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FX = 2sinx (cosx SiNx) minimum positive period Monotonically increasing interval
f（x）=2sinx（cosx-sinx）
=2sinxcosx-2（sinx）^2
=sin2x+1-2（sinx）^2-1
=sin2x+cos2x-1
=√2sin(2x+π/4)-1
The minimum positive period T = 2 π / 2 = π
2Kπ-π/2≤2x+π/4≤2Kπ+π/2
Then the monotone increasing interval is [K π - 3 π / 8, K π + π / 8] (k is an integer)
It is known that SiNx + cosx = - 7 / 13,0 Math homework help users 2016-12-02 report Use this app to check the operation efficiently and accurately!
Because SiNx + cosx = - 7 / 13, ①
The square of both sides gives 1 + 2sinxcosx = 49 / 169,
2sinxcosx=-120/169＜0,
So (SiNx cosx) ^ 2 = 1-2sinxcosx = 289 / 169,
From 0 ﹤ x ﹤ π, sinxcosx ﹤ 0, we obtain π / 2 ﹤ x ﹤ π,
So SiNx > 0, cosx < 0, SiNx cosx = 17 / 13, ②
SiNx = 5 / 13 is obtained from ① and ②,
cosx=-12/13 ,
So cosx + 2sinx = - 12 / 13 + 10 / 13 = - 2 / 13
Given SiNx + cosx = - 7 / 13, find the value of cosx + 2sinx
(sinx)^2+(cosx)^2=1
sinx*cosx=[(sinx+cosx)^2-((sinx)^2+(cosx)^2)]=-120/169
So SiNx and cosx are the two roots of the equation a ^ 2 + 7a / 13-120 / 169 = 0, - 15 / 13 and 8 / 13
So there are two possibilities - 24 / 13 and 1 / 13
Known - π / 2
(sinx+cosx=)^2=1+2sinxcosx=1/25
2sinxcosx=-24/25
sinxcosx=-12/25
(sinx-cosx) ^2=1-2 sinxcosx
=1+2*24/25=49/25
-π/20
SiNx cosx < 0
sinx-cosx=-7/5
The image of y = 2Sin x on [0,2 π] by "five point method"
Since it is a five point method, the five points are as follows:
x 0 π/2 π 3π/2 2π
y 0 2 0 －2 0
Make the following function image Y = 2x y = - 2x y = 3x x = - 3x four
List when x = 1, y = several. Make a coordinate system, trace points and then connect lines
What are the two basic methods for making function images
1. General method: list, trace point and connect line;
2. Based on the correlation between images, it is obtained by translation, stretching and so on
Make a function image of y = x ^ 2-2 | x | - 1
If f (x) = x ^ 2-2x-1, it is better to fold the positive half of X to the negative half
X ^ 2-2x-1 is a parabola with a fixed point at 1, - 2) opening
Using the five point method to make the image of function: y = SiNx x x ∈ [0,2 π]
That is, SiNx = 0 and ± 1
be
(0,0),(π/2,1),(π,0),(3π/2,-1),(2π,0)
Describe the subset
The image of function y = SiNx + 1 on [0,2 π] is made by five point method
1: When x = 0, y = 1
2: When x = π / 2, y = 2
3: When x = π, y = 1
4: When x = 3 π / 2, y = 0
5: When x = 2 π, y = 1
Draw the five points (0,1), (π / 2,2), (π, 1), (3 π / 2,0), (2 π, 1) on the coordinate system and connect them with smooth curves
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