1. ## trigonometry Addition Formulae (ii)

As you may see, I am struggling at this.

Show that, for triangle RST, t = r sin (alpha + beta) / sin alpha

iv tried using the sine rule,
t / sin(180 - alpha - beta) = r / sin alpha
t = r / sin alpha x sin(180 - alpha - beta)

After that, i get confused because usually i would expand the sin bracket. ie:
sin(45+45) = sin 45 cos 45 + cos 45 sin 45

but in this case, i have 180, alpha, and beta in one bracket.......

Am i even taking the right steps?

2. Hello, r_maths!

Show that, for $\Delta\!:\;\;t \:= \:\frac{r\sin(\alpha + \beta)}{\sin\alpha}$
Code:
              T
*
*  *
*     *
s *        * r
*           *
*              *
*                 *
* α                β *
R * * * * * * * * * * * * * S
t

From the Law of Sines, we have: . $\frac{t}{\sin T} \:=\:\frac{r}{\sin\alpha}$ [1]

Note that: . $T \;=\;180^o - (\alpha + \beta)$
. . and: . $\sin T \;= \;\sin[180^o - (\alpha + \beta)] \;=\;\sin(\alpha + \beta)$

So [1] becomes: . $\frac{t}{\sin(\alpha + \beta)} \:=\:\frac{r}{\sin\alpha}\quad\Rightarrow\quad t \:=\:\frac{r\sin(\alpha + \beta)}{\sin\alpha}$

3. Originally Posted by Soroban
Hello, r_maths!
So [1] becomes: . $\frac{t}{\sin(\alpha + \beta)} \:=\:\frac{r}{\sin\alpha}\quad\Rightarrow\quad t \:=\:\frac{r\sin(\alpha + \beta)}{\sin\alpha}$

[/size]
$\frac{t}{\sin(\alpha + \beta)}$

SinT = sin[180-(alpha+beta) = sin (alpha + beta)

Where did the 180 go? (is it sin 180 = 0)?
and the minus sign? wouldnt that make it sin(-aplha - beta)
:S

thanks for your help, i really appreciate.

4. Nevermind, looked back at my work, worked out why.
Thanks!!

5. Just for completeness:
$sin(180 - [ \alpha + \beta]) = sin(180) \cdot cos(\alpha + \beta) - sin(\alpha + \beta) \cdot cos(180)$

Since $sin(180) = 0$ and $cos(180) = -1$ we get
$sin(180 - [ \alpha + \beta]) = 0 - - sin(\alpha + \beta) = sin(\alpha + \beta)$

-Dan