Let a be an acute angle, if Sina =? Then a =? If Tan a = root 3 / 3, then a =?

Let a be an acute angle, if Sina =? Then a =? If Tan a = root 3 / 3, then a =?

Sina =?, a = arcsin?. Tana = radical 3 / 3, a = 30du
In △ ABC, it is known that sina: sinc = 5:2, tanb = - root 3, triangle area is 10 root 3, and the lengths of a and C are obtained
Sqrt is the root sign
According to tanb = - sqrt (3), B = 120 ° sine law
A / Sina = C / sinc introduces Sina / sinc = A / C = 5 / 2
S=acsinB=acsin60°=10sqrt(3)
∴c=2sqrt(2) a=5sqrt(2)
If the image of function y = - 4 / 3x-4 intersects X-axis at a and y-axis at B, then the distance between two points AB is?
x=0,y=0-4=-4
y=0,-4/3x-4=0,x=-3
So a (- 3,0), B (0, - 4)
The origin is o
Then Ao = | - 3 | = 3
BO=|-4|=4
From Pythagorean theorem
AB=√(AO²+BO²)=5
Let f (x) = (x + 1) ln (x + 1) - ax get the extremum at x = 0. (1) find the value of a and the monotone interval of F (x); (2) prove the inequality nlnn ≥ (n-1) ln (n + 1) for any positive integer n
(1) ∵ f (x) = (x + 1) ln (x + 1) - ax, ∵ f '(x) = ln (x + 1) + 1-A, ∵ f (x) obtains the extremum at x = 0, ∵ f' (0) = 0, ∵ a = 1, so f '(x) = ln (x + 1). When x + 1 ＞ 1, i.e. x ＞ 0, f' (x) ＞ 0, when 0 ＜ x + 1 ＜ 1, i.e. - 1 ＜ x ＜ 0, f '(x) ＜ 0, ∵ f ∵ f (x) ＜ 0, i.e. x ＜ 0
Let Sn = 2n2 be the sum of the first n terms of the sequence {an}, and {BN} be the equal ratio sequence, and A1 = B1, B2 (a2-a1) = B1. (I) find the general formula of the sequence {an} and {BN}; (II) let cn = anbn, find the first n terms and TN of the sequence {CN}
(1) : when n = 1, A1 = S1 = 2; when n ≥ 2, an = sn-sn-1 = 2n2-2 (n-1) 2 = 4n-2, so the general formula of {an} is an = 4n-2, that is, {an} is the arithmetic sequence of A1 = 2, tolerance d = 4. Let the common ratio of {BN} be q, then b1qd = B1, d = 4, q = 14
Two problems of quadratic equation with one variable. ① x square + (√ 2 + √ 3) x + √ 6 = 0, ② (2x + 1) square + 3 (2x + 1)
Two problems of quadratic equation of one variable
① X square + (√ 2 + √ 3) x + √ 6 = 0
② (2x + 1) square + 3 (2x + 1) + 2 = 0
1. Multiply by cross to get (x + √ 2) (x + √ 3) = 0
2 also get [(2x + 1) + 1] [(2x + 1) + 2] = 0
If high school set a = {[x, y] ly = alxl}, B = {[x, y] ly = x ^ 2, X ∈ r}, C = a intersection B, and set C is a single element set, then the value range of real number a is
C is a set of single elements
So a / B has only one common element
So there is only one intersection point between y = a|x|and y = x & #178
When a
Non negative integer. Y = alxl in set a and y = LXL LXL in set B. because a intersects B is a single element set, a = LXL, so a is a non negative integer
Using monotonicity of function to prove inequality: ln (1 + x) < x (x > 0)
Let f (x) = x-ln (1 + x)
Then f '(x) = 1-1 / (1 + x) = x / (x + 1) is always positive when x ≥ 0
So f (x) is a strictly monotone increasing function when x ≥ 0
So when x > 0, f (x) > F (0) = 0-ln1 = 0, that is ln (1 + x)
It is known that the sum of the first n terms of the equal ratio sequence {an} is Sn = 2n + C, find the value of C and the general term formula of the sequence {an}; [2] BN = Sn + 2n + 1, find the sum of the first n terms of the sequence {BN}
1. According to this, the relation between an and Sn is obtained as follows: an-s 1
If the sum of the first n terms of the sequence {an} is Sn = 2n + C, then A1 = S1 = 2 + C, an = SN-S (n-1) = 2n + C-2 (n-1) - C = 2,
Because the sequence {an} is an equal ratio sequence, so (2 + C) / 2 = 1, that is, C = 0, an = 2
2. BN = Sn + 2n + 1 = 2n + 2n + 1 = 4N + 1, which is an arithmetic sequence with 5 first term and 4 tolerance, so the sum of the first n terms of the sequence {BN} is Sn = [(5 + 4N + 1) * n] / 2 = n * (2n + 3)
An = SN-S (n-1) = 2, so {an} is a constant column, C = 0
bn=sn+2n+1=4n+1
Let tn be the sum of the first n terms of {BN}
Then TN = 4 + 1 + 4 * 2 + 1 + 4 * 3 + 1 +... 4 * n + 1
=4+4*2+4*3+......+4*n+n
=4(1+2+3+......+n)+n
=2n(n+1)+n
=2n^2+3n
The solution of quadratic equation of one variable with x square + 2x-3 = 0
x^2+2x-3=0
(x-1)(x+3)=0
X = 1 or x = - 3