If f (x) = 2tanx + 2sin2x 2−1 sinx 2cosx 2, then f (π) 12) The value of is______ .

If f (x) = 2tanx + 2sin2x 2−1 sinx 2cosx 2, then f (π) 12) The value of is______ .

∵f（x）=2tanx+2sin2x
2−1
sinx
2cosx
2=2tanx+cosx
One
2sinx=2tanx+2cotx=4
sin2x，
Then f (π)
12）=4
sinπ
6=8，
So the answer is: 8

If f (x) = 2tanx + (2Sin 2 (x / 2) - 1) / (sin (x / 2) cos (x / 2), then f (π / 12) =? I see the same question in knowing, and the answer is 8 The teacher told me that it was too fast to sit on the wrong side. The teacher said - 4 √ 3 f(x)=2tanx+(2sin²(x/2)-1)/(sin(x/2)cos(x/2) f(x)=2tanx-2cosx/sinx f(x)=2(sinx/cosx-cosx/sinx) f(x)=2(sinx^2+cosx^2)/(cosxsinx) F (x) = 4 / sin2x - how did this come about? F (π / 12) = 4 / (1 / 2) = 8 -- how did this come about?

Identity sin? X + cos? X = 1
This is very useful. Remember it
And sin2x = 2sinxcosx
SO 2 (sin? X + cos? X) / sinxcosx
=4×1/(2sinxcosx)
=4/sin2x
f(x)=4/sin2x
So f (π / 12) = 4 / sin (2 × π / 12)
=4/sin(π/6)
=4/(1/2)
=8

If f (x) = 2tanx + 2sin2x 2−1 sinx 2cosx 2, then f (π) 12) The value of is______ .

∵f（x）=2tanx+2sin2x
2−1
sinx
2cosx
2=2tanx+cosx
One
2sinx=2tanx+2cotx=4
sin2x，
Then f (π)
12）=4
sinπ
6=8，
So the answer is: 8

F (x) = 2Sin (2x pi / 3) + 1, X belongs to [pi / 4, PI / 2]. Find the maximum and minimum values of F (x)

Pi/4=

f(x)=2sin(pi/2x+pi/3),f(1)+f(2)+…… +The value of F (2009)

1, the period is 4, f (1) + F (2) + F (3) + F (4) = 0, so f (x) = f (2009) = f (1) = 1

Let f (x) = SiNx + 2Sin (π / 4 + X / 2) cos (π / 4 + X / 2). If f (a) = √ 2 / 2, a ∈ (- π / 2,0), find the value of A Please tell me the answer and solution process! Thank you! If sin (x / 2) = 4 / 5, X ∈ (π / 2, π), f (x) =?

The formula F (x) = SiNx + sin [2 (π / 4 + X / 2)] = SiNx + sin (π / 2 + 2 + x) = SiNx + SiNx (π / 2 + 2 + x) = SiNx + cosx = 2 (√ 2 / 2sinx + √ 2 / 2cosx) = √ 2 (COS π / 4sinx + sin π π / 4 cosx x x) = √ 2Sin (x + π / 4) f (a) = √ 2 / 2, namely, sin (x + π / 4) = 1 / 2A ∈ (- π / 2,0) a + π / 2,0) a + π / 2,0) a + π / 2,0) a + π / 2,0) a + π / 2,0) a + π / 2,4 (- π / 4, π / 4) so a + π

If f (x) = SiNx + cos π / 4, then f '(π / 4) =?

If your topic is right, then
f'(x)=cosx
Qi
F '(π / 4) = cos π / 4 = radical 2 / 2
If f (x) = SiNx + cosx / 4
f'(x)=cosx-sinx/4
Substituting x = π / 4

F (x) = SiNx, find {f (1 + H) - f (1)} / h. how to get {2Sin H / 2cos 2 + H / 2} / h I did not finish high school, many do not understand, hope to give a detailed process, or point out the use of knowledge

[Sinh] / Sinh (1) = {Sinh (1) / Sinh (1) / Sinh (2) = {Sinh (1) / Sinh (1) / Sinh (2) = {Sinh (1) / Sinh (2) = {Sinh (1) / Sinh (2)}

If 0 ≤ α≤ π, the function f (x) = 2Sin (x + α) cos (x + α) - 2 √ SiNx If 0 ≤ α≤ π, the function f (x) = 2Sin (x + α) cos (x + α) - 2 √ 3sin ^ 2 (x + α) + √ 3 (x belongs to R) is even function (1) find the value of α and the minimum positive period of the function (2) finding the maximum value of a function and the X set that makes the function get the maximum value

f(x)=2sin(x+α)cos(x+α)-2√3sin^2(x+α)+√3
=sin2x+2a)+√3/cos(2x+2a)
=2sin(2x+2a+π/3)
It's even function
a=π/12
f(x)=2cos2x
T=π
Maximum value of function 2
x=kπ

Simplify 2Sin ^ 2 [(π / 4) + x] + radical 3 (sin ^ x-cos ^ x) - 1

2Sin ^ 2 [(π / 4) + x] + radical 3 (sin ^ x-cos ^ x) - 1
=-(1-2sin^2[(π/4)+x)-√3cos2x
=-cos(π/2+2x)-√3cos2x
=sin2x-√3cos2x
=2[1/2*sin2x-√3/2*cos2x]
=2sin(2x-π/3)