ctgx,\ tgx,\ \sin x,\ \cos x>0\\ tgx=\frac1{ctgx}=\frac1{0,25}=4;\\
\left \{ {{\frac{\sin x}{\cos x}=tgx=4;} \atop \sin^2x+\cos^2x=1}} \right.
\sin x=4\cos x;\\
16\cos^2x+\cos^2x=1;\\
17\cos^2x=1;\\
\cos^2x=\frac1{17};\ \ \cos x=\frac1{\sqrt{17}} ;\\
\sin^2x=1-\cos^2x=1-\frac{1}{17}=\frac{16}{17};\\
\sin x=\frac4{\sqrt17}" alt="ctgx=0,25;\\ x\in\left(0;\frac\pi2\right)==>ctgx,\ tgx,\ \sin x,\ \cos x>0\\ tgx=\frac1{ctgx}=\frac1{0,25}=4;\\
\left \{ {{\frac{\sin x}{\cos x}=tgx=4;} \atop \sin^2x+\cos^2x=1}} \right.
\sin x=4\cos x;\\
16\cos^2x+\cos^2x=1;\\
17\cos^2x=1;\\
\cos^2x=\frac1{17};\ \ \cos x=\frac1{\sqrt{17}} ;\\
\sin^2x=1-\cos^2x=1-\frac{1}{17}=\frac{16}{17};\\
\sin x=\frac4{\sqrt17}" align="absmiddle" class="latex-formula">
ответ:
