Feladat:	1130. matematika feladat	Korcsoport: 16-17	Nehézségi fok: nehéz
Megoldó(k):	Fodor H. ,  Haar A. ,  Hajdu P. ,  Krampera Gy. ,  Schuster Gy. ,  Székely József ,  Tóth A.
Füzet:	1904/szeptember, 30 - 32. oldal		PDF  |  MathML
Témakör(ök):	Trigonometriai azonosságok, Feladat
Hivatkozás(ok):	Feladatok: 1903/február: 1130. matematika feladat

A szöveg csak Firefox böngészőben jelenik meg helyesen. Használja a fenti PDF file-ra mutató link-et a letöltésre. $\begin{array}{c}{\displaystyle 4\text{ctg}\varphi =\text{ctg}\frac{\gamma }{2}\left[\text{ctg}\frac{\alpha }{2}\text{ctg}\frac{\beta }{2}+\text{tg}\frac{\alpha }{2}\text{tg}\frac{\beta }{2}\right]+}\end{array}$ $\begin{array}{c}{\displaystyle +\text{tg}\frac{\gamma }{2}\left[\text{ctg}\frac{\alpha }{2}\text{tg}\frac{\beta }{2}+\text{tg}\frac{\alpha }{2}\text{ctg}\frac{\beta }{2}\right]=}\end{array}$ $\begin{array}{c}{\displaystyle =\text{ctg}\frac{\gamma }{2}\left[\frac{\text{cos}{}^2\frac{\alpha }{2}\text{cos}{}^2\frac{\beta }{2}+\text{sin}{}^2\frac{\alpha }{2}\text{sin}{}^2\frac{\beta }{2}}{\text{sin}\frac{\alpha }{2}\text{cos}\frac{\alpha }{2}\text{sin}\frac{\beta }{2}\text{cos}\frac{\beta }{2}}\right]+}\end{array}$ $\begin{array}{c}{\displaystyle +\text{tg}\frac{\gamma }{2}\left[\frac{\text{cos}{}^2\frac{\alpha }{2}\text{sin}{}^2\frac{\beta }{2}+\text{sin}{}^2\frac{\alpha }{2}\text{cos}{}^2\frac{\beta }{2}}{\text{sin}\frac{\alpha }{2}\text{cos}\frac{\alpha }{2}\text{sin}\frac{\beta }{2}\text{cos}\frac{\beta }{2}}\right]\mathrm{,}}\end{array}$ vagy $\begin{array}{c}{\displaystyle \text{ctg}\varphi =\frac{\text{ctg}\frac{\gamma }{2}\left[\text{cos}{}^2\left(\frac{\alpha }{2}+\frac{\beta }{2}\right)+2\text{sin}\frac{\alpha }{2}\text{cos}\frac{\alpha }{2}\text{sin}\frac{\beta }{2}\text{cos}\frac{\beta }{2}\right]}{\text{sin}\alpha \text{sin}\beta }+}\end{array}$ $\begin{array}{c}{\displaystyle +\frac{\text{tg}\frac{\gamma }{2}\left[\text{sin}{}^2\left(\frac{\alpha }{2}+\frac{\beta }{2}\right)-2\text{sin}\frac{\alpha }{2}\text{cos}\frac{\alpha }{2}\text{sin}\frac{\beta }{2}\text{cos}\frac{\beta }{2}\right]}{\text{sin}\alpha \text{sin}\beta }}\end{array}$ $\begin{array}{c}{\displaystyle =\frac{\text{ctg}\frac{\gamma }{2}\text{sin}{}^2\frac{\gamma }{2}+\text{tg}\frac{\gamma }{2}\text{cos}{}^2\frac{\gamma }{2}}{\text{sin}\alpha \text{sin}\beta }+}\end{array}$ $\begin{array}{c}{\displaystyle +\frac{2\text{sin}\frac{\alpha }{2}\text{sin}\frac{\beta }{2}\text{cos}\frac{\alpha }{2}\text{cos}\frac{\beta }{2}\left(\text{ctg}\frac{\gamma }{2}-\text{tg}\frac{\gamma }{2}\right)}{\text{sin}\alpha \text{sin}\beta }=}\end{array}$ $\begin{array}{c}{\displaystyle =\frac{\text{sin}\gamma }{\text{sin}\alpha \text{sin}\beta }+\frac{\text{ctg}\frac{\gamma }{2}-\text{tg}\frac{\gamma }{2}}{2}\mathrm{.}}\end{array}$ De $\begin{array}{c}{\displaystyle \frac{\text{ctg}\frac{\gamma }{2}-\text{tg}\frac{\gamma }{2}}{2}=\frac{1-{\text{tg}}^2\frac{\gamma }{2}}{2\text{tg}\frac{\gamma }{2}}=\text{ctg}\gamma \mathrm{,}}\end{array}$ tehát $\begin{array}{c}{\displaystyle \text{ctg}\varphi =\frac{\text{sin}\gamma }{\text{sin}\alpha \text{sin}\beta }+\text{ctg}\gamma =\frac{\text{sin}\alpha \text{cos}\beta +\text{cos}\alpha \text{sin}\beta }{\text{sin}\alpha \text{sin}\beta }+\text{ctg}\gamma =}\end{array}$ $\begin{array}{c}{\displaystyle =\text{ctg}\alpha +\text{ctg}\beta +\text{ctg}\gamma \mathrm{,}}\end{array}$ vagyis $\begin{array}{c}{\displaystyle 4\left(\text{ctg}\alpha +\text{ctg}\beta +\text{ctg}\gamma \right)=\text{ctg}\frac{\alpha }{2}\text{ctg}\frac{\beta }{2}\text{ctg}\frac{\gamma }{2}+\text{ctg}\frac{\alpha }{2}\text{tg}\frac{\beta }{2}\text{tg}\frac{\gamma }{2}+}\end{array}$ $\begin{array}{c}{\displaystyle +\text{ctg}\frac{\beta }{2}\text{tg}\frac{\alpha }{2}\text{tg}\frac{\gamma }{2}+\text{ctg}\frac{\gamma }{2}\text{tg}\frac{\alpha }{2}\text{tg}\frac{\beta }{2}\mathrm{.}}\end{array}$ (Székely József, Budapest.)