Преобразуйте выражение

$f(x)=5sinx+12cosx\\\\5^2+12^2=25+144=169=13^2\\\\f(x)=5sinx+12cosx=[\frac{\cdot 13}{\cdot 13}]=13\cdot (\frac{5}{13}sinx+\frac{12}{13}cosx)\\\\(\frac{5}{13})^2+(\frac{12}{13})^2=\frac{25+144}{169}=\frac{169}{169}=1\; \; \Rightarrow \; \; \frac{5}{13}=cos \alpha ,\; \frac{12}{13}=sin \alpha \; ,\; tak\; kak\\\\sin^2 \alpha +cos^2 \alpha =1\\\\tg \alpha =\frac{sin \alpha }{cos \alpha }=\frac{12}{5}\\\\f(x)=13\cdot (cos \alpha \cdot sinx+sin \alpha \cdot cosx)=13\cdot sin(x+ \alpha ),$

где $tg \alpha =\frac{12}{5}$ , $\alpha =arctg\frac{12}{5}.$