Sunday, April 23, 2017

trigonometry - Proving that $frac{sin alpha + sin beta}{cos alpha + cos beta} =tan left ( frac{alpha+beta}{2} right )$

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?

No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...