The value of the vector sin (2 / 2) of the given function sin (2) x (1 / 2) is defined

The value of the vector sin (2 / 2) of the given function sin (2) x (1 / 2) is defined

f(x)=sin(x+ pai/6)+2sin pai/2*sinx/2=sin(x+ pai/6)+sinx=√3 * sin(x+pi/3)
Zero

The function y = SiNx + cosx, X belongs to the range of [0, Pai]

y=sinx＋cosx
=√2sin(x＋π/4)
Because x ∈ [0, π], then x + π / 4 ∈ [π / 4,5 π / 4], then sin (x + π / 4) ∈ [- √ 2 / 2,1],
Then y ∈ [- 1, √ 2]

Given that the value range of F (x) = a (2Sin ^ 2 x / 2 + SiNx) + B in the interval [0, Pai] is [2,3]

The value range of F (x) = a (2Sin ^ 2 x / 2 + SiNx) + B on the interval [0, Pai] is [2,3],
F (x) = a (1-cosx + SiNx) + B = a radical 2Sin (x - π / 4) + A + B; X - π / 4 ∈ [- π / 4,3 π / 4]
a> When 0, a radical 2 + A + B = 3, & B = 2; a = radical 2-1, B = 2;
A

It is known that f (x) = root sign 3sin2x-2sin square x + 2, X belongs to R (1) to find the maximum value of function f (x) and the corresponding value range of X (2) Draw the image of y = f (x) on the closed interval 0, π Just a rough image

f(x)=√3sin2x-(1-cos2x)+2
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
The maximum value is 2 + 1 = 3
In this case, 2x + π / 6 = 2K π + π / 2
X = k π + π / 6, where k is an arbitrary integer

If the point P (1, - radical 3) is on the final edge of the angle α, find the value of F (a)

Sina = - radical 3 / 2, cosa = 1 / 2F
F (a) = radical 3 * (2 * Sina * COSA) - 2 * (Sina) ^ 2 = radical 3 * [2 * (- radical 3 / 2) * (1 / 2)] - 2 * (- radical 3 / 2) ^ 2
=-3/2-3/2= -3

Let f (x) = sinxcosx root 3cos (π + x) * cos if the function y = f (x) shifts π / 4 units to the right and 3 / 2 units to the upper Get the function y = g (x) and find the maximum value of G (x) on [0, π / 4]

It is reduced to y = sin (2x + π / 3) + radical 3 / 2
G (x) = sin2 [(x + π / 6) + π / 4)] + radical 3
G (x) = sin (2x + 5 π / 6) + radical 3
When x is on [0, π / 4], 2x + 5 π / 6 is in [5 π / 6,8 π / 6]
Its maximum value is root 3 + 1 / 2
The translation method is: add and subtract from the top, add and subtract from the left and right

The known function f (x) = sinxcosx - (radical 3) cos ^ 2 + (radical 3) / 2 When 0 ＜ = x ＜ = π / 2, the value range of FX is 0 ＜ = x ＜ = π / 2

2cos^2-1=cos2x
cos^2=(1+cos2x)/2
F (x) = sinxcosx - (radical 3) cos ^ 2 + (Radix 3) / 2
=Sin2x / 2-radical 3 * (cos2x + 1) / 2
=Sin2x / 2-radical 3 * / 2 * cos2x
=sin2xsin30-cos2xcos30
=-cos(2x+30)
=cos(2x-150)
1, the minimum positive period π
Image symmetry center coordinate x = 5 π / 12
2 when 0 ＜ = x ＜ = π / 2
FX range [- √ 3 / 2,1]

The known function f (x) = root sign 3sinxcosx + cos ^ 2 + m, where m is a constant. Find the minimum positive period of F (x), monotonically increasing interval, all symmetric axis equations, range of values I'm asking again~ Come on chenyuwei1994, 0.5√3sin2x+0.5（1+cos2x）+m =sin（2x+1/6π）+m I think there is something wrong with this step~ I can't find you in Baidu Hi~ So here it is~

According to the formula of double angle and the formula of decreasing power: F (x) = √ 3 / 2sin22x + 1 / 2cos2x + 1 / 2cos2x + 1 / 2 + m, then according to the auxiliary angle formula: F (x) = sin (2x + π / 6) + 1 / 2 + m, so t = π 2x + π / 6 ∈ [2K π - π / 2,2k π + π / 2] (K ∈ z) x ∈ [K π - π / 3, K π + π / 6] (K ∈ z) monotone interval is [K π - π / 3, K K π + π / 6] (K ∈ z) monotone interval is [K π - π / 3, k k k k k k k k π / π +

What is the minimum positive period of the function y = SiNx / [SiNx + 2Sin (x / 2)]?

From F (x) = SiNx / (SiNx + 2sinx / 2), and SiNx = 2Sin (x / 2) * cos (x / 2), we obtain the following results
f(x)
=[2sin(x/2)*cos(x/2)]/[2sin(x/2)*cos(x/2)+2sin(x/2)]
=cos(x/2)/[cos(x/2)+1]
So 1 / F (x) = 1 + 1 / cos (x / 2)
That is [1 / F (x)] - 1 = 1 / cos (x / 2)
Because the period of the function y = 1 / cos (x / 2) is 4 π
[1 / F (x + 4 π)] - 1 = [1 / F (x)] - 1, so the period of the original function is 4 π

F (x) = sin2x - Double radical 3sin ^ x + radical 3 + 1, find the minimum positive period and single increasing interval

f(x) = sin2x - 2√3 sin²x + √3 +1
= sin2x - √3 (1 - cos2x) + √3 + 1
= sin2x + √3 cos2x + 1
= 2 sin(2x + π/3) + 1
Minimum positive period T = 2 π / 2 = π
- π/2 + 2kπ < 2x+ π/3 < π/2 + 2kπ
The increasing interval (- 5 π / 12 + K π, π / 12 + K π) is obtained