4
$\begingroup$

Using double angle identities a total of four times, one for each expression in the left hand side, I acquired this.

$$\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \frac{\sin \left ( \frac{\alpha}{2}\right ) \cos \left ( \frac{\alpha}{2}\right ) + \sin \left ( \frac{\beta}{2}\right ) \cos \left ( \frac{\beta}{2}\right )}{\cos^2 \left ( \frac{\alpha}{2} \right) - \sin ^2 \left ( \frac{\beta}{2} \right )}$$

But I know that if $\alpha$ and $\beta$ are angles in a triangle, then this expression should simplify to

$$\tan \left ( \frac{\alpha + \beta}{2} \right )$$

I can see that the denominator becomes $$\cos \left ( \frac{\alpha + \beta}{2} \right ) $$

But I cannot see how the numerator becomes

$$\sin \left ( \frac{\alpha + \beta}{2} \right )$$

What have I done wrong here?

$\endgroup$
3

3 Answers 3

7
$\begingroup$

$$\sin\alpha + \sin\beta = 2\sin(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2}).$$

$$\cos\alpha + \cos\beta = 2\cos(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2}).$$

So, you get the conclusion.

$\endgroup$
1
  • $\begingroup$ This can be thought as $$\sin(A+B)+\sin (A-B)=2\sin A\cos B.$$ with $\alpha= \frac{\alpha+\beta}{2} +\frac{\alpha-\beta}{2}$ and $\beta= \frac{\alpha+\beta}{2} - \frac{\alpha-\beta}{2}$ $\endgroup$ Commented Sep 28, 2022 at 5:12
3
$\begingroup$

Another approach:

Put: $\tan (\alpha/2)=a$ and $ \tan (\beta/2)=b$.

Than: $$ \sin \alpha= \dfrac{2a}{1+a^2} \qquad \cos \alpha=\dfrac{1-a^2}{1+a^2} $$ $$ \sin \beta= \dfrac{2b}{1+b^2} \qquad \cos \beta=\dfrac{1-b^2}{1+b^2} $$

and: $$ \frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \dfrac{2a(1+b^2)+2b(1+a^2)}{(1-a^2)(1+b^2)+(1-b^2)(1+a^2)} $$ that, after a bit of algebra, becomes: $$ =\dfrac{a+b}{1-ab}=\dfrac{\tan \alpha/2+\tan \beta/2}{1-\tan \alpha/2 \tan \beta/2}= \tan \left(\dfrac{\alpha +\beta}{2} \right) $$

$\endgroup$
1
$\begingroup$

Assume that $$P=(\cos\ a,\sin\ b),\ Q=(\cos\ b,\sin\ b)$$ Then what is slope of line passing through $-Q,\ P$ :

For convenience $0<a<b<\pi/2$, let $R=(0,1)$. Then $$\angle\ ROQ=\frac{\pi}{2}-b,\ O=(0,0)$$

and $$\angle\ Q(-Q)P = \frac{b-a}{2} $$

When $\angle\ PP'(-Q) = \pi/2$, then $\angle\ P(-Q)P'= \frac{a+b}{2}$

Hence $$ \tan\ \frac{a+b}{2} = \frac{{\rm difference \ of}\ y-{\rm coordinates\ of}\ P,\ -Q }{{\rm difference \ of}\ x-{\rm coordinates \ of}\ P,\ -Q } $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .