simple..
do this
as A,B,C is an acute triangle this implies all angles are less than 90.
Now well start...
fr a triangle u hv a property tht
1>cosA + cosB + cosc>1.5
and
1.5>sinA + sinB + sinC>2
Hence ur result follows..
the proof of these
here
cosA + cosB + cosC
=2cos((A+B)/2).cos((A-B)/2) + 1 - 2sin^2 (C/2)
=1 + 2sin(c/2)cos(A-B /2) - 2sin^2 (C/2)
=1 + 2sin(c/2){cos(A-B /2) - sin (C/2)}
=1 + 2sin(c/2){cos(A-B /2) - cos (A+B/2)}
=1 + 4sinC/2 sinB/2 sinA/2
as
1>sinC/2 sinB/2 sinA/2>0
this implies
1>cosA + cosB + cosC>1.5
u can proof the other similarly!
2007-03-18 09:17:02
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answer #1
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answered by Second Newton... 2
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yes this possible SIN(A+B)= SINACOSB+COSASINB SAME way u put the equation of SIN(A+B+C) and compare you will get the result still if you have any maths doubts then contact me on my mobile no 09824776781 but you have to pay me 100 rs for each question
2007-03-17 22:55:01
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answer #2
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answered by MONTU L 1
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sinA+sinB+sinC
solve ho gaya na
2007-03-18 02:31:17
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answer #3
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answered by Vijay Kandpal 2
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For ? ABC, A + B + C = ? => A/2 + B/2 + C/2 = ?/2 => A/2 + B/2 = ?/2 - C/2 => tan(A/2 + B/2) = tan(?/2 - C/2) => [tan(A/2) + tan(B/2)] / [one million - tan(A/2) tan(B/2)] = cot(C/2) = one million/tan(C/2) => tan(C/2) tan(A/2) + tan(B/2) tan(C/2) = one million - tan(A/2) tan(B/2) => tan(A/2) tan(B/2) + tan(B/2) tan(C/2) + tan(C/2) tan(A/2) = one million ... ( one million ) additionally, A/2 + B/2 + C/2 = ?/2 => sin(A/2 + B/2 + C/2) = sin(?/2) => sin(A/2) cos(B/2) cos(C/2) + sin(B/2) cos(C/2) cos(A/2) + sin(C/2) cos(A/2) cos(B/2) - sin(A/2) sin(B/2) sin(C/2) = one million ... ( 2 ) From ( one million ) and ( 2 ), sin(A/2) cos(B/2) cos(C/2) + sin(B/2) cos(C/2) cos(A/2) + sin(C/2) cos(A/2) cos(B/2) - sin(A/2) sin(B/2) sin(C/2) = tan(A/2) tan(B/2) + tan(B/2) tan(C/2) + tan(C/2) tan(A/2) => sin(A/2) cos(B/2) cos(C/2) + sin(B/2) cos(C/2) cos(A/2) + sin(C/2) cos(A/2) cos(B/2) = sin(A/2) sin(B/2) sin(C/2) + tan(A/2) tan(B/2) + tan(B/2) tan(C/2) + tan(C/2) tan(A/2).
2016-12-18 16:38:45
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answer #4
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answered by ? 4
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hellllloooooo.........u can use the trigo rules to solve tht......
2007-03-17 22:17:20
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answer #5
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answered by nehucool 2
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can u please provide more details
2007-03-17 21:18:58
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answer #6
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answered by sadiq 1
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