If cos a + 2cos b + cos c = 2 then a, b, c
are in
Answers
ANSWER
Given:cosA+2cosB+cosC=2
⇒cosA+cosC=2−2cosB
Using transformation angle formulae, we have
⇒2cos(
2
A+C
)cos(
2
A−C
)=2(1−cosB)
Using multiple angle formula, we get
⇒2cos(
2
A+C
)cos(
2
A−C
)=2.2sin
2
(
2
B
)
We have A+B+C=π⇒.A+C=π−B⇒
2
A+C
=
2
π
−
2
B
⇒2cos(
2
π
−
2
B
)cos(
2
A−C
)=2.2sin
2
(
2
B
)
⇒2sin(
2
B
)cos(
2
A−C
)=2.2sin
2
(
2
B
)
⇒cos(
2
A−C
)=2sin(
2
B
)
⇒cos(
2
A−C
)=2sin(
2
π
−
2
A+C
)
⇒cos(
2
A−C
)=2cos(
2
A+C
)
⇒
cos(
2
A+C
)
cos(
2
A−C
)
=
1
2
By componendo and dividendo method,
⇒
cos(
2
A−C
)+cos(
2
A+C
)
cos(
2
A−C
)−cos(
2
A+C
)
=
2+1
2−1
⇒
2cos(
2
A−C+A+C
)cos(
2
A−C−A−C
)
−2sin(
2
A−C+A+C
)sin(
2
A−C−A−C
)
=
3
1
⇒
cos(
2
A
)cos(
2
C
)
sin(
2
A
)sin(
2
C
)
=
3
1
⇒tan(
2
A
)tan(
2
C
)=
3
1
⇒
s(s−a)
(s−b)(s−c)
s(s−c)
(s−a)(s−b)
=
3
1
⇒
s
s−b
=
3
1
⇒3s−3b=s
⇒2s=3b
We know that a+b+c=2s
∴a+b+c=3b
Hence,a+c=2b
∴a,b,c are in A.P
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