Given that SiNx = cos2x, X is between 90 'and 180', find TaNx

Given that SiNx = cos2x, X is between 90 'and 180', find TaNx

cos2x= 1- 2sin*2 X
2sin^2 X +sinx -1=0
(sinx+1)(2sinx-1)=0
It's between 90 and 180, so SiNx = - 1 is rounded off
So SiNx = 0.5, x = 150 degrees
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Tan150 = negative one third root sign three

f(x)=(1+cos2x)(sinx)^2
Is the maximum value of F (x) - 4 / 3

F (x) = (1 cos2x) (2 / 1-2 / 1cos2x) = 4 / 1-4 / 1cos4x. The minimum value of cos4x is - 1, so the maximum value of F (x) is 5 / 4

Let f (x) = cos2x + SiNx find the value of F (π / 3)

F (π / 3) = cos (2 π / 3) + sin (π / 3) = - cos (π / 3) + sin (π / 3) = - 1 / 2 + (radical 3) / 2

Given SiNx = cos2x, X ∈ (Π / Z, Π), then TaNx

sinx=1-2sin²x,
2sin²x+sinx-1=0
(2sinx-1)(sinx+1)=0
Since x belongs to (π / 2, π), SiNx > 0
So 2sinx-1 = 0, SiNx = 1 / 2,
We get cosx = - 3 / 2, so TaNx = - 3 / 3
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If a tangent is drawn from a point on the straight line y = x + 2 to the circle (x-3) 2 + (y + 1) 2 = 2, then the minimum length of the tangent ()
A. 4B. 3C. 22D. 1

In order to minimize the tangent length, the distance from point C to the straight line must be the minimum. At this time, the distance from the center C (3, - 1) to the straight line y = x + 2 D = | 3 + 1 + 2 | 2 = 32  is the minimum PM = (32) 2 − (2) 2 = 4, so select a

Oa-ob + ad =? NQ + QP + mn-mp =? These two problems are the process of vector calculation

As follows:
OA-OB+AD=BA+AD=BD
NQ+QP+MN-MP=(MN+NQ)+QP-MP=MQ+QP-MP=MP-MP=0

As shown in the figure, P is a point in the rectangle ABCD, PA = 3, PD = 4, PC = 5, then Pb is ()
A. 4.5B. 23C. 32D. 4

Let Ag = DH = a, BG = ch = B, AE = BF = C, de = CF = D, then ap2 = A2 + C2, CP2 = B2 + D2, bp2 = B2 + C2, DP2 = D2 + A2, so ap2 + CP2 = bp2 + DP2, and because PA = 3, PD = 4, PC = 5

When x = 3 is known, the value of the algebraic expression PX & sup3; + QX + 1 is 2010. When x = - 3, the value of the algebraic expression PX & sup3; + QX + 1 is obtained
Given that x = 3, the value of the algebraic formula PX & sup3; + QX + 1 is 2010. When x = - 3, the value of the algebraic formula PX & sup3; + QX + 1?
How did you get that + 2? -（27P+3Q+1）+2=-2010+2

Substituting x = 3 into PX & sup3; + QX + 1 = 2010, we get 27P + 3Q + 1 = 2010
27P + 3Q = 2009
When x = - 3 is substituted into PX & sup3; + QX + 1: - 27p-3q + 1 = - (27P + 3Q) + 1 = - 2009 + 1 = - 2008

It is known that the two end points a and B of the line segment AB with length 1 + radical (2) slide on the x-axis and y-axis respectively, P is the point on AB, and the vector AP = radical 2-2 vector Pb
Find the P trajectory equation

Let a (a, 0), B (0, b), P (x, y),
Because | ab | = 1 + √ 2,
So | ab | ^ 2 = 3 + 2 √ 2,
That is, a ^ 2 + B ^ 2 = 3 + 2 √ 2. (1)
And because the vector AP = √ 2 / 2 * Pb,
So (x-a, y) = √ 2 / 2 * (0-x, B-Y),
That is, x-a = √ 2 / 2 * (- x), y = √ 2 / 2 * (B-Y),
The solution is a = (1 + √ 2 / 2) x, B = (1 + √ 2) y,
Substitute (1) to get (1 + √ 2 / 2) ^ 2 * x ^ 2 + (1 + √ 2) ^ 2 * y ^ 2 = 3 + 2 √ 2,
It is reduced to x ^ 2 / 2 + y ^ 2 = 1. This is the trajectory equation of P

Factorization of (1-A's Square) (1-B's Square) - 4AB

(the square of 1-A) (the square of 1-B) - 4AB = 1-A & # 178; - B & # 178; + A & # 178; B & # 178; - 2ab-2ab = 1 + A & # 178; B & # 178; - 2Ab - (A & # 178; + B & # 178; + 2Ab) = (AB-1) &# 178; - (a + b) &# 178; = (AB-1 + A + b) (ab-1-a-b) hope to help you. If you have any questions, please ask