Answers
Solution:-LHS=sin2a +Sin2b+sin2c
as we know that
Sina+sinb=2sin(a+b)/2.cos(a-b)/2
using this here we can write
LHS=2sin2(a+b)/2.cos2(a-b)/2+2sinc.cosc
LHS=2sin(a+b).cos(a-b)+2sinc.cosc
also we know that
a+b+c=180,as they are the angles of triangle,so
a+b=180-c
LHS=2sin(180-c).cos(a-b)+2sinc.cosc
LHS=2sinc.cos(a-b)+2sinc.cos{180-(a+b)}
LHS=2sinc.cos(a-b)+2sinc.-cos(a+b) {cos(180-a)=-cosa}
LHS=2sinc{cos(a-b)-cos(a+b)}
LHS=2sinc{2sin(a-b+a+b)/2×sin(b+a-a+b)/2
LHS=4sinc×sin(2a/2)×sin(2b/2)
LHS=4sina.sinb.sinc
now,we can write sinc=sin(180-a-b)
LHS=4sina.sinb.sin(a+b)
LHS=4sina.sinb.{sinacosb+sinbcosa}
LHS=4sin^2asinbcosb+4sin^2bsinacosa
let
Z=4sin^2asinbcosb+4sin^2bsinacosa
dZ/da=8sina.cosa.sinbcosb+4sin^2b(cos^2a-sin^2a)
dZ/da=8sina.cosa.sinb.cosb+4sin^2bcos^2a-4sin^2bsin^2a
equating it to zero we get
4(2Sina.cosa.sinb.cosb+sin^2bcos^2a-
sin^2bsin^2a)=4sinb(2sina.cosa.cosb-sinbsin^2a+sinb.cos^2a)=0
2sina.cosb.cosa+sinb.cos^2a-sinb.sin2^a=0 .i)
now
dZ/db=4sin^2a(cos^2b-sin^2b)+8sinb.cosb.sina.cosa
dZ/db=4sin^2acos^b-4sin^2asin^2b+
8sinb.cosb.sina.cosa
dZ/db=4sina(sina.cos^2b-sina.sin^2b+
2sinb.
cosa.cosb)=0
sina.cos^2b-sina.sin^2b+2sinb.cosa.cosb=0. .ii)
now equating i) and ii) we get
2sina.cosb.cosa+sinb.cos^2a-sinb.sin^2a=2sinb.cosa.cosb+sina.cos^2b-sina.
sin^2b
now, comparing the terms of first and second, we get
a=b
similary differentiating the term with respect to c and b we will get
b=c
so,we can say a=b=c
also,a+b+c=180
3a=180
a=60
a=b=c=60
this shows that, maximum value of sin2a.sin2b.sin2c will be at a=b=c=60
putting this we get
max. value=4×sin60×sin60×sin60=3√3/2
hence
Sin2a.sin2b.sin2c < or=3√3/2
Can you explain the differentiation process a bit more.