What is the problem solving process of (SiNx + sin2x) / (1 + cosx + cos2x) = TaNx

What is the problem solving process of (SiNx + sin2x) / (1 + cosx + cos2x) = TaNx

SiNx + sin2x = SiNx + 2sinxcosx = SiNx (1 + 2cosx) 1 + cosx + cos2x = 2cosx ^ 2 + cosx = cosx (1 + 2cosx)

Given the function f (x) = - radical 3sin ^ 2x + sinxcosx. (2) find the value range of function f (x) on (0, π / 2) (2) ∵ 0 < x < π/2 ∴ π/3 < π/3 + 2x < π + π/3 ν - √ 3 / 2 < sin (π / 3 + 2x) ≤ 1, how does this step come out The result is - √ 3 < - √ 3 / 2 + sin (π / 3 + 2x) ≤ - √ 3 / 2 + 1 That is - √ 3 < y ≤ - √ 3 / 2 + 1 The value range of the function on (0, π / 2) is (- √ 3, - 3 / 2 + 1]

Let π / 3 + 2x be t, and the second formula becomes π / 3
Homework help users 2017-09-20
report

Given the function f (x) = - √ 3sin ^ 2x + sinxcosx, let α∈ (0, π), f (α / 2) = 1 / 4 - √ 3 / 2,

f(x)=-√3sin^2x+sinxcosx
f(x)=-√3(1-cos2x)/2+sin2x/2
f(x)=-√3/2+√3/2*cos2x+1/2*sin2x
f(x)=-√3/2+sin60*cos2x+cos60*sin2x
f(x)=-√3/2+sin(60+2x)
f（α/2）=1/4-√3/2
f(α/2)=-√3/2+sin(30+α)=1/4-√3/2
sin(30+α)=1/4
sin(30+α) = sin30cosα+cos30sinα
1/4 = 1/2cosα+√3/2sinα
1/2=cosα+√3sinα
1/2=√(1-sin^2α)+√3sinα
(1/2-√3sinα)^2=1-sin^2α
1/4-2√3sinα+4sin^2α=1
4sin^2α-2√3sinα-3/4=0
4sin^2α-2√3sinα+3-3-3/4=0
(2sinα-√3)^2-15/4=0
(2sinα-√3-√15/2)(2sinα-√3+√15/2)=0
2sinα-√3-√15/2=0
2sinα-√3-√15/2=0
Sin α = √ 3 / 2 + √ 15 / 4 > 1 (omitted)
(2sinα-√3+√15/2)=0
sinα=√3/2-√15/4

The known function f (x) = cos 2 x-sinxcosx - ψ (1) Find the minimum positive period sum range of function f (x), (2) if f (A / 2) = (3 times the root sign 2) / 10, find the value of sin2a

(1) (1 + cos2x) / 2 - (1 / 2) sin2x-1 / 2 (1 / 2) sin2x-1 / 2 = (1 / 2) (cos2x-sin2x) = (√ 2 / 2) [(2 / 2) cos2x - (√ 2 / 2) sin2x - (√ 2 / 2) sin2x] = (√ 2 / 2) sin (π / 4-2x) therefore, the minimum positive period is t = 2 π / 2 = π, the domain is [- √ 2 / 2, √ 2 / 2] (2) f (A / 2) = (√ 2 / 2) sin (π / 4-A) = 3 3 3 (3) f (A / 2) = (2 / 2 / 2 / 2) sin (π / 4-A) = 3 3 3 3 3 3 2 / 10, so

Solve the known function f (x) = sinxcosx + cos? X Find the maximum and minimum value of the periodic function of the function and obtain the maximum and minimum value of the set function of X monotone increasing interval!

