X belongs to (0, half PAI), compared with cos (SiNx), cosx.sin (cosx) size

X belongs to (0, half PAI), compared with cos (SiNx), cosx.sin (cosx) size

If f (x) = x-sinx, then f (x) is odd function, f '(x) = 1-cos (x) > 0, and f (x) increases monotonically. Because f (0) = 0, then f (x) > 0, when x > 0, f (x) is monotonically increasing

To get the image of the function y = cos (x / 2-pai / 4), simply translate the image of function y = SiNx / 2 to () by () units

Shift (PAI / 2) units to (left) y = SiNx / 2 = cos (x / 2-pai / 2) = cos (PAI / 2-x / 2) shift principle: left plus right minus

It is known that sin α / 2-cos α / 2 = - radical 5 / 5 α is the second quadrant angle, and Tan α / 2 is obtained

Sin α / 2-cos α / 2 = - radical 5 / 5 problem
(sinα/2)^2+(cosα/2)^2=1
sinα/2

It is proved that [1 + sin α) / (1 + cos α)] * [(1 + sec α) / (1 + CSC α)] = Tan α

Left = [(1 + sin α) / (1 + cos α)] * [(1 + 1 / cos α) / (1 + 1 / sin α)]
=[(1+sinα)/(1+cosα)]*[(cosα+1)/cosα]/[(sinα+1)/sinα]
=[(1+sinα)/(1+cosα)]*[(cosα+1)sinα]/[(sinα+1)cosα]
=sinα/cosα
=tanα

Vector α = (COS α, sin α), B = (COS β, sin β) | A-B | 2 root sign 2 / 5 1. Find the value of COS (α - β)? 2. If 0 < α < π / 2, - π < β < 0, sin β = - 5 / 13, calculate the value of sin α?

|A-B ^ 2 = (a-b) ^ 2 = (COSA CoSb) ^ 2 + (Sina SINB) ^ 2 = 2-2cos (a-b), so cos (a-b) = 17 / 25
SINB and CoSb can be calculated by using the formula of double angle and combining with sin ^ 2 (a) + cos ^ 2 (a) = 1

Given the vector a = (COS α, sin α), vector b = (COS β, sin β) / A-B / = two root sign five / 5, find the value of COS (α - β)

a-b=(cosα-cosβ,sinα-sinβ)
|a-b|^2=(cosα-cosβ)^2+(sinα-sinβ)^2=(sinα^2+cosα^2)+(sinβ^2+cosβ^2)-2(cosαcosβ+sinαsinβ)=2-2cos(α-β)=4/5
Cos (α - β) = 3 / 5

It is known that sin (A / 2) + cos (A / 2) = - 3, root number 5 / 5, and 450

Four hundred and fifty

(2-root 2 times cos *, 2 + root 2 times sin *) For graphics!

If this is a coordinate, then you can find that the sum of squares of (abscissa-2) and (ordinate-2) is always 2. Therefore, the graph is a circle with (2,2) as the center and root 2 as the radius

It is known that the period of 3 sin ω xcos ω x-cos 2 ω x (ω 0) is π / 2

The original formula = √ 3sin ω xcos ω x-cos ^ 2 ω x, whose period T = π / 2
The original formula = 2cosx [√ 3 / 2 (sin ω X - (1 / 2) cos ω x]
=2cosωx[sinωxcos(π/6)-cosωxsin(π/6)]
That is, f (x) = 2cos ω x * sin (ω X - π / 6)
It is found that the period of CO x + ω s is ω 2
T=|2π/ω|=π/2
∴ω=4.
Similarly, sin (ω x + 2 π - π / 6) = sin ω (x + 2 π / ω - π / 6)
T=|2π/ω|=π/2
∴ω=4.
The ω = 4 of the function f (x)

The known function f (x) = sinxcosx radical 3cos ^ 2x radical 3

f(x)=sinxcosx-√3cos^2x-√3
=1/2sin2x-√3(1+cos2x)/2-√3
=1/2sin2x-√3/2-√3/2cos2x-√3
=1/2sin2x-√3/2cos2x-3√3/2
=sin2xcosπ/3-cos2xsinπ/3-3√3/2
=sin(2x-π/3)-3√3/2
T=2π/2=π
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