in any triangle ABC if sinA: sinB:sinC is equal to 4: 5: 6 then prove that cos A :cosB is: cosC =12:9:2
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Home»Forum»Trigonometry»trignometric identites
In triangle ABC prove that cosA+cosB+cosC is always positive
7 years ago
Answers : (1)
cosA+cosB+cosC=1+4sin(a/2)sin(b/2)sinc(c/2)
.. LHS
= ( cos A + cos B ) + cos C
= { 2 · cos[ ( A+B) / 2 ] · cos [ ( A-B) / 2 ] } + cos C
= { 2 · cos [ (π/2) - (C/2) ] · cos [ (A-B) / 2 ] } + cos C
= { 2 · sin( C/2 ) · cos [ (A-B) / 2 ] } + { 1 - 2 · sin² ( C/2 ) }
= 1 + 2 sin ( C/2 )· { cos [ (A -B) / 2 ] - sin ( C/2 ) }
= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - sin [ (π/2) - ( (A+B)/2 ) ] }
= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - cos [ (A+B)/ 2 ] }
= 1 + 2 sin ( C/2 )· 2 sin ( A/2 )· sin( B/2 ) ... ... ... (2)
= 1 + 4 sin(A/2) sin(B/2) sin(C/2)
= RHS ..
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Home»Forum»Trigonometry»trignometric identites
In triangle ABC prove that cosA+cosB+cosC is always positive
7 years ago
Answers : (1)
cosA+cosB+cosC=1+4sin(a/2)sin(b/2)sinc(c/2)
.. LHS
= ( cos A + cos B ) + cos C
= { 2 · cos[ ( A+B) / 2 ] · cos [ ( A-B) / 2 ] } + cos C
= { 2 · cos [ (π/2) - (C/2) ] · cos [ (A-B) / 2 ] } + cos C
= { 2 · sin( C/2 ) · cos [ (A-B) / 2 ] } + { 1 - 2 · sin² ( C/2 ) }
= 1 + 2 sin ( C/2 )· { cos [ (A -B) / 2 ] - sin ( C/2 ) }
= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - sin [ (π/2) - ( (A+B)/2 ) ] }
= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - cos [ (A+B)/ 2 ] }
= 1 + 2 sin ( C/2 )· 2 sin ( A/2 )· sin( B/2 ) ... ... ... (2)
= 1 + 4 sin(A/2) sin(B/2) sin(C/2)
= RHS ..
jheelam:
This is not the answer to the question I asked.I'm sorry to say.
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