12, CosA=a SinC/c can be known from the proportional formula, asinc = c/sinc can be known from the sine theorem, a sinC/c=sinA, so sinA=cosA, so A = 45 and B = 45, so △ABC is an isosceles right triangle.
13, the symmetry axis of trigonometric function is the value of function minus the maximum or minimum value of independent variable, and trigonometric function of asinx+bcosx type can be transformed into the form of root sign (A 2+B 2) sin (X+S), where S is determined by ab. So the maximum value is the root sign (a 2+b 2). In the problem, the maximum value is root number 5, and the minimum value is-root number 5, that is, sinθ/2-2cosθ/2= root number 5 or-root number 5. In order to find sin(θ), we can square this formula and then solve it with the double angle formula. It's complicated The best way is to understand and remember how the trigonometric function of asinx+bcosx is transformed into the root sign (A 2+B 2) SIN (X+S), and the relationship between S and A and B ... We start to look for this relationship from the root sign (A 2+B 2) SIN (X+S). Using the trigonometric summation formula,
Root number (a 2+b 2) sin (x+s) = root number (a2+B2)sinx cos s+ root number (a 2+b 2) cosx sin s,
Compared with the original formula 1, cos s=a/ radical sign (a 2+b 2) and sins = b/ radical sign (a 2+b 2) sin (x+s).
According to this, the original formula can be simplified as root number 5sin(x/2+s), cos s= 1/ root number 5, sins=-2/ root number 5.
The symmetry axis requires θ/2 =π/2s or -π/2s, so θ=π-2s or -π-2s.
sin(θ)=sin2s=2sinscoss=-4/5
14, the area defined by three straight lines is the triangle they enclose. The so-called existence point satisfies this constraint, that is, the circumference and this triangle have a common point. When drawing, it is easy to find that the minimum radius is 2, so R =2 the minimum value of 2.
15, the distance from m to y axis = the distance from m to directrix minus p/2 = the distance from m to focus -p/2.
So the distance from m to y axis+the distance from m to straight line l = the distance from focus to m+the distance from m to straight line l-p/2 >; = distance from focus to straight line -p/2. If and only if the point m is the intersection of the vertical line from the focus to the straight line L and the parabola, the equal sign holds. Therefore, the minimum value is the distance from the focus to the straight line -p/2, and the focus coordinate is (p/2,0), which can be calculated by the formula of the distance from the point to the straight line. The minimum value is 1/2, and p=5.
16, (1) Because cosx decreases on (0, π), and 1, 2, 3 are all less than π, the inequality holds; (2) When n = 1, 2, 3, respectively, find a 1=4, a2=5, a3=6, then a4=a 1=4, a5=a2=5, a6=a3=6, and so on. A20 13 = A3, A20 14 = A 1, A20 15 = A2, so the inequality does not hold. (3) A of the three hyperbolas are 1, 2, 2, c is 2, the number of roots is 7, the number of roots is 5, the eccentricity is 2, the number of roots is 7/2, and the number of roots is 5/2, so the inequality holds. (4) If any of X 1, X2 and X3 is less than zero, the following inequality is meaningless, so it cannot be established.
17, x symbol (x+3-a)=(x-5)/(a- 1-x) Solve this fractional inequality, and then compare it with the following set. If you can't see the following assembly clearly, I won't explain it in detail.
Mathematics classroom games
1, forest game
In the "forest game" class, I designed a r