If the angle f1pf2 = α, the area of △ pf1f2 is proved to be B ^ 2tan α / 2 Speed ~ ~ ~ ~ online~~~~~~~~~~~~~~~ The process should be detailed 2xy+2cosα*xy=4b^2 The area is 1 / 2Sin α * xy = B ^ 2 * sin α / (1 + cos α) = B ^ 2tan α / 2 How did you get this????? 、、

If the angle f1pf2 = α, the area of △ pf1f2 is proved to be B ^ 2tan α / 2 Speed ~ ~ ~ ~ online~~~~~~~~~~~~~~~ The process should be detailed 2xy+2cosα*xy=4b^2 The area is 1 / 2Sin α * xy = B ^ 2 * sin α / (1 + cos α) = B ^ 2tan α / 2 How did you get this????? 、、

F1P=x,F2P=y,F1F2=2c
x^2+y^2-2cosα*xy=4c^2
(x+y)^2=4a^2
2xy+2cosα*xy=4b^2
The area is 1 / 2Sin α * xy = B ^ 2 * sin α / (1 + cos α) = B ^ 2tan α / 2
Let f1p and F2P on both sides use cosine theorem and sine theorem
How to find 2 sin (x + π / 3) in 2 (COS π / 3sinx + sin π / 3cosx) = 2 sin (x + π / 3)?
because
sinacosb+cosasinb
=sin(a+b)
The sine formula of the sum of two angles
F1 and F2 are the two focal points of the ellipse x ＾ 2 / 4 + y ＾ 2 / 3 = 1. A point P on the ellipse satisfies ∠ pf1f2 = 120 ° to find the area of △ pf1f2 (to be solved)
F1 and F2 are the two focal points of the ellipse x ＾ 2 / 4 + y ＾ 2 / 3 = 1. A point P on the ellipse satisfies ∠ pf1f2 = 120 ° to find the area of △ pf1f2 (to be solved,
Let: F1, F2 as the left and right focus, make PM ⊥ F1F2 in M, F1 (- 1,0), C = 1, | F1F2 | = 2, the slope of straight line Pf1 is - √ 3, the equation of 〈 Pf1 is y = - √ 3 (x + 1), substitute into the elliptic equation to get: X & sup2 / / 4 + (x + 1) & sup2; = 1, the solution is: x = - 8 / 5 〉 MF1 = | - 8 / 5 - (- 1) | = 3 / 5 〉 PM = √ 3 * MF1 = 3 √ 3 / 5 〉 s △ pf1f2 = F1F2
The known function f (x) = 2 √ 3sinx (x + π / 4) cos (x + π / 4) - sin (2x + π)
f(x)=2√3sinx(x+π/4)cos(x+π/4)-sin(2x+π)
=√3*2sinx(x+π/4)cos(x+π/4)-sin(2x+π)
=√3sin(2x+π/2)+sin2x
=√3cos2x+sin2x
=2(1/2*sin2x+√3/2*cos2x)
=2(sin2xcosπ/3+cos2xsinπ/3)
=2sin(2x+π/3)
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If the distance difference between the point P on the ellipse x216 + y212 = 1 and the two focal points F1 and F2 is 2, then △ pf1f2 is ()
A. Acute triangle B. right triangle C. obtuse triangle D. isosceles right triangle
From the definition of ellipse, we can get: | Pf1 | + | PF2 | = 2A = 8, we also know that | Pf1 | - | PF2 | = 2, two forms can get | Pf1 | = 5, | PF2 | = 3, and | F1F2 | = 2C = 4, so | PF2 | 2 + | F1F2 | 2 = | Pf1 | 2, so △ pf1f2 is a right triangle
sin(X-π/3)-cos(x+π/6)+√3cosx
sin(x-π/3))-cos(x+π/6)+√3cosx
=sinxcosπ/3-cosxsinπ/3-cosxcosπ/6+sinxsinπ/6+√3cosx
=1/2*sinx-√3/2*cosx-√3/2*cosx+1/2*sinx+√3cosx
=sinx
If we know any point P on the ellipse x ^ 2 / 16 + y ^ 2 / 4 = 1 and the left and right focal points are F1 and F2, then the maximum value of the triangle pf1f2 is
I use the cosine theorem and the solution of the basic inequality
cosa=(pf1)^2+(pf2)^2-4c^2/2pf1pf2
=4b^2-2pf1pf2/2pf1pf2
Because Pf1 + F2 ≥ 2 radical (Pf1 + PF2) (Pf1 + PF2) ^ 2 / 4 ≥ pf1pf2
So I have the maximum value of copfsa = 1,
I want to use s = B ^ 2tan (A / 2) to solve it, but I don't think it has anything to do with Tan (A / 2) to find the maximum value of cosa, because cosx is a decreasing function. The larger the value is, the smaller the degree is. Isn't it related to tan (A / 2)?
What question are you asking? I don't understand,
1) What's the maximum value of triangle pf1f2?
2) The maximum of cosa does not exist, it can be infinitely close to 1
The center of symmetry of the image of the function y = sin α + cos α is
Be more detailed
Function y = sin α + cos α
=Radical 2Sin (a + π / 4)
a+π/4=kπ
a=kπ-π/4 k∈Z
So all the symmetry centers of the function are (K π - π / 4,0)
K = 0 a symmetry center is (- π / 4,0)
(2011 · Anhui simulation) the area of △ pf1f2 is ()
A. 20B. 22C. 24D. 28
From the meaning of the title, we can get a = 7, B = 26, C = 5, two focuses F1 (- 5, 0), F2 (5, 0), and point P (m, n), then we can get & nbsp; nm + 5 · nm − 5 = - 1, M249 + n224 = 1, N2 = 24225, n = ± 245, then the area of △ pf1f2 is & nbsp; 12 × 2C ×| n | = 12 × 10 × 245 = 24, so we choose C
What are cos Wu / 3 and sin Wu / 3, respectively
a half