sinα+sinβ=1/4cosβ+cosα=1/3,tanα*tanβ=

sinα+sinβ=1/4cosβ+cosα=1/3,tanα*tanβ=

Sina = 1 / 4-sinb sin ^ 2 a = (1 / 4-sinb) ^ 2cos ^ 2 a = (1 / 3-cosb) ^ 2 add to get 1 = 1 / 16-1 / 2 SINB + sin ^ 2 B + 1 / 9 - 2 / 3 CoSb + cos ^ 2 b = 1 / 16 + 1 / 9 + 1-1 / 2sinb-2 / 3cosb1 / 16 + 1 / 9-1 / 2sinb-2 / 3cosb = 01 / 2sinb + 2 / 3cosb = 1 / 16 + 1 / 9... (1) sin ^ 2 b = (1

If the distance from a point P on the bisector of angle AOB to OA is equal to 5com and Q is any point of ray ob, then the length of PQ is related to 5cm____________

Because P is a point on the bisector of angle AOB
And the distance from P to OA is equal to 5com
And to shape
So the minimum distance of PQ is 5cm
So the length of PQ is related to 5cm
PQ≥5cm

It is known that the period of the function f (x) = 3in ω xcos ω x + 1 − sin2 ω x is 2 π, where ω & gt; 0. (I) find the value of ω and the monotone increasing interval of the function f (x); (II) in △ ABC, let the lengths of the opposite sides of the inner angles a, B and C be a and B respectively. If a = 3, C = 2, f (a) = 32, find the value of B

(1) Function f (x) = 3sinwxcoswx + 1-sin2wx = 32sin2 ω x + 12cos2 ω x + 12 = sin (2 ω + π 6) + 12 ∵ t = 2 π 2 ω = 2 π, ∵ ω = 12 ∵ f (x) = sin (x + π 6) + 12 ∵ the monotone increasing interval of function f (x) is [2K π - 2 π 3, 2K π + π 3], K ∈ Z; (II) ∵ f (x) = sin (x

Find the equation of the line L1 which is parallel to the line L: y = 2x + 1 and whose intercept is 6 on the two coordinate axes

y=2x+4

Simple calculation of 58 * 102-58-58

58*102-58-58
=58*102-58*2
= 58*（102-2）
= 58*100
=5800

A straight line passes through the point P (3 / 4,2) and intersects with the positive half axis of x-axis and y-axis at two points a and B respectively. O is the coordinate origin. Does there exist such a straight line that satisfies the following conditions: the perimeter of triangle AOB is 12?

Let the linear equation passing through point P (3 / 4,2) be Y-2 = K (x-3 / 4)
The shortest perimeter of the triangle appears when k = - 1,
At this time, the length of the two right angles of the triangle is 11 / 4, and the length of the hypotenuse is (11 / 4) (√ 2)
The perimeter of the triangle is
11/4+11/4+(11/4)(√2) =11/2+(11/4)(√2)
because
12-［11/2+(11/4)(√2)］=13/2-(11√2)/4
=(26-11√2)/4>0
That is, the minimum perimeter of the triangle AOB is less than 12,
So there is a straight line like this

Let the sum of the first n terms of the sequence {an} be Sn, and the points (n, Sn / N) and (n ∈ n +) are all on the image of the function y = 3x-2. Note: N in SN is the subscript
(1) The general term formula of the sequence {an}
(2) Let BN = 3 / (an × an + 1), tn be the sum of the first n terms of the sequence {BN}, and find the minimum positive integer m that makes TN < m / 20 hold for all n ∈ n +
Note: the subscript of an + 1 in (2) is n instead of N + 1, that is, an + 1,

(1) If n = 1, A1 = S1 = 3-2 = 1, and N ≥ 2, an = SN-S (n-1) = (3N ^ 2-2n) - [3 (n-1) ^ 2-2 (n-1)] = 6n-5, the above formula also holds for n = 1, an = 6n-5 (2) BN = 3 /

There are two point charges in the vacuum, and the amount of charge is 2 * 10 to the negative sixth power c. if the charge converges 0.2m, the force between them is much larger

F=kq^2/r^2=9*(10^9)/0.2^2N=2.25*(10^11)N

Given proposition p: for any x ∈ R, there exists m ∈ r such that 4 ^ x + 2 ^ x + 1 + M = 0. If P is not a false proposition, then the range of real number m is?

Is it 2 ^ x + 1 or 2 ^ x

One third of the product of two numbers is - 6, one of which is 8 times of the absolute value of - 4

If a number is eight times the absolute value of - 4, then the number is: 8 ×| - 4 | = 8 × 4 = 32
If one third of the product of two numbers is - 6, then the product of two numbers is - 6 × 3 = - 18
So another number: - 18 △ 32 = - 9 / 16