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

4 Answers
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

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

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".

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