Can You Prove This Trigonometry Equation?

[Mr.Krypton]

I'm new to this forum, anyhow thanks for the admin making such forum, doesn't matter !

I've been trying this calculation for one day still no answer,

Homework Statement

cos (A/2)+cos (B/2)+cos (C/2)=4cos [(B+C)/4]*cos [(C+A)/4]*cos [(A+B)/4]

A+B+C = pi

Homework Equations

The Attempt at a Solution

Cos C+Cos D=2cos[(C+D)/2]*cos[(C-D)/2]

with above equation tried no use !

 

[Filip Larsen]

Welcome to PF!

Perhaps you will find [1] useful.

[1] http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

 

[Mr.Krypton]

Welcome to PF!

Perhaps you will find [1] useful.

[1] http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

still didn't !

 

[Mr.Krypton]

no one to help me?

 

[Mark44]

I'm new to this forum, anyhow thanks for the admin making such forum, doesn't matter !

I've been trying this calculation for one day still no answer,

Homework Statement

cos (A/2)+cos (B/2)+cos (C/2)=4cos [(B+C)/4]*cos [(C+A)/4]*cos [(A+B)/4]

A+B+C = pi

Homework Equations

The Attempt at a Solution

Cos C+Cos D=2cos[(C+D)/2]*cos[(C-D)/2]

with above equation tried no use !

What is D?

Although you didn't say this, it looks like you are supposed to show that the equation above is an identity. So there is no "calculation" to do - just show that the left and right sides of the equation are the same for any values of A, B, and C that add up to \pi radians.

Since the assumption is that A + B + C = \pi, you can replace A + B on the right side with \pi - C, and do something similar with C + A and with B + C.

 

[Mr.Krypton]

What is D?

Although you didn't say this, it looks like you are supposed to show that the equation above is an identity. So there is no "calculation" to do - just show that the left and right sides of the equation are the same for any values of A, B, and C that add up to \pi radians.

Since the assumption is that A + B + C = \pi, you can replace A + B on the right side with \pi - C, and do something similar with C + A and with B + C.

D is nothing it just a formula to solve the above ! btw thanks for ur help, I got a simple clue
let me try !

 

[NascentOxygen]

Ah! I missed that line with A+B+C=Pi
That changes things!

 

[SammyS]

I'm new to this forum, anyhow thanks for the admin making such forum, doesn't matter !

I've been trying this calculation for one day still no answer,

Homework Statement

cos (A/2)+cos (B/2)+cos (C/2)=4cos [(B+C)/4]*cos [(C+A)/4]*cos [(A+B)/4]

A+B+C = pi

The Attempt at a Solution

Cos C+Cos D=2cos[(C+D)/2]*cos[(C-D)/2]

with above equation tried no use !

There may be a simpler method. The way I did it is somewhat involved.

cos x + cos y = 2cos[(x+y)/2]*cos[(x-y)/2] is a place to start.

Then, cos (A/2)+cos (B/2) = 2cos[(A/2+B/2)/2]*cos[(A/2-B/2)/2] = 2cos(A/4+B/4)*cos(A/4-B/4)

Also, notice that A/2 + B/2 + C/2 = pi/2 , so that C/2 = π/2 - (A/2 + B/2) .

Therefore, cos (C/2) = cos(π/2 - (A/2 + B/2)) = sin(A/2 + B/2) = sin(2(A/4 + B/4)) .

And sin(2(A/4 + B/4)) = 2sin(A/4 + B/4)*cos(A/4 + B/4) .

Now you have cos (A/2)+cos (B/2)+cos (C/2) = 2cos(A/4+B/4)*cos(A/4-B/4) + 2sin(A/4 + B/4)*cos(A/4 + B/4) .

Factor out 2cos(A/4+B/4) .

That's a start. Still a ways to go.

 

[ehild]

I would do it just backwards... The proof is very simple if you apply the identity cos(x)cos(y)=cos(x+y)+cos(x-y) to the right-hand side twice.

ehild

 

[ehild]

I go ahead for those who read this thread. I will rename the variables, so A/2=a, B/2=b, C/2=c.

Prove that

cos(a)+cos(b)+cos(c)=4cos(0.5(a+b))*cos(0.5(a+c))* cos(0.5 (c+b)), if a+b+c=pi/2.Using the identity cosxcosy=(cos(x+y)+cos(x-y))/2

cos(0.5(a+c))* cos(0.5 (b+c))=0.5[cos( 0.5(a+b+2c))+cos(0.5(a-b))],

multiplying with 4cos(0.5(a+b)):

2cos(0.5(a+b))*[ cos(0.5(a+b+2c))+cos(0.5(a-b))]=

=[cos(0.5(2a+2b+2c))+ cos(0.5(a+b+2c-(a+b))]+[ cos(0.5(a+b)+0.5(a-b))+ cos(0.5(a+b)-0.5(a-b))]=

=cos(a+b+c)+cos(c)+cos(a)+cos(b)=cos(pi/2)+cos(a)+cos(b)+cos(c) ,

equal to the LHS of the original equation.
ehild