Find the maximum value of SiNx / 2 (1 + cosx), 0

Find the maximum value of SiNx / 2 (1 + cosx), 0

sinx(1+cosx)
Let t = x / 2
Original formula = 4sint (cost) ^ 3
=4 / radical 3 * (radical 3sint) * cost * cost * cost
From the fundamental inequality to the fourth root (ABCD)
It is known that sinxcosx = 1 / 8 and π / 4
1+2sinxcosx=1+2/8=5/4
sin²x+cos²x+2sinxcosx=5/4
(sinx+cosx)²=5/4
From X range, SiNx > cosx > 0
So SiNx + cosx > 0
sinx+cosx==√5/2
sinxcosx=1/8
By Weida theorem
SiNx and cosx are the roots of the equation A & sup2; - √ 5 / 2 A + 1 / 8 = 0
And SiNx > cosx
a=(√5±√3)/4
So SiNx = (√ 5 + √ 3) / 4, cosx = (√ 5 - √ 3) / 4
π/40
sinxcosx=1/8 ,sin^2x+cos^2x=1
The solution is SiNx = (√ 5 + √ 3) / 4, cosx = (√ 5 - √ 3) / 4
It is known that SiNx = 1 / 8 and 0 degree
0°＜x＜45°
cosx>0
From sin & sup2; X + cos & sup2; X = 1
cos²x=63/64
So cosx = (3 √ 7) / 8
So cosx SiNx = [(3 √ 7) - 1] / 8
(sinx+cosx)^2=1/16
(sinx)^2+(cosx)^2+2sinxcosx=1/16
1+sin2x=1/16
sin2x=-15/16
0°＜x＜45°
cosx>0
From sin & sup2; X + cos & sup2; X = 1
So cosx = √ (1-1 / 64) = 3 √ 7 / 8
So cosx SiNx = (3 √ 7-1) / 8
Sinxcosx = 1 / 8, then the value of cosx SiNx is equal to
(cosX-sinX)^2=(cosx)^2+(sinx)^2-2sinxcosx=1-1/4=3/4
So cosx SiNx = √ 3 / 2 or - √ 3 / 2
sin²+cos²=1
sin²-2sinXcosX+cos²=1-1/4
(sinx-cosx)²=3/4
sinx-cosx±√ 3/2
The first n sum of the equal ratio sequence is SN. It is known that S1, S3 and S2 are equal ratio sequence. Find 1, common ratio Q
The first n sum of the equal ratio sequence is SN. It is known that S1, S3 and S2 are equal ratio sequence, find 1, common ratio Q. 2, if A1-A3 = 3, find SN
(1)S3-S1=S2-S3a1+a2+a3-a1=a1+a2-a1-a2-a3 a2+2a3=0 a2+2a2q=0 q=-1/2(2)a1-a3=3 a1-a1/4=3 a1=4Sn=a1(1-q^n)/(1-q)=4[1-(-1/2)^n]/(3/2)=8[1-(-1/2)^n]/3
Proof (or reason) of the chain rule of derivation
Differential is the linear mapping of tangent space, and the chain rule means that the differential of composite function is the composition of differential. For one-dimensional calculus, linear mapping is number multiplication, and the composition of number multiplication is simple multiplication
140!
