103 Trigonometry Problems

1. Trigonometric Fundamentals 41
In general, because 4[ABC] =2ab sin C = 2bc sin A = 2ca sin B, wehave
cot α = 1
tan α = a2 + b 2 + c 2 a 2
=
4[ABC] 2bc sin A + b 2
2ca sin B + c 2
2ab sin C
sin 2 A
=
2 sin B sin C sin A + sin 2 B
2 sin C sin A sin B + sin 2 C
2 sin A sin B sin C
= sin2 A + sin 2 B + sin 2 C
,
2 sin A sin B sin C
by the law of sines.
There is another symmetric identity:
csc 2 α = csc 2 A + csc 2 B + csc 2 C.
Because ̸ PCA+ ̸ PAC = ̸ PAB + ̸ PAC = ̸ CAB, it follows that ̸ CPA =
180 ◦ − ̸ CAB, and so sin ̸ CPA = sin A. Applying the law of sines to triangle
CAP gives
x
sin α =
b
b sin α
sin ̸ , or x =
CPA sin A .
Similarly, by working with triangles ABP and BCP, we obtain y = c sin sin B
α and
z = a sin sin C α . Consequently,
[CAP]= 1 zx sin ̸ CPA
2
ab sin C
=
2
= 1 2 · a sin α
sin C · b sin α
sin A · sin A
· sin2 α
sin 2 C =[ABC]· sin2 α
sin 2 C .
Likewise, we have [ABP ]=[ABC]· sin2 α
sin 2 A and [BCP]=[ABC]· sin2 α
sin 2 B . Adding
the last three equations gives
( sin 2 )
α
[ABC] =[ABC]
sin 2 C + sin2 α
sin 2 A + sin2 α
sin 2 ,
B
implying that csc 2 α = csc 2 A + csc 2 B + csc 2 C.
Vectors
In the coordinate plane, let A = (x 1 ,y 1 ) and B = (x 2 ,y 2 ). We define the vector
−→
AB =[x 2 − x 1 ,y 2 − y 1 ], the displacement from A to B. We use a directed segment
to denote a vector. We call the starting (or the first) point (in this case, point A) the