tan22.5°1-tan222.5°= ___ ．

tan22.5°1-tan222.5°= ___ ．

∵ 45 ° = 2 × 22.5 °, ∵ tan45 ° = 1, that is, Tan (2 × 22.5 °) = 1. According to the sine formula of double angle, we get: 2tan22.5 ° 1 - & nbsp; tan222.5 ° 1, we can get tan22.5 ° 1 - & nbsp; tan222.5 ° 12

Change √ 3 / 2cosx + 1 / 2sinx to ACOS (ω x + φ) (a > 0, ω > 0, | φ)|

√3/2cosx+1/2sinx=sin60°*cosx+cos60°*sinx
=sin(x+60°)=cos(x+30°)

Let a be an inner angle of a triangle. Which of sina, cosa, Tana, Tan (A / 2) are negative? Please write the answer and explain the reason

The maximum internal angle of triangle is 0 ~ 180, when a > 90, cosa

A proof of solid geometry in senior one
If a line outside the plane is parallel to a line in the plane, the line is parallel to the plane
How is this theorem proved?

Counter evidence
Let the line be parallel to the plane
(1) A straight line is in the plane (contradicting a straight line out of the known plane)
(2) A line intersects a plane
Let a line intersect at point a and pass through a point outside the line to form a line parallel to the line in the plane
(there is only one line parallel to the known line when passing through a point outside the line) contradicts the known line
So we have to prove it

As shown in the figure, in the pyramid v-abcd, the bottom surface ABCD is a square, the side VaD is an equilateral triangle, the plane VAD ⊥ the bottom surface ABCD

It is proved that: in ∩ pyramid v-abcd, the bottom surface ABCD is a square, ∨ ab ⊥ ad, ∨ plane VAD ⊥ bottom surface ABCD, plane VAD ∩ bottom surface ABCD = ad, ab ⊂ plane ABCD, ∨ ab ⊥ plane VAD (the property of plane perpendicular to plane)

As shown in the figure, it is known that ABCD is a spatial quadrilateral, ab = ad, CB = CD, and the proof is BD ⊥ AC

It is proved that Ao, Co. ∩ co = O, ∩ BD ⊥ plane ACO, AC ⊂ plane ACO, ∨ BD ⊥ AC are connected by the midpoint o of BD

It is difficult to collect all the solid geometry proof questions in Chapter 2 of mathematics compulsory for senior one~

2. (2007) as shown in the figure, all the edges of the regular triangular prism abc-a1b1c1 are 2, and D is the midpoint of CC1
(1) Verification: Ab1 ⊥ face a1bd; (2) the size of dihedral angle a-a1d-b;
(3) Find the distance from point C to plane a1bd

As shown in the figure, in the quadrilateral ABCD, ad ∥ BC, ad = AB, ∠ BCD = 45 ° and ∠ bad = 90 °. Fold △ ADB along BD to make the plane abd ⊥ plane BCD to form the triangular pyramid a-bcd. Then in the triangular pyramid a-bcd, the following proposition is correct ()
A. Planar abd ⊥ planar ABCB. Planar ADC ⊥ planar bdcc. Planar ABC ⊥ planar bdcd. Planar ADC ⊥ planar ABC

∵ in the quadrilateral ABCD, ad ∩ BC, ad = AB, ∠ BCD = 45 °, and ∩ BD ⊥ CD is planar abd ⊥ planar BCD, and planar abd ∩ planar BCD = BD, so CD ⊥ planar abd, then CD ⊥ AB, and ad ⊥ AB, so ab ⊥ planar ADC, so planar ABC ⊥ planar ADC

Proof: the triangle with a (4,1,9), B (10, - 1,6), C (2,4,3) as vertex is isosceles right triangle

It is proved that a (4,1,9), B (10, - 1,6), C (2,4,3), ab = (4 − 10) 2 + (1 + 1) 2 + (9 − 6) 2 = 7, AC = (4 − 2) 2 + (1 − 4) 2 + (9 − 3) 2 = 7, BC = (10 − 2) 2 + (− 1 − 4) 2 + (6 − 3) 2 = 72, AB2 + ac2 = BC2, ab = AC, so △ ABC is an isosceles right triangle

Divide a triangle into two isosceles by a straight line passing through a vertex of the triangle?
At least two
Isosceles and right triangles don't count!

It is conditional to divide a triangle into two isosceles triangles by a straight line passing through a vertex of the triangle. If one of the three conditions is met, it can be divided into two isosceles triangles
1. Right triangle
2. One angle of a triangle is twice as big as the other
3. One corner of a triangle is three times the other