Proving Sine Formula in Triangle ABC

[whkoh]

By using the Sine formula in triangle ABC, show that:

$$\frac{a+b}{c} = \frac{cos\frac{A-B}{2}}{sin\frac{c}{2}}$$.

I've tried:
$$\frac{2 sin C}{c} = \frac{sin A}{a} + \frac{sin B}{b}$$
$$\frac{2 sin C}{c} = \frac{b sin A + a sin B}{a+b}$$
$$\frac{a+b}{c} = \frac{b sin A + a sin B}{2 sin C}$$
$$\frac{a+b}{c} = \frac{b sin A + a sin B}{4sin\frac{c}{2}cos\frac{c}{2}}$$

Am I on the right track? Don't really know how to continue.

 

[VietDao29]

By using the Sine formula in triangle ABC, show that:

$$\frac{a+b}{c} = \frac{cos\frac{A-B}{2}}{sin\frac{c}{2}}$$.

I've tried:
$$\frac{2 sin C}{c} = \frac{sin A}{a} + \frac{sin B}{b}$$
$$\frac{2 sin C}{c} = \frac{b sin A + a sin B}{a+b}$$
$$\frac{a+b}{c} = \frac{b sin A + a sin B}{2 sin C}$$
$$\frac{a+b}{c} = \frac{b sin A + a sin B}{4sin\frac{c}{2}cos\frac{c}{2}}$$

Am I on the right track? Don't really know how to continue.

No, you are not.
There's an error when you go from line #1 to line #2. Line #2 should read:
$$\frac{2 sin C}{c} = \frac{b sin A + a sin B}{ab}$$ not $$\frac{2 sin C}{c} = \frac{b sin A + a sin B}{a+b}$$.
You may want to try this way:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{a + b}{\sin A + \sin B} = \frac{c}{\sin C}$$.
Can you go from here?

 

[Architeuthis Dux]

Hello,

I understand the importance of using mathematical formulas to prove theories and concepts. In this case, we are trying to prove the Sine formula in triangle ABC, which states that:

\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}

To begin, let's start with the given equation:

\frac{a+b}{c} = \frac{cos\frac{A-B}{2}}{sin\frac{c}{2}}

We can rewrite the left side of the equation as:

\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}

Now, let's use the Sine formula to rewrite the right side of the equation:

\frac{cos\frac{A-B}{2}}{sin\frac{c}{2}} = \frac{cos\frac{A}{2}cos\frac{B}{2} - sin\frac{A}{2}sin\frac{B}{2}}{sin\frac{c}{2}}

Next, we can use the double angle formula for cosine to simplify the equation:

\frac{cos\frac{A}{2}cos\frac{B}{2} - sin\frac{A}{2}sin\frac{B}{2}}{sin\frac{c}{2}} = \frac{cos\frac{A}{2}cos\frac{B}{2} - sin\frac{A}{2}sin\frac{B}{2}}{2sin\frac{c}{2}cos\frac{c}{2}}

Now, we can use the Pythagorean identity: cos^2x + sin^2x = 1 to simplify the equation even further:

\frac{cos\frac{A}{2}cos\frac{B}{2} - sin\frac{A}{2}sin\frac{B}{2}}{2sin\frac{c}{2}cos\frac{c}{2}} = \frac{\frac{1}{2}(cosA + cosB) - \frac{1}{2}(sinA + sinB)}{2sin\frac{c}{2}cos\frac{c}{2}}

Finally, we can substitute the values of cosA, cosB, sinA, and sinB using the S