If f (TaNx) = sin2x, then the value of F (- 1) is () A. -sin2B. -1C. 12D. 1

If f (TaNx) = sin2x, then the value of F (- 1) is () A. -sin2B. -1C. 12D. 1

Because Tan (K π - π 4) = - 1, (K ∈ z), f (- 1) = f [Tan (K π - π 4)] = sin2 (K π - π 4) = sin (2k π - π 2) = - sin π 2 = - 1

If f (TaNx) = sin2x, then the value of F (- 1) is ()
A. 1B. -1C. 12D. 0

∵ f (TaNx) = sin2x, ∵ f (- 1) = f (tan135 °) = sin270 ° = - 1

Given 2F (- TaNx) + F (TaNx) = sin2x, find f (x)

Let x = TaNx, then sin2x = 2sinx * cosx = 2sinx * cosx / ((SiNx) ^ 2 + (cosx) ^ 2)
=2tanx / ((TaNx) ^ 2 + 1) numerator and denominator divide by (cosx) ^ 2
So the original formula is 2F (- x) + F (x) = 2x / (x ^ 2 + 1) -- (1)
Let x = - x, so 2F (x) + F (- x) = - 2x / (x ^ 2 + 1) -- (2)
Simultaneous equations (1), (2)
The solution is f (x) = - 3x / (x ^ 2 + 1)

If f (TaNx) = sin2x, then the value of F (3 / 4) is

tanx=3/4
sinx/cosx=3/4
sinx=(3/4)cosx
Bring in the identity Sin & # 178; X + cos & # 178; X = 1
cos²x=16/25
That is, cosx = ± 4 / 5, SiNx = 3 / 5
2sinxcosx=±24/25
f(3/4)=±24/25

If f (TaNx) = sin2x, then f (radical 2 + 1) =

1/sin2x=1/(2sinxcosx)
=(sin²x+cos²x)/(2sinxcosx)
Divide up and down by cos & sup2; X
=(sin²x/cos²x+1)/(2sinx/cosx)
=(tan²x+1)/(2tanx)
So f (TaNx) = sin2x = 2tanx / (Tan & sup2; X + 1)
So f (x) = 2x / (X & sup2; + 1)
So f (√ 2 + 1) = 2 (√ 2 + 1) / (2 + 2 √ 2 + 1 + 1) = 1

If f (TaNx) = sin2x, then f (- 1) =?

sin2x=2sinxcosx=2tanxcosx^2=2tanx/(1+tanx^2)
f(tanx)=sin2x=2tanx/(1+tanx^2)
f(x)=2x/(1+x^2)
f(-1)=-2/2=-1

Let sin2x = 2 / 3, then the value of TaNx + Cotx is

tanx+cotx=sinx/cosx + cosx/sinx =1/( sinx.cosx )=2/sin2x=3

Given the function f (x) = 3cos & # 178; X + 2cosxsinx + Sin & # 178; X (1), find the analytic expression of the function (2) and write F
Given the function f (x) = 3cos & # 178; X + 2cosxsinx + Sin & # 178; X (1), find the analytic expression (2) and write the monotone increasing interval of F (x)

F (x) = 3cos & # 178; X + 2cosxsinx + Sin & # 178; X = 2cos & # 178; X + 2sinxcosx + 1 = cos2x + sin2x + 2 = √ 2Sin (2x + π / 4) + 2 the minimum positive period is t = 2 π / 2 = π monotone interval, monotone increasing interval is 2x + π / 4 ∈ [2K π - π / 2,2k π + π / 2]

The minimum positive period of function f (x) = sin (π / 6-2x) - 1 / 2

f(x)=sin(π/6-2x)-1/2
T=2π/|ω|=π

Given the function f (x) = - X & # 178; + 2x, G (x) = 1 / X
Find f (f (1)), G (1), f (g (x))

1,1,(-1/x^2)+(2/x)