Муодилаи тригонометриро ҳал кунед:
\(\operatorname{ctg}t-\sin t=2\sin^2\frac{t}{2}\).
Ҳал.
