Simplify cos (X-30 °) + sin (x + 30 °) + cosx + 1

Simplify cos (X-30 °) + sin (x + 30 °) + cosx + 1

Cos (X-30 °) + sin (x + 30 °) + cosx + 1 = cosx * cos30 ° + SiNx * sin30 ° + SiNx * cos30 ° + cosx * sin30 ° + cosx + 1 = [(radical 3 + 1) / 2] SiNx + = [(radical 3 + 2) / 2] cosx + 1

Simplify f (x) = sin (x / 2 + π / 12) times cos (x / 2 + π / 12) - cosx / 2 times cosx / 2

Sin [x / 2 + π / 12] cos [x / 2 + π / 12] - cos [x / 2] cos [x / 2] = 1 / 2 sin [π / 6 + x] - cos ^ 2 [x / 2] = 1 / 2 sin [π / 6 + x] - cos [x] / 2-1 / 2 of course, the last step can continue to be transformed into a trigonometric function mode, but I don't think that is necessary, so I don't want to simplify it

It is known that sin θ = (1 - α) / (1 + α), cos θ = (3 α - 1) / (1 + α). If θ is the second quadrant angle, then the real number α is______

((1-a)/(1+a))^2+((3a-1)/(1+a))^2=1
9A ^ 2-10a + 1 = 0 a = 1 / 9 A = 1 (rounding)

If the focus F1, F2 and point P of the ellipse x squared / 9 + y squared / 2 = 1 are on the ellipse, | Pf1 | = 4, then | PF2 | =? Angle f1pf2 =?

a=3
|PF1|+|PF2|=2a=6
|PF2|=2
In the triangle pf1f2
|PF1|=4
|PF2|=2
|F1F2|=2c=2√7
Using cosine theorem
Cos ∠ f1pf2 = - 1 / 2
F1PF2=120°

It is known that the projection of a point of intersection P of a straight line y = √ 2 / 2x and an ellipse x ＾ 2 / 16 + y ＾ 2 / M = 1 on the X axis is exactly the right focus F of the ellipse, then the value of M is

Let x = C, then y = B ^ 2 / A
So (b ^ 2 / a) / C = √ 2 / 2
a^2-c^2=√2/2ac
e^2+√2/2e－1＝0
e＝√2/2

The ellipse equation is x ^ 2 / 2 + y ^ 2 / 8 = 1, the ray y = 2x (x ≤ 0) intersects with the ellipse m, and the two complementary straight lines with inclination angle through M intersect with the ellipse at two points ab
Prove that the slope of line AB is 2
Finding the maximum area of triangle Amber

It is known that the ellipse C: x2 / 8 y ^ 2 / m ^ 2 = 1 (M > 0) and the straight line L: y = √ 2 / 2x, if the projection of a point of intersection a of the straight line L and the ellipse C on the X axis is exactly the same as that of the ellipse C
Focus, then the value of M is

The projection of a on x-axis is exactly the focus of ellipse C, that is, the abscissa of a and C are the same
The rest is to solve the equations
1. Abscissa of a, x ^ 2 / 8 + x ^ 2 / 2m ^ 2 = 1
Abscissa of C, x = root (8 - m ^ 2)

If M (3, a) is known to be on the parabola y ^ 2 = 4x, then the distance from m to the focus of the parabola is

On the parabola y ^ 2 = 4x
Substituting into the equation:
a^2=12
The focal coordinates of the parabola with y ^ 2 = 4x are: (1,0)
therefore
What is the distance from m to the focus of the parabola
=√(3-1)^2+a^2=√4+12=√16=4

A parabolic equation with the center of the ellipse X & # 178 / 25 + Y & # 178 / 16 = 1 as the vertex and the vertex of the major axis as the focus

A = 5, the vertex of the major axis is (5,0) or (- 5,0)
p/2=5,2p=20
The parabolic equation is y ^ 2 = 20x or Y ^ 2 = - 20x

The parabolic equation with the center of the ellipse X & # 178 / 16 + Y & # 178 / 4 = 1 as the vertex and the right vertex as the focus is

So the parabola is x ^ 2 = 2PY, P / 2 = 4, P = 8,
Parabolic equation: x ^ 2 = 16y