f（x）＝sinxcosx＋cos²x=sin2x/2+cos2x/2+1/2
=1 / radical (2) * sin (2x + pi / 4) + 1 / 2
T=2pi/w=pi
F (x) max = radical (2) / 2 + 1 / 2 at this time, 2x + pi / 4 = pi / 2 + 2K * PI ν x = pi / 8 + k * PI
F (x) min = - radical (2) / 2 + 1 / 2 at this time, 2x + pi / 4 = - pi / 2 + 2K * PI ν x = - 3pi / 8 + k * PI
When - pi / 2 + 2K * pi=

Given the function f (x) = cos? X + sinxcosx. (1) find the maximum value of function f (x); (2) In △ ABC, ab = AC = 3, the angle a satisfies f (A / 2 + π / 8) = 1, and calculates the area

(1)
f(x)＝(1+cos2x)/2+(1/2)sin2x
＝(√2/2)sin(2x+π/4)+1/2.
When sin (2x + π / 4) = 1,
y|min＝(1+√2)/2.
(2)
f(A/2+π/8)＝1,
∴(√2/2)sin[2(A/2+π/8)+π/8]+1/2＝1
→sin(A+3π/8)＝√2/2
→A+3π/8＝3π/4
→A＝3π/8.
The triangle area s = (1 / 2) · 3 · 3 · sin (3 π / 8)

The known function y = cos? X + √ (3) sinxcosx + 1 / 2, X ∈ R 1) Determine the period of the function and write the monotone interval of the function 2) Find the maximum value of the function, and find the set of X when y takes the maximum value 3) The image of this function can be obtained from the image of y = SiNx. (x ∈ R) Can you answer me... Please

When y = cos 2x + √ 3sinxcosx + 1 / 2 = 1 / 2 (cos2x + 1) + √ 3 / 2 sin2x + 1 / 2 = √ 3 / 2 sin2x + 1 / 2 cos2x + 1 = sin2xcos π / 6 + cos2xsin π / 6 + 1 = sin (2x + π / 6) + 1 minimum positive period = 2 π / 2 = π 2x + π / 6 ∈ (2k π - π / 2,2k π + π / 2)

Let F X cos (π / 3 + x) cos (π / 3-x) - sinxcosx + 1 / 4 find monotone increasing interval

Let f (x) = g (x) = g (x) + H (x) + 1 / 4G (x) = cos (π / 3 + x) cos (π / 3-x) = [cos π / 3cosx-sin π / 3sinx] [cos π / 3cosx + sin π / 3sinx] [cos π / 3cosx + sin π / 3sinx] = [1 / 2cosx - √ 3 / 2sinx] [1 / 2cosx + √ 3 / 2sinx] = 1 / 4cos x-3 / 4sin / x = 1 / 4cos X / 3 / 4sin / x = 1 / 4cos / x-3 / 3 / 4 (1-cos) x-3 / 4 (1-cos) x-3 / 4 (1-cos) 1-cos (17) 1-cos (17) 1

Given the function f (x) = 1 + sinxcosx, G (x) = cos ^ 2 (x + π / 12), (1) Let x = x0 be a symmetric axis on the image of function y = f (x), and find the value of G (x0). (2) make h (x) = f (Wx / 2) + G (Wx / 2), (W is greater than 0), and is the maximum value of W of increasing function on the interval [- 2 π / 3, π / 3]

(1)f(x)=(1/2)sin2x+1,x0=kπ+π/4.g(x0)=[cos(kπ+π/4+π/12)]^2=[cos(kπ+π/3)]^2=(+-cosπ/3)^2=1/4.(2)h(x)=(1/2)sinwx+1+[cos(wx/2+π/12)]^2=(1/2)sinwx+(1/2)cos(wx+π/6)+3/2=(1/2)sinwx+(1/2)coswxcosπ/...

Given the function f (x) = cos ^ 2x + sinxcosx (x ∈ R), find the value of F (π / 8)

f(x)=cos^2x+sinxcosx=(1+cos2x)/2+0.5sin2x
Therefore, f (π / 8) = (1 + cos π / 4) / 2 + 0.5sin π / 4 = 0.5 + (radical 2) / 2