9.1.
d).
9.2.
You can calculate the value sin(/6)
= ½. Because /6
< ½ , the point (/6,
½) lies above the line y = x.
9.3.
It is easy to see that sin(/2
– )
= sin(/2
+)
since its left and right parts are equal to cos.
The points (/2
–)
and (/2
+)
are symmetric on the number line over the point
/2.
The identity says that the values of sin x at
that points are equal. It means that the graph
of sine is symmetric over the line x = /2.
9.4.
In problems 1) – 5) take the graph of the
parabola y = x^{2} and make the following
transformations:
1) Shift three units up;
2) Shift three units down;
3) Shift three units to the left;
4) Shift three units to the right;
5) Stretch three times along the yaxis.
In problems 6) and 7), take the graph of the function
y = sin x and make the following transformations:
6) Shift two units to the right;
7) Shift two units down;
9.5.
The graph has the shape of the sine with the amplitude
of 2 (y runs from – 2 to 2), has the period
of 2/3
2.1, starts at the point (0, 2sin1)
(0, 1.7) and ends at the point (10, 2sin31)
(10, – 0.8).
