

-1/2
One edge of the default conditional angle α is required to be parallel to the X axis
tanα=√3/(-1)=-√3
a=2π/3
cos2π/3=-1/2

The known function f (x) = (radical 3 / 2) sin2x - cos ^ 2 x + 1 / 4 (1) Find the minimum positive period; (2) When x is the value, the function takes the maximum value and finds the maximum value

F (x) = (radical 3 / 2) sin2x cos ^ 2x + 1 / 4 = (radical 3 / 2) sin2x - (2cos ^ 2x-1) / 2 + 3 / 4 = (radical 3 / 2) sin2x - (cos2x) / 2 + 3 / 4 = sin ((2x - π / 6)) + 3 / 4 minimum positive period = 2 π / 2 = π when (2x - π / 6) = 2K π + π / 2, the function takes the maximum value, when x = k π + π / 3, the function takes the maximum value

Known function f = [radical 3] / 2 sin2x cos ^ 2 (x) - 1 / 2 (x ∈ R) When x ∈ [(- π) / 12, (5 π) / 12], find the minimum and maximum value of function F 2. Let the corresponding sides of the inner angles a, B, C of △ ABC be a, B, C respectively, and C = radical 3, f = 0. If vector M = (1, Sina) and vector n = (2, SINB) are collinear, then find the value of A.B

（1）
F = [radical 3] / 2 sin2x cos ^ 2 (x) - 1 / 2
=√3/2*sin2x-1/2(1+cos2x)-1/2
=√3/2sin2x-1/2cos2x-1
=sin(2x-π/6)-1
∵x∈[-π/12,5π/12]
∴2x∈[-π/6,5π/6]
2x-π/6∈[-π/3,2π/3]
When 2x - π / 6 = π / 2, the maximum value of F (x) is 0
When 2x - π / 6 = - π / 3, f (x) gets the minimum value - √ 3 / 2-1
(2)
∵f(C)=sin(2C-π/6)-1=0
∴sin(2C-π/6)=1
∵-π/6