-1 2 + (- 2) 3 × 1-27 cube root × - 1 / 3 absolute value + 2 × - root 4

-1 2 + (- 2) 3 × 1-27 cube root × - 1 / 3 absolute value + 2 × - root 4

-1 2 + (- 2) 3 × 1-27 cube root × - 1 / 3 absolute value + 2 × - root 4
=-1 + (- 8) × 1-3 × 1 / 3 + 2 △ 2
=-1-1-1+1
=-2

Given the absolute value A-4 + radical B-9 = 0, find the value of a + B in B 2 ^ [(B in a-b) · (a + B in ab)]

|A-4 | + radical B-9 = 0
a-4=0,b-9=0
a=4,b=9
Original formula = B ^ 2 / (a + b) △ AB ^ 2 / (a-b) * (a + b)
=b^2/(a+b)*(a-b)*(a+b)/ab^2
=(a-b)/a
=(4-9)/4
=-5/4

(1) Given TaNx = 2, find cosx + SiNx Value of cosx − SiNx (2) It is known that SiNx + cosx = 2 3. Find the value of sin4x + cos4x

（1）∵tanx=2，
The original formula = 1 + TaNx
1−tanx=1+2
1−2=-3；
(2) Square the two sides of the known equation to get: (SiNx + cosx) 2 = 1 + 2sinxcosx = 4
9, namely sinxcosx = - 5
18，
Then sin4x + cos4x = 1-2sin2xcos2x = 1-2 × 25
18×18=137
162．

If sin (30 ° + a) = (radical 3) / 2, then the value of COS (60 ° - a) is known,

cos(60°-a)
=cos[90°-(30°+a)]
=sin(30°+a)
=√3/2

Help to simplify {radical (1-sin4 * Cos4)}

{radical (1-sin4 * Cos4)}
First, SiN4 * Cos4 is converted into 1 / 2 * 2 * sin4cos4
In this way, it can be converted into {radical (1-1 / 2 * sin (2 * 4))}
The final result is {radical (1-1 / 2 * sin8)}

Given the vector a = (COS θ, sin θ) vector b = (√ 3, - 1), then the maximum and minimum values of | 2a-b | are respectively?

B = 3 + 1 = 4 AB = 4 + 4 + 4-4 + 4-4-4-4-4-4-4-4-4-4-4 (3 / 3 / 2cos θ - 1 / 2Sin θ) = 8 + 8 + 8sin (θ - 1 / 2Sin θ) = 8 + 8sin (θ - π / 3) the maximum value is 8 + 8 = 16, the minimum value is 8-8 = 8 = 8 = 8 = 8 = 8, the maximum value is 8 + 8 = 16, the minimum value is 8-8 = 8 = 8 = 0, so, so | 2a-b |, B | 4 | 4 | 4-4 | 4 | 4 | 4 | 4 | 4 | 4 it's a good idea

Known vector a=(cosθ，sinθ)， B= ( 3,1)| A- The maximum value of b|is () A. 1 B. Three C. 3 D. 9

A kind of
A-
b|2= 
A2+
b2-2
a•
b=1+4-2（
3cosθ+sinθ）=5-4sin（θ+π
3) When 4sin (θ + π)
3) When = - 1|
A-
B|2 has a maximum value of 9|
A-
The maximum value of b|is 3
Therefore, C is selected

Given the vector a = (COS θ, sin θ), vector b = (1 / 2, root sign 3 / 2), then the maximum value of | 3 vector A-4 vector b|is

3a-4b = (3cos θ, 3sin θ) - (2,2 radical sign 3) = (3cos θ - 2,3sin θ - 2 radical sign 3)
The square of the module is (3cos θ - 2) ^ 2 + (3sin θ - 2 radical 3) ^ 2 = 25-12sin (θ + 30 °)
So the maximum value is 37
Therefore, the maximum value of | 3 vector A-4 vector B | is the root 37

Given that α and β are acute angles, Tan α = 1 / 7, sin β = root 10 / 10, find the value of Tan (α + 2 β) Sin β = root 10 / 10, so cos β = 3 root 10 / 10 sin2β=2sinβcosβ=3/5 Sin β = root 10 / 10

∵cos2β=4/5,(sina)^2+(cosa)^2=1∴sin2β=3/5,
∴tan2β=sin2β/cos2β=3/4

Given that sin α = root 5 / 5, sin (α - β) = negative root 10 / 10, α, β are acute angles, calculate β

sinβ=-sin（-β）=-sin（α-β-α）=-[sin(α-β)cosα-cos(α-β)sinα]=cos(α-β)sinα-sin(α-β)cosα ①
α. All β are acute angles, so cos α = 2 root number 5 / 5, cos (α - β) = 3 root number 10 / 10
Substituting into Formula 1, we can get: sin β = radical 2 / 2, so β = π / 4