Find the period and maximum value of the function f (x) = sin sixth power X cos sixth power X In the middle is the plus sign

Find the period and maximum value of the function f (x) = sin sixth power X cos sixth power X In the middle is the plus sign

f(x)=(sinx)^6+(cosx)^6
=[(sinx)^2]^3+[(cosx)^2]^3
=[(sinx)^2+(cosx)^2][(sinx)^4-(sinxcosx)^2+(cosx)^4]
=[(sinx)^2+(cosx)^2]^2-3(sinxcosx)^2
=1-(3/4)(2sinxcosx)^2
=1-(3/4)(sin2x)^2
=5/8+(3/8)cos4x
Therefore, the minimum positive period of function f (x) is t = 2 π / 4 = π / 2, the period is k π / 2, and K is a non-zero integer
The maximum value is 5 / 8 + 3 / 8 = 1, and the minimum value is 5 / 8-3 / 8 = 1 / 4
T=pi/2
max=1
min=1/4
Using: sin square x = 1-cos square x
And COS squared x = 1-sin squared X
Let f (x) = 1 - (3 / 4) sin square (2x) = 5 / 8 + (3 / 8) cos (4x)
Is there a formula for the tangent equation of a circle? What is it?
How to find out his slope?
Perpendicular to the line passing through the tangent point and the center of the circle
So you know the slope
The product of the slopes of two vertical lines is - 1
s=1/2rl
Find the golden section of a line
As shown in the figure: then point C is the golden section point of line ab
Proof of tangent equation formula of circle
The tangent equation of point P (x0, Y0) on (x-a) ^ 2 + (y-b) ^ 2 = R ^ 2 is (x0-a) (x-a) + (yo-b) (y-b) = R ^ 2
The tangent equation of a point P (x0, Y0) passing through the circle x ^ 2 + y ^ 2 + DX + ey + F = 0 is x0x + y0y + D [(x + x0) / 2] + E [(Y0 + y)] + F = 0
The tangent length of P (x0, Y0) circle passing through a point outside the circle is √ [(x0-a) ^ 2 + (y0-y) ^ 2-r ^ 2} or √ (x0 ^ 2 + Y0 ^ 2 + dx0 + ey0 + F)
The tangent equation of a point P (x0, Y0) passing through the circle x ^ 2 + y ^ 2 + DX + ey + F = 0 is x0x + y0y + D [(x + x0) / 2] + E [(Y0 + y) / 2] + F = 0
1. The slope of the center (a, b) and the tangent point (x0, Y0) is (y0-b) / (x0-a), so the slope of the tangent is - (x0-a) / (y0-b) because the tangent passes (x0, Y0), so the tangent is y = - (x0-a) / (y0-b) (x-x0) + Y0, which is (x0-a) (x-x0) + (yo-b) (y-yo) = 0, because (x0-a) ^ 2 + (y0-b) ^ 2 = R ^ 2
How to draw the golden section point with ruler?
Make a line segment AB, make its middle vertical line De, intersect AB at point E, then pass through point B, make CB vertical AB, perpendicular foot B, cut BF = be on BC, connect AF, then make an arc with point a as the center of the circle, AE as the radius, intersect AF at point P, then make an arc with a as the center of the circle, PF as the radius, intersect AB at point Q, then q is the golden section point of ab
Solutions of cubic and quartic univariate equations in Mathematics
+ 7X^2 + 12 = 0
X^6 - 7X^3 - 8 = 0
（X+4）（X^2-7x+49）= 5
X^3 + 12 = 3X^2 + 4X
X ^ 3 - 5.3x ^ 2 + 1.46x + 0.2 = 0 (where X1 = 5)
X^4+7X²+12=0
(x²+3)(x²+4)=0
x1=√3i,x2=-√3i,x3=2i,x4=-2i
X^6-7X³-8=0
(x³-8)(x³+1)=0
x1=-1,x2=2
X³+12=3X²+4X
x³-4x=3x²-12
x(x+2)(x-2)=3(x+2)(x-2)
(x-3)(x-2)(x+2)=0
x1=3,x2=2,x3=-2
X³- 5.3X²+1.46X+0.2=0
x³-5x²-0.3x²+1.5x-0.04x+0.2=0
x²(x-5)-0.3x(x-5)-0.04(x-5)=0
(x²-0.3x-0.04)(x-5)=0
(x+0.1)(x-0.4)(x-5)=0
x1=-0.1,x2=0.4,x3=5
What you're going to do now is a quadratic equation with two variables, right?
