Prove that 2*sqrt3*(sinA+sinB+sinC)/9 < 1
Solution
We know that sinA+sinB+sinC = 4cosA/2*cosB/2*cosC/2..........(1), where A .B and C are
angles of a triangle.
{ cosA/2*cosB/2*cosC/2}^(1/3) <= {[cosA/2+cosB/2+cosc/3]/3}..(2) as GM < AP holds good
as cosA/2 , cosB/2 and cosC/2 are all postive for A/2, B/2 and C/2 each of which are less than
180/2 =90 deg.
The equality sign holds only when A/2 = B/2 = C/2 = 30 deg when RHS of eq (2) is
{sqr3/2+sqrt3/2+sqrt3/2}/3 = sqrt3/2. So,
cosA/2*cosB/2*cosC/2 < ((sqrt3)/2)^3. Substitute this in (1):
sinA+sinB+sinC < 4 ((sqrt3)/2)^3 = (3sqrt3)/2 = (9/2)/sqrt3.
Divide both sides by (9/2) /sqrt3 and get:
2sqrt3(sinA+sinB+sinC) /9 < 1..
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Prove that 2sqrt3(sinA+sinB+sinC)9 1SolutionWe know that .pdf
1. Prove that 2*sqrt3*(sinA+sinB+sinC)/9 < 1
Solution
We know that sinA+sinB+sinC = 4cosA/2*cosB/2*cosC/2..........(1), where A .B and C are
angles of a triangle.
{ cosA/2*cosB/2*cosC/2}^(1/3) <= {[cosA/2+cosB/2+cosc/3]/3}..(2) as GM < AP holds good
as cosA/2 , cosB/2 and cosC/2 are all postive for A/2, B/2 and C/2 each of which are less than
180/2 =90 deg.
The equality sign holds only when A/2 = B/2 = C/2 = 30 deg when RHS of eq (2) is
{sqr3/2+sqrt3/2+sqrt3/2}/3 = sqrt3/2. So,
cosA/2*cosB/2*cosC/2 < ((sqrt3)/2)^3. Substitute this in (1):
sinA+sinB+sinC < 4 ((sqrt3)/2)^3 = (3sqrt3)/2 = (9/2)/sqrt3.
Divide both sides by (9/2) /sqrt3 and get:
2sqrt3(sinA+sinB+sinC) /9 < 1.
![Prove that 2*sqrt3*(sinA+sinB+sinC)/9 < 1
Solution
We know that sinA+sinB+sinC = 4cosA/2*cosB/2*cosC/2..........(1), where A .B and C are
angles of a triangle.
{ cosA/2*cosB/2*cosC/2}^(1/3) <= {[cosA/2+cosB/2+cosc/3]/3}..(2) as GM < AP holds good
as cosA/2 , cosB/2 and cosC/2 are all postive for A/2, B/2 and C/2 each of which are less than
180/2 =90 deg.
The equality sign holds only when A/2 = B/2 = C/2 = 30 deg when RHS of eq (2) is
{sqr3/2+sqrt3/2+sqrt3/2}/3 = sqrt3/2. So,
cosA/2*cosB/2*cosC/2 < ((sqrt3)/2)^3. Substitute this in (1):
sinA+sinB+sinC < 4 ((sqrt3)/2)^3 = (3sqrt3)/2 = (9/2)/sqrt3.
Divide both sides by (9/2) /sqrt3 and get:
2sqrt3(sinA+sinB+sinC) /9 < 1.](https://image.slidesharecdn.com/provethat2sqrt3sinasinbsinc91solutionweknowthat-230326103657-ef7b97ca/85/Prove-that-2sqrt3-sinA-sinB-sinC-9-1SolutionWe-know-that-pdf-1-320.jpg)
