The known function f (x) = sinxcosx - √ 3sin? X 1. Find the minimum positive period of F (x) 2. Find the maximum and minimum of F (x) on the interval [0, PI / 2]

The known function f (x) = sinxcosx - √ 3sin? X 1. Find the minimum positive period of F (x) 2. Find the maximum and minimum of F (x) on the interval [0, PI / 2]

The minimum positive period of F (x) = sinxcosx - √ 3 sin? X  = 1 / 2 sin2x - √ 3 / 2 (1 - cos2x)  = 1 / 2 sin2x + √ 3 / 2cos2x - √ 3 / 2  = sin (2x + π / 3) -√ 3 / 2F (x) is 2 π / 2 = π 0 ≤ x ≤ π / 20 ≤ 2x ≤ π π / 3 ≤ 2x + π / 3 ≤ 4 π / 3

The known function f (x) = - √ 3sin? X + sinxcosx Find the value range of function in FX ∈ [0, π / 2]

(x) = - 3 (1-cos2x) / 2 + sin2x / 2 = - √ 3 (1-cos2x) / 2 + sin2x / 2 = - √ 3 / 2 + 3 / 2 + (3 / 2cos2x + sin2x / 2 = - √ 3 / 2 + sin (2x + π / 3) ∵ x ∈ [0, π / 2]  x ∈ [0, π / 2], 2x + π / 3 ∈ [π / 3,4 π / 3], sin (2x + π / 3) ∈ [- √ 3 / 2,1], the value domain is [- √ 3, (3), (3, (3), (1)] the domain is [- √ 3, (3), (3), (3), (1,1], the domain is [-] 3, (2 - √ 3) / 2] Hello, I'm very happy for you

Given the function f (x) = SiNx + cosx, G (x) = 2sinx, the moving line x = t intersects the point P and Q respectively with the image of F (x), G (x), the value range of | pq|is () A. [0，1] B. [0，2] C. [0， 2] D. [1， 2]

It can be seen from the meaning of the title
|PQ|=|f（t）-g（t）|=|sint+cost-2sint|
=|sint-cost|=
2|sin（t-π
4）|
∴0≤|PQ|≤
Two
Therefore, C

Find the definition domain of function y = radical 1-cosx / 2sinx-1 + LG (2cosx + Radix 2)

If the function is meaningful:
(1-cosx) / (2sinx-1) ≥ 0 (1-cosx ≥ 0 always holds)
｛2cosx+√2>0
= = >
{sinx>1/2
{cosx>-√2/2
= = >
{π/6+2kπ

Find the definition domain of the following functions: 1. Y = LG (2sinx radical 2) - 1-2cosx under radical sign

2sinx-√2>0 , 2kπ+π/4

y＝lg(2sinx−1)+ The domain of 1 − 2cosx is______ .

∵ y = LG (2sinx − 1) + 1 − 2cosx ᙽ 2sinx-1 ﹥ 0 ᙽ 1-2cosx ≥ 0; ② from ①, SiNx ﹥ 12, from ② cosx ≤ 12,

Given the function f (x) = 2cosx * sin (x + π / 3) - radical 3 / 2, find the minimum positive period

The minimum positive period is "Pai", which is original = 2cx (1 / 2sx + root 3 / 2cx) - root 3 / 2 = cxsx + root 3 / 2 (Cx) ^ 2 - root 3 / 2 = 1 / 2S (2x) + root 3 / 2C (2x) = s (2X + Pai / 3); therefore, t = 2 Pai / 2 = Pai note: SX: SiNx; CX = cosx; ( )^2:(… ）The second power of

Given vector a = (2cosx, - 1), vector b = (sin (x + π / 3), (radical 3) / 2), the function f (x) = vector a * vector B 1. Find the minimum positive period of F (x); 2. If x belongs to [0, π / 2], find the range of F (x)

f(x)=a*b
=2cosxsin(x+π/3)-√3/2
=sin(2x+π/3)
(1) T=2π/2=π
(2) x∈[0,π/2],y∈[-√3/2,1]

The least positive period of the function FX is obtained by subtracting the square of the root sign 3sinxcosx 2cosx (x belongs to R)

FX = 2 √ 3sinxcosx + 2cos ^ 2 x - 1 = √ 3sin2x + cos2x = 2 (√ 3 / 2 sin2x + 1 / 2 cos2x) = 2Sin (2x + π / 6), so the minimum positive period is π. We suggest you take a look at the formula of double angle

Given the function f (x) = 2 radical 3sinxcosx + 2cosx ^ 2x-1, if the final edges of angles α and β are not collinear, and f (α) = f (β), find the value of Tan (α + β) I've reduced f (x) = 2Sin (2x + π / 6)

Now sin (2 α + π / 6) = sin (2 β + π / 6), so it can only be (1) 2 α + π / 6 = 2 β + π / 6 + integer * 2 π (2) 2 α + π / 6 + 2 β + π / 6 = integer * 2 π + π