Question

How to prove cosA + cosB + cosC =< 3/2​

Answer 1

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.