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1994 Korea National Olympiad
Problem 3
cos\alpha : cos\beta : cos\gamma = 39 : 33 : 25
cos\alpha : cos\beta : cos\gamma = 39 : 33 : 25
Source: Korean Mathematical Olympiad 1994, Final Round P6 FKMO
July 21, 2018
Trigonometric Identities
trigonometry
geometry
Problem Statement
Let
α
,
β
,
γ
\alpha,\beta ,\gamma
α
,
β
,
γ
be the angles of
△
A
B
C
\triangle ABC
△
A
BC
. a) Show that
c
o
s
2
α
+
c
o
s
2
β
+
c
o
s
2
γ
=
1
−
2
c
o
s
α
c
o
s
β
c
o
s
γ
cos^2\alpha +cos^2\beta +cos^2 \gamma =1-2cos\alpha cos\beta cos\gamma
co
s
2
α
+
co
s
2
β
+
co
s
2
γ
=
1
−
2
cos
α
cos
β
cos
γ
. b) Given that
c
o
s
α
:
c
o
s
β
:
c
o
s
γ
=
39
:
33
:
25
cos\alpha : cos\beta : cos\gamma = 39 : 33 : 25
cos
α
:
cos
β
:
cos
γ
=
39
:
33
:
25
, find
s
i
n
α
:
s
i
n
β
:
s
i
n
γ
sin\alpha : sin\beta : sin\gamma
s
in
α
:
s
in
β
:
s
inγ
.
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