Sinx first moves π/6 units to the right, and obtains:

Sine (x-π/6)

Then change the abscissa to the original 1/2 to get:

Sine (2x-π/6)

Then expand the ordinate to twice the original, and get:

2 inches (2x-π/6)

①A, when 2x-π/6 = π/2+kπ, the symmetry axis can be found;

② The symmetry point can be found when b, 2x-π/6 = kπ;

③C, sin(π/2+t)= cost.

④D, which can be obtained through images.

( 1 1)

①A, connect B 1D 1, because the diagonal of the square is divided vertically, you can get b1d1⊥ a1,and because

B 1B⊥ plane A 1b 1d 1, so b1b will be perpendicular to any straight line on this plane, so b1b ⊥ a/kloc-0.

Since B 1D 1 and B 1B are both ⊥A 1C 1 ⊥, a1⊥ faces b/kloc-

So A 1C 1⊥BD 1, we can get C 1D⊥BD 1.

The last available line BD1⊥ a1c1d.

②B, for a triangular pyramid, take C 1 as the vertex and PA 1D as the bottom, and find its volume. The distance from the point of C 1 to the bottom is C 1 to the point of A 1B 1CD, which is a constant value.

③C, the angle between AP and A 1D is the angle between AP and b1c. For equilateral triangle ACB 1, when point p is at point.

B 1 or c, the minimum angle is 60, when p arrives,

At the midpoint of B 1C, AP bisects B 1C vertically, forming a maximum angle of 90, so the angle range formed by AP and A 1D is [60,90].

④D, intersecting with the perpendicular OP of plane A 1C 1D, according to OPtion B, the length of op is constant, and sinα = OP/C 1P, so when C 1P takes the minimum value, sinα has the maximum value.

( 12)

For odd functions: f (-x) =-f (x)

For even functions, f (-x) = f (x)

F(2x+2) is odd function, and F (x) =-f (2x+2) equals -f (4-x), so.

(2,0) is the symmetry point of f(x);

F(x- 1) is an even function, so f (-x- 1) = f (x- 1), so the symmetry axis of f (x) is x =- 1.

So the period t of f(x) = 4× (2-(-1)) =12.

Analog trigonometric function image

Similarly, the period of g(x) is 8.