So you turn the topic into what you can do.
For example, the first question can be transformed into (x ^ 2) ^ 2 + 7x ^ 2 + 12 = 0, and then cross method
So, (x ^ 2 + 3) (x ^ 2 + 4) = 0, so you can see at a glance that there is no solution
The following you see, specific will not ask. One of the principles is to transform what you don't know into what you do
What you're going to do now is a quadratic equation with two variables, right?
So you turn the topic into what you can do.
For example, the first question can be transformed into (x ^ 2) ^ 2 + 7x ^ 2 + 12 = 0, and then cross method
So, (x ^ 2 + 3) (x ^ 2 + 4) = 0, so you can see at a glance that there is no solution
The following you see, specific will not ask. One principle is to turn what you won't into what you will put away
X^4+7X²+12=0
(x²+3)(x²+4)=0
x1=√3i,x2=-√3i,x3=2i,x4=-2i
X^6-7X³-8=0
(x³-8)(x³+1)=0
x1=-1,x2=2
X³+12=3X²+4X
X & # 179; - 4x = 3x... Expansion
X^4+7X²+12=0
(x²+3)(x²+4)=0
x1=√3i,x2=-√3i,x3=2i,x4=-2i
X^6-7X³-8=0
(x³-8)(x³+1)=0
x1=-1,x2=2
X³+12=3X²+4X
x³-4x=3x²-12
x(x+2)(x-2)=3(x+2)(x-2)
(x-3)(x-2)(x+2)=0
x1=3,x2=2,x3=-2
X³- 5.3X²+1.46X+0.2=0
x³-5x²-0.3x²+1.5x-0.04x+0.2=0
x²(x-5)-0.3x(x-5)-0.04(x-5)=0
(x²-0.3x-0.04)(x-5)=0
(x+0.1)(x-0.4)(x-5)=0
X1 = - 0.1, X2 = 0.4, X3 = 5, put away
Ruler drawing: the method of golden section point
1、 2. Find a point C on the vertical line to make BC = 2Ab and connect AC3. Take C as the center of the circle and ab as the radius to make the circular intersection line AC at D4, take ad as the midpoint M5, take a as the center of the circle and am as the radius to make the circular intersection line AB at P, that is AP / AB ≈ 0.618 = (radical 5-1) / 2, 1, take ray AB2, take B as the center of the circle a
Mathematical integral equation, need to solve the problem process and answer in detail!
(1) A (x + 2010) = 2009x (a is not equal to 2009)
(2)b^2x^2+(b^2x)^2=b^2+(b^2)^2
(3)c^2x+5x=1-2d^2x
(4) MX ^ 2 + 1 = 9-mx ^ 2 (M is not equal to 0)
(5)(2a-1)x=(2x-1)a
(6)(x^2/n^2)+2=4-n^2x^2
(7) (1 / 2 KX) ^ 2-P ^ 2 = 0 (k is not equal to 0)
(8)hx^2+1=2hx^2
(1)ax+2010a=2009x
(2009-a)x=2010
x=2010/(2009-a)
(2)b²x²+b^4x²=b^2+b^4
(b^2+b^4)x^2=b^2+b^4
x^2=1
X = 1 or x = - 1
(3)(c^2+5+2d^2)x=1
x=1/(c^2+5+2d^2)
(4)(m+m)x^2=9-1
x^2=4/m
X = 2 / √ m or x = - 2 / √ M
(5)(2a-1)x=2ax-a
(2a-2a-1)x=-a
X=a
(6)(1/n^2+n^2)x^2=4
[(1+n^4)/n^2]x^2=4
x^2=4n^2/(1+n^4)
X = 2 | n | / √ (1 + n ^ 4) or x = - 2 | n | / √ (1 + n ^ 4)
(7)1/4 k^2x^2=p^2
x^2=4p^2/k^2
X = 2 | P / K | or x = - 2 | P / K|
(8)1=hx^2
x^2=1/h
X = 1 / √ h or x = - 1 / √ H
The answer is very hard, please give me a "satisfactory answer", I can "graduate from primary school"!
