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Bogdan Stan
2013 Bogdan Stan
3
cos(a)+cos(b)+cos(c)=sin(a)+sin(b)+sin(c)=0
cos(a)+cos(b)+cos(c)=sin(a)+sin(b)+sin(c)=0
Source:
July 5, 2020
trigonometry
inequalities
Trigonometric inequality
algebra
Problem Statement
Let
a
,
b
,
c
a,b,c
a
,
b
,
c
be three real numbers such that
cos
a
+
cos
b
+
cos
c
=
sin
a
+
sin
b
+
sin
c
=
0.
\cos a+\cos b+\cos c=\sin a+\sin b+\sin c=0.
cos
a
+
cos
b
+
cos
c
=
sin
a
+
sin
b
+
sin
c
=
0.
Prove that i)
cos
6
a
+
cos
6
b
+
cos
6
c
=
3
cos
(
2
a
+
2
b
+
2
c
)
\cos 6a+\cos 6b+\cos 6c=3\cos (2a+2b+2c)
cos
6
a
+
cos
6
b
+
cos
6
c
=
3
cos
(
2
a
+
2
b
+
2
c
)
ii)
sin
6
a
+
sin
6
b
+
sin
6
c
=
3
sin
(
2
a
+
2
b
+
2
c
)
\sin 6a+\sin 6b+\sin 6c=3\sin (2a+2b+2c)
sin
6
a
+
sin
6
b
+
sin
6
c
=
3
sin
(
2
a
+
2
b
+
2
c
)
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