25 -15 -80 = 10 -80 = -70
26 -6 -64 = 20 -64 = -44
27 + 3 -48 = 30 -48 = -18
28 + 12 -32 = 40 -32 = 8
29 + 21 -16 = 50 -16 = 34
30 + 30 + 0 = 60 + 0 = 60
31 + 39 + 16 = 70 + 16 = 86
32 + 48 + 32 = 80 + 32 = 112
33 + 57 + 48 = 90 + 48 = 138
34 + 66 + 64 = 100 + 64 = 164
35 + 75 + 80 = 110 + 80 = 190
36 + 84 + 96 = 120 + 96 = 216
37 + 93 + 112 = 130 + 112 = 242
38 + 102 + 128 = 140 + 128 = 268
39 + 111 + 144 = 150 + 144 = 294
40 -30 -140 = 10 -140 = -130
41 -21 -124 = 20 -124 = -104
42 -12 -108 = 30 -108 = -78
43 -3 -92 = 40 -92 = -52
44 + 6 -76 = 50 -76 = -26
45 + 15 -60 = 60 -60 = 0
46 + 24 -44 = 70 -44 = 26
47 + 33 -28 = 80 -28 = 52
48 + 42 -12 = 90 -12 = 78
49 + 51 + 4 = 100 + 4 = 104
50 + 60 + 20 = 110 + 20 = 130
51 + 69 + 36 = 120 + 36 = 156
52 + 78 + 52 = 130 + 52 = 182
53 + 87 + 68 = 140 + 68 = 208
54 + 96 + 84 = 150 + 84 = 234
55 -45 -200 = 10 -200 = -190
56 -36 -184 = 20 -184 = -164
57 -27 -168 = 30 -168 = -138
58 -18 -152 = 40 -152 = -112
59 -9 -136 = 50 -136 = -86
60 + 0 -120 = 60 -120 = -60
61 + 9 -104 = 70 -104 = -34
62 + 18 -88 = 80 -88 = -8
63 + 27 -72 = 90 -72 = 18
64 + 36 -56 = 100 -56 = 44
65 + 45 -40 = 110 -40 = 70
66 + 54 -24 = 120 -24 = 96
67 + 63 -8 = 130 -8 = 122
68 + 72 + 8 = 140 + 8 = 148
69 + 81 + 24 = 150 + 24 = 174
70 -60 -260 = 10 -260 = -250
71 -51 -244 = 20 -244 = -224
72 -42 -228 = 30 -228 = -198
73 -33 -212 = 40 -212 = -172
74 -24 -196 = 50 -196 = -146
11 3^3-5
12 4^2-34%
13 3.25-315%
14 7^3+445%
15 12+5268.32-2569
16 123+456-52*8
17 45%+6325
18 1/2+1/3+1/4
19 789+456-78
20 45%+54%-36%
Let me add
Send up the questions
no problem
Because only fractions with the same denominator can be added, the addition requires general division. General division is to find out the common divisor of the two denominators and turn the two denominators into the same term
The sentence "general division is to find out the common divisor of two denominators and change them into the same item" is wrong. When two numbers are added, general division is to find the least common multiple of two fractions, change the denominator into the denominator of the same denominator, and the sum of the numerators of two fractions is the numerator, and finally it is the simplest fraction
Given that the second, third and sixth terms of an arithmetic sequence with nonzero tolerance are three consecutive terms of an arithmetic sequence, then the common ratio of the arithmetic sequence is equal to ()
A. 34B. −13C. 13D. 3
∵ the second, third, and sixth items of the arithmetic sequence with non-zero tolerance are three consecutive items of an arithmetic sequence in turn, ∵ (a1 + 2D) 2 = (a1 + D) (a1 + 5d). After sorting out, we get A1 & nbsp; = − D2. ∵ the common ratio of this arithmetic sequence is q = a1 + 2da1 + D = − D2 + 2D − D2 + D = 3
On derivation of chain rule
y=sin（3e＾（2x）+1）
y（2x-1/x+3）＾4
These are basic, but I can't
I'm not familiar with you, and I'm not familiar with it, I'm not familiar with it, I'm not familiar with it, I'm not familiar with it, I'm not familiar with it. 1.y = sin (3E (3e ＾ (3e ＾ (2x) (2x (2x) + 1) (3E (3E (3e ＾＾ (2x) (2x) (2x) +) 1) 1)) y '= cos (3E (3e ＾ (3e ? (2x (2x (2x ? (2x) (2x (2x (2x) (2x) (2x) (2x) (2x) (2x) (2x) (2x) (2x) (2x) (2x) (2x) (2x) (2x^ 3 /
y=sin（3e＾（2x）+1）
y'=cos（3e＾（2x）+1）*（3e＾（2x）+1）'
=cos（3e＾（2x）+1）*（3e＾（2x）*2=6e＾（2x）*cos（3e＾（2x）+1）
y=（2x-1/x+3）＾4
y'=4（2x-1/x+3）＾3*（2x-1/x+3）'
=4（2x-1/x+3）＾3*（2+1/x^2)
=4*（2+1/x^2)*（2x-1/x+3）＾3
Derivative = cos (3e ＾ (2x) + 1) * (6e ^ 2x)
It is to find the derivative of compound function layer by layer
Let u = 3E ＾ (2x) + 1, the derivative of u = 6e ^ 2x
The derivative of the original formula = the derivative of COSU * u
1.y'=cos（3e＾（2x）+1）*（3e＾（2x）+1)'
=cos（3e＾（2x）+1）* [3e＾（2x)*ln3e]*（2x)'
=cos（3e＾（2x）+1）* [3e＾（2x)*(1+ln3)]*2
2.y=4*(2x-1/x+3）＾3 * (2x-1/x+3)'
y=4*(2x-1/x+3）＾3 * (2+1/x^2)