Golden section point
Longer vs. shorter = fraction
Shorter = longer times fractions
Something like that
The calculation formula of golden section point: length * 0.618 = golden section point. The shorter is 0.382, because the shape designed according to this proportion is very beautiful, so it is called golden section
1. A project can be completed in 63 days by Party A alone, and then in 28 days by Party B. If Party A and Party B cooperate, it will take 48 days to complete. Now Party A will do it in 42 days, and then Party B will complete it alone. How many more days will it take?
2. For a project, team a can complete it in 12 days, team a can complete it in 3 days, and team B can complete half of it in 2 days. Now team a and team B can cooperate for several days, and team B can complete it alone. After finishing, it is found that the two sections take the same time. How many days do they share?
3. The swimming pool uses three water injection pipes (a, B and C). If a single pipe needs 20 hours to fill the pool; if a and B pipes are closed, it takes 8 hours to fill the pool; if B and C pipes are closed, it takes 6 hours to fill the pool?
1. If the cooperation between Party A and Party B takes 48 days to complete, the efficiency of Party A and Party B's cooperation is 1 / 48; if Party A does it alone for 63 days, and then Party B does it alone for 28 days, which is equivalent to Party A's cooperation for 28 days. If Party A does it alone for 63-28 = 35 days, 28 * (1 / 48) = 7 / 12, the remaining 1-7 / 12 = 5 / 12, and Party A does it in 35 days, the efficiency of Party A will be improved
1. The sum of the work efficiency of Party A and Party B is 1 / 48. Suppose the work efficiency of Party B is x, then the work efficiency of Party A is 1 / 48-x
63 (1 / 48-x) + 28x = 1, solving the equation x = 1 / 112, so the efficiency of a is 1 / 48-1 / 112 = 1 / 84
A finished 42 / 84 in 42 days, 1 / 2 left, B needed 56 days
2. Let's share X days. According to the meaning of the question, a's work efficiency is 1 / 12, so a can complete 1 / 4 in 3 days, B can start in 2 days
1. The sum of the work efficiency of Party A and Party B is 1 / 48. Suppose the work efficiency of Party B is x, then the work efficiency of Party A is 1 / 48-x
63 (1 / 48-x) + 28x = 1, solving the equation x = 1 / 112, so the efficiency of a is 1 / 48-1 / 112 = 1 / 84
A finished 42 / 84 in 42 days, 1 / 2 left, B needed 56 days
2. Let X days be shared. According to the meaning of the question, if the efficiency of a is 1 / 12, then a can complete 1 / 4 in 3 days and B needs to complete 1 / 4 in 2 days, so the efficiency of B is 1 / 8
(1 / 12 + 1 / 8) (x / 2) + (1 / 8) (x / 2) = 1, x = 6
3. From the meaning of the question, the work efficiency of a is 1 / 20, the work efficiency sum of a and B is 1 / 8, the work efficiency of B is 1 / 8-1 / 20 = 3 / 40, let the work efficiency of C be x, then there is (x + 3 / 40) × 6 = 1, and the solution equation is x = 11 / 120
1. A project can be completed by Party A in 63 days, and then by Party B in 28 days. If Party A and Party B cooperate, it will take 48 days to complete. How many more days will it take for a to do it alone for 42 days, and then for B to do it alone?
It will take X more days
(1-28 / 48) / (63-28) = 5 / 12 / 35 = 1 / 84 (A's work efficiency)
(1-63 / 84) △ 28 = 1 / 4 △ 28 = 1 / 112 (working efficiency of Party B)
Equation: 42 / 84 + X / 112 = 1