How to prove cosA + cosB + cosC =< 3/2
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Step-by-step explanation:
E=2cos
2
A+B
cos
2
A−B
+1−2sin
2
2
C
−
2
3
=−2sin
2
2
C
+2cos
2
A−B
sin
2
C
−
2
1
=−2x
2
+2xcos
2
A−B
−
2
1
,x=sin
2
C
..............(1)
Now we know that sign of a quadratic is the same as that of the first term provided △<0.
Here △=4cos
2
2
A−B
−4, which is clearly -ive as
cos
2
2
A−B
≤1
∴E≤0.
∴cosA+cosB+cosC≤3/2
Equality will be possible when cos
2
2
A−B
=1
(i.e. △=0) or A=B.
Then from (1),
4x
2
−4x+1=0 or (2x−1)
2=0
∴x=sin 2C
= 2
1
∴C=60
0
∴A=B=C=60
0
Hence the triangle is equilateral.
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