Ask for help in high school math and ask for help in high school math

There is an equation about tan(2) that can be solved, pay attention to the range, and estimate that a negative root is discarded.

Then there is a universal formula: let tan( 2)=t, then sin =2t (1+t 2)cosα=（1-t^2)/(1+t^2)

Then open sin( - 3) and bring it in.

I don't bother to forget the specifics.

Set a; x^2/4≤1,2≤x≤2;

b: [2x-1)-(x-3)] x-3) 0, (x+2) (x-3) 0, solution x -2 or x>3

So cub={x|-2 then a (cub)={x|-2

Brother, brother, brother, brother.

sin50 (cos10 cos10 + root number 3sin10 cos10) = sin50 * 2sin40 Li Ji cos10 = sin50 * 2cos50 cos10 = sin100 sin80 = 1

f(x)=(1+sinx+cosx)(sinx 2-cosx 2) 2cosx which with socks 2=2cosx 2(sinx 2+cosx 2))(sinx 2-cosx 2) 2cosx 2

sin 2 (x 2) -cos 2 (x 2) = -cosx primitive = -1 2

Hope for simplicity.

Because (b-c) 2 is greater than or equal to 0

So b 2 + c 2-2bc is greater than or equal to 0

So B2+C2 is greater than or equal to 2BC

And because a is greater than 0, a(b 2+c 2) is greater than or equal to 2abc

In the same way, b(c 2+a 2) is greater than or equal to 2abc and the sum of the two formulas:

a(b 2 + c 2) + b (c 2 + a 2) is greater than or equal to 4abc (certified).

When t>2, [t,t+2] is to the right of the axis of symmetry x=2, when x=t, the function value is the smallest, g(t)=t2(the square of t)-4t-4, when t+1<2, it means that [t,t+2] is to the left of the axis of symmetry x=2, when x=t+1, the function value is the smallest, g(t)=(t+1) 2-4(t+1)-4, i.e., g(t)=t 2-2t-7, when t<=2 and t+1>=2, that is, 1<=t<=2, it means that x=2 is in the interval, at this time, x=2, the function value is the smallest, -8, and then write the answer into a segment function, divided into three segments, and the image is also three segments, compare it in ** minimum can be.