Proving Trig Identity: csin(A-B/2)=(a-b)cos(C/2)

In summary: MFB's method is to use the Euler's Formula: \frac{d}{dx}\left( {1+x^2} \right)However, I'm still stuck. I tried mfb's method but couldn't progress any further with it. I feel like I'm just trying random tautologies in the hope it leads somewhere, I'm reminded of monkeys and typewriters.I don't know what you're trying to do, but it sounds like you're not following the instructions.The Attempt at a SolutionIt looks rather similar to a formula mentioned in my book's lead-in to this exercise:\frac{a-b}{a+
  • #1
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Homework Statement


Given a triangle ABC prove that

[itex]c sin \frac{A-B}{2} = (a-b)cos \frac{C}{2}[/itex]

Homework Equations

The Attempt at a Solution



It looks rather similar to a formula mentioned in my book's lead-in to this exercise:

[itex]\frac{a-b}{a+b}=tan\frac{A-B}{2}tan\frac{C}{2}[/itex]

Which can be rearranged to

[itex](a+b)\frac{sin\frac{C}{2}}{cos\frac{A-B}{2}}sin\frac{A-B}{2}=(a-b)cos\frac{C}{2}
[/itex]

So from this it would seem I need only prove that

[itex](a+b)\frac{sin\frac{C}{2}}{cos\frac{A-B}{2}}= c[/itex]

However this doesn't seem much of a simplification and I feel I'm still at square one trying to prove another identity. I also felt my approach was a bit unorthodox. Any suggestions would be appreciated.
 
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  • #2
a,b,c are side lengths, A,B,C are the angles?

I didn't find a solution, but expressing C in terms of A and B and converting the cos to a sine makes the two sides more similar. There are also formulas for the sine of a sum/a difference.
 
  • #3
Thanks for your reply, yes, a,b,c are side lengths and A,B,C are angles.
 
  • #4
Appleton said:

Homework Statement


Given a triangle ABC prove that

[itex]c sin \frac{A-B}{2} = (a-b)cos \frac{C}{2}[/itex]

Homework Equations

The Attempt at a Solution



It looks rather similar to a formula mentioned in my book's lead-in to this exercise:

[itex]\frac{a-b}{a+b}=tan\frac{A-B}{2}tan\frac{C}{2}[/itex]

Which can be rearranged to

[itex](a+b)\frac{sin\frac{C}{2}}{cos\frac{A-B}{2}}sin\frac{A-B}{2}=(a-b)cos\frac{C}{2}
[/itex]

So from this it would seem I need only prove that

[itex](a+b)\frac{sin\frac{C}{2}}{cos\frac{A-B}{2}}= c[/itex]

However this doesn't seem much of a simplification and I feel I'm still at square one trying to prove another identity. I also felt my approach was a bit unorthodox. Any suggestions would be appreciated.

What do you mean by "csin"?
 
  • #5
Sorry for any confusion, I ommitted a space. c is the opposite side length to the angle C and is being multiplied by sin ((A-B)/2).
 
  • #6
Appleton said:
Sorry for any confusion, I ommitted a space. c is the opposite side length to the angle C and is being multiplied by sin ((A-B)/2).
To lessen ambiguity, you can use an asterisk to indicate multiplication -- c * sin((A - B)/2).
 
  • #7
Please use the respective TeX syntax for trig functions. \sin instead sin, \tan instead of tan etc. trig functions are operators and not variables. It's bad style and uncomfortable to read.
 
  • #8
nuuskur said:
Please use the respective TeX syntax for trig functions. \sin instead sin, \tan instead of tan etc. trig functions are operators and not variables. It's bad style and uncomfortable to read.
Phooey.
As far as I'm concerned, ##\sin(x)## is only marginally better than ##sin(x)##.

Also, trig functions are not operators - they are functions.
 
  • #9
Sorry, I meant operators in TeX terms. I did write my work without proper syntax and got booed on by Eeryone. things like [itex]sin f(x) , \sin{f(x)}, \frac{d}{dx}f(x) , \frac{\mathrm{d}}{\mathrm{d}x}f(x) [/itex]. Now I boo on it, myself :D Curiously, no one is complaining about the improper use of f(x), but it just hurts my eyes if everything is italic - difficult to keep track of where a function, operator is or a variable.
 
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  • #10
nuuskur said:
Sorry, I meant operators in TeX terms. I did write my work without proper syntax and got booed on by Eeryone. things like [itex]sin f(x) , \sin{f(x)}, \frac{d}{dx}f(x) , \frac{\mathrm{d}}{\mathrm{d}x}f(x) [/itex]. Now I boo on it, myself :D Curiously, no one is complaining about the improper use of f(x), but it just hurts my eyes if everything is italic - difficult to keep track of where a function, operator is or a variable.
It's not difficult if you use parentheses, as in sin(f(x)). People are going to comment if what you write is unclear.

This one really seems like overkill: \frac{\mathrm{d}}{\mathrm{d}x}f(x). I use as little LaTeX as I can - pages with tons of LaTeX sometimes take much longer to render in a browser, so there's a cost to using it unnecessarily.
 
  • #11
OK I'll bear that in mind next time I post. I'm still no closer to solving it. I tried mfb's method but couldn't progress any further with it. I feel like I'm just trying random tautologies in the hope it leads somewhere, I'm reminded of monkeys and typewriters.
 

1. What is the definition of a trigonometric identity?

A trigonometric identity is a mathematical equation that is true for all values of the variables involved. In other words, it is an equality that holds true regardless of the specific values of the angles or sides involved.

2. Why is it important to prove trigonometric identities?

Proving trigonometric identities allows us to verify the accuracy of equations and build a solid understanding of the relationships between trigonometric functions. It also helps in solving complex trigonometric equations and in applications such as engineering and physics.

3. How do you prove a trigonometric identity?

To prove a trigonometric identity, we use algebraic manipulation and various trigonometric identities and properties such as the Pythagorean identities, sum and difference identities, and double angle identities. We aim to transform one side of the equation to match the other side through these manipulations.

4. How do you prove the identity csin(A-B/2)=(a-b)cos(C/2)?

To prove this identity, we can start by using the double angle identity for sine: sin(2x)= 2sin(x)cos(x). We can then substitute A-B for x, giving us sin(2(A-B/2))= 2sin(A-B/2)cos(A-B/2). We can then use the sum and difference identities for sine and cosine to expand this expression and manipulate it until we reach the desired identity.

5. Are there any tips for proving trigonometric identities?

One tip for proving trigonometric identities is to always start with the more complicated side of the equation and simplify it using known identities and properties. It is also helpful to work with one side of the equation at a time and make sure to carefully keep track of each step of the manipulation process. It is also important to practice and familiarize yourself with common trigonometric identities to make the process easier.

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