In a #DeltaABC#, prove that the value of #cosA/(sinBsinC)+cosB/(sinCsinA)+cosC/(sinAsinB)# is a prime?

4 Answers
Sonnhard
Jul 17, 2018

#2# and #2# is a prime

Explanation:

Proof: we need the theorem of cosines
#cos(alpha)=(b^2+c^2-a^2)/(2bc)# etc

and the area of a triangle

#A=1/2a*bsin(gamma)# etc and the formula by Heron

#A=sqrt(s(s-a)(s-b)(s-c))# where #s=(a+b+c)/2# so we get

#((b^2+c^2-a^2)/(2bc))/(2A/(ac)*2A/(ab))+((a^2+c^2-b^2)/(2ac))/(2A/(ab)*2A/(bc))+((a^2+b^2-c^2)/(2ab))/(2A/(bc)*2A/(ac))#

simplifying and expanding

#2*(2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4)/((a+b+c)(-a+b+c)(a-b+c)(a+b-c))#
Now it is
#(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4#
so our result is #2#

maganbhai P.
Jul 17, 2018

Please see below.

Explanation:

We know that ,

#(F_1)sin(x+y)=2sin((x+y)/2)cos((x-y)/2)#

#(F_2)cos(x-y)=-2sin((x+y)/2)sin((x-y)/2)#

We take ,

#X=cosA/(sinBsinC)+cosB/(sinCsinA)+cosC/(sinAsinB#

#=>X=(sinAcosA+sinBcosB+sinCcosC)/(sinAsinBsinC)#

#=>X=(2sinAcosA+2sinBcosB+2sinCcosC)/(2sinAsinBsinC)#

#=>X=(sin2A+sin2B+sin2C)/(2sinAsinBsinC)to(1)#

Now , In a #triangle ABC#, #A+B+C=pi#

#=>color(red)(A+B=pi-C and C=pi-(A+B)#

So ,

#diamondsin2A+sin2B+sin2CtoApply(F_1)#

#=2sin((2A+2B)/2)cos((2A-2B)/2)+sin2C#

#=2sincolor(red)((A+B))cos(A-B)+sin2C#

#=2sincolor(red)((pi-C))cos(A-B)+2sinCcosC#

#=2sinCcos(A-B)+2sinCcosC#

#=2sinC[cos(A-B)+cos color(red)(C)]#

#=2sinC[cos(A-B)+coscolor(red)((pi-(A+B)))]#

#=2sinC[cos(A-B)-cos(A+B)]toApply(F_2)#

=#2sinC[-2sin((A-B+A+B)/2)sin((A-B-A-B)/2)]#

#=2sinC[-2sinAsin(-B)]#

#=2sinC[2sinAsinB]#

#=4sinAsinBsinC#

#i.e.2A+sin2B+sin2C =4sinAsinBsinCto(2)#

From #(1) and (2)#

#X=(4sinAsinBsinC)/(2sinAsinBsinC)#

#=>X=2#

Ratnaker Mehta
Jul 17, 2018

Please refer to another Proof in Explanation.

Explanation:

In #DeltaABC, A+B+C=pi#.

#:. B+C=pi-A#.

#:. cos(B+C)=cos(pi-A)=-cosA#,

# i.e., cosBcosC-sinBsinC=-cosA#.

#:. cosA/(sinBsinC)=(sinBsinC-cosBcosC)/(sinBsinC), #

#=(sinBsinC)/(sinBsinC)-(cosBcosC)/(sinBsinC)#.

#=1-cotBcotC#.

#:."The Exp."=3-{cotBcotC+cotCcotA+cotAcotB}#,

#=3-{1/(tanBtanC)+1/(tanCtanA)+1/(tanAtanB)}#,

#=3-(tanA+tanB+tanC)/(tanAtanBtanC)..............(star_1)#,

But, #B+C=pi-A#.

#:. tan(B+C)=tan(pi-A)#.

#:. (tanB+tanC)/(1-tanBtanC)=-tanA#.

#:. tanB+tanC=-tanA+tanAtanBtanC#.

#:. tanA+tanB+tanC=tanAtanBtanC............(star_2)#.

Therefore, from #(star_1) and (star_2)#

#"The Exp."=3-1=2," a prime".#

P dilip_k
Jul 17, 2018

Given

#A+B+C=pi#

#=>cot(A+B)=cot(pi-C)#

#=>(cotAcotB-1)/(cotB+cotA)=-cotC#

#=>cotAcotB+cotBcotC+cotCcotA=1#

Now 1st part of the given expression
#=cosA/(sinBsinC)#

#=cos(pi-(B+C))/(sinBsinC)#

#=(-cosBcosC+sinBsinC)/(sinBsinC)#

#=1-cotBcotC#

Similarly 2nd part

#=1-cotAcotB#

And 3rd part

#=1-cotCcotA#

So whole LHS

#=3-(cotAcotB+cotBcotC+cotCcotA)#

#=3-1=2#,which is a prime

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