Simplify e ^ in (1 / 2) function

Simplify e ^ in (1 / 2) function

e^In(1/2)=1/2

Simplification of mathematical functions 2cos10°-sin20°/cos20°=? sin40°+2cos70°/sin50°=?

It's OK to change cos10 into cos (30-20). The root is 3
Change cos70 into sin20, sin20 into sin (60-40) is OK, root 3

According to the following conditions: (what I want to know is whether it is easier to replace the problem directly or to simplify it first and then bring it in) f(x)=sin(x+π/4)+2sin(x-π/4)-4cos2x+3cos(x+3π/4) (1)x=π/4 (2)x=3π/4 If you simplify first, how should you simplify it?

General trigonometric function problem known angle to find the value of the function is first simplified and then replaced by simple, but this problem directly into the simple

On simplifying function in Senior High School Given Tana = radical 2 / 2, find the value of COS ^ 2 (Wu - a) + sin (Wu + a) * cos (Wu - a) + 2Sin ^ 2 (a - Wu)

[cos(π-a)]^2+sin(π+a)cos(π-a)+2[sin(a-π)]^2
=(-cosa)^2+(-sina)(-cosa)+2(-sina)^2
=(cosa)^2+sinacosa+2(sina)^2
=[(cosa)^2+sinacosa+2(sina)^2]/[(cosa)^2+(sina)^2]
=[1+tana+2(tana)^2]/[1+(tana)^2]
=(1 + radical 2 / 2 + 1] / [1 + 1 / 2] = (4 + Radix 2) / 3

Simple simplification questions! Is 6 + 2 radical 7 / 6 the simplest?

no
Original formula = 6 + 2 * (Radix 42 / 6)
=6 + (Radix 42 / 3)
This is the simplest

For example, how to simplify the function in question 1 1. (2011 · new curriculum standard nationwide) let the minimum positive period of the function f (x) = sin (ω x + φ) + cos (ω x + φ) be π, and f (- x) = f (x), then () A. F (x) decreases monotonically B. F (x) decreases monotonically C. F (x) increases monotonically D. F (x) increases monotonically 2. (investigation in Xi'an, 2013) Let f (x) be an odd function defined on R, and when x ≥ 0, f (x) monotonically decreases. If X1 + x2 > 0, then the value of F (x1) + F (x2) () A. Constant positive value B. equal to zero C. Constant negative D. cannot determine positive and negative Please write down the process,

One
f(x)=sin(ωx+φ)+cos(ωx+φ)
=√2[√2/2*sin(ωx+φ)+√2/2cos(ωx+φ)]
=√2sin(wx+φ+π/4)
From 2 π / w = π, w = 2
∴f(x)=√2sn(2x+φ+π/4)
∵f(-x)=f(x)
/ / F (x) is an even function, and the image is symmetric about the y-axis
Then when x = 0, f (0) is the maximum
That is sin (φ + π / 4) = ± 1
∴φ+π/4=kπ+π/2,kπ∈Z
∴φ=kπ+π/4,k∈Z
When k = 0, take φ = π / 4
∴f(x)=√2sin(2x+π/2)=√2cos2x
As for the option, the condition input is not complete and cannot be judged
Two
Let f (x) be an odd function defined on R,
When x ≥ 0, f (x) decreases monotonically,
So x0
∴x1>-x2
∴f(x1)

The known function f (x) = cos2x / [sin (π / 4-x)]

cos2x=sin(π/2-2x)=2sin(π/4-x)cos(π/4-x)
cos2x/[sin(π/4-x)]
=2sin(π/4-x)cos(π/4-x)/[sin(π/4-x)]
=2cos(π/4-x)/

Given the function f (x) = (sin2x cos2x + 1) / (2sinx), find the domain of F (x) The known function f (x) = (sin2x cos2x + 1) / 2sinx (1) Find the definition domain of F (x) (2) Let α be an acute angle and Tan α = 4 / 3, then find the value of F (α)

1. SiNx ≠ 0,  x ≠ K π.  Tan α = 4 / 3, sin? α + cos? α = 1sin α / cos

The logic function f = AB + a'c + b'c is simplified by formula method,

The logic function f = AB + a'c + b'cf = AB + a'c + b'c = AB + a'c (B + B ') + b'c (a + a') = AB + a'bc + a'b'c + a b'c + a'b'c = AB + a'bc + a b'c + a'b'c = AB + (a'B + ab ') C + a'b'c = AB + C + a'b'c = AB + C + a'b'c = AB + C

The following logic functions are simplified by formula method

F = AB'+BD+CDE+A'D
= AB'+(A'+B)D+CDE
= (A'+B)'+(A'+B)D+CDE
=(A'+B)'+D+CDE
=(A'+B)'+D(1+CE)
= (A'+B)'+D
=AB'+D