103 Trigonometry Problems

148 103 Trigonometry Problems
formulas, wehave
s
= 2 sin (x + y)cos (x − y) − 1 + 2(cos (x + y) − cos (x − y))sin z
2R
= 2 cos z cos (x − y) − 1 + 2(sin z − cos (x − y))sin z
= 2 cos (x − y)(cos z − sin z) − cos 2z
= 2 cos (y − x) · cos2 z − sin 2 z
− cos 2z
cos z + sin z
[ ]
2 cos (y − x)
=
cos z + sin z − 1 cos 2z,
where we may safely introduce the quantity cos z + sin z because it is positive
when 0 max { cos z, cos ( π
2
− z )} =
max{ cos z, sin z }. Hence 2 cos(y − x) > cos z + sin z, or
2 cos (x − y)
cos z + sin z − 1 > 0.
Thus, s = p cos 2z for some p>0. It follows that s = a + b − 2R − 2r
is positive, zero, or negative if and only if angle C is acute, right, or obtuse,
respectively.
26. Let ABC be a triangle. Points D, E, F are on sides BC,CA,AB, respectively,
such that |DC|+|CE|=|EA|+|AF |=|FB|+|BD|. Prove that
|DE|+|EF|+|FD|≥ 1 2 (|AB|+|BC|+|CA|).
Solution: As shown in Figure 5.7, Let E 1 and F 1 be the feet of the perpendicular
line segments from E and F to line BC.
F
A
E
F
E
A
B
F1
D
E1
C
B
Figure 5.7.
F1
D
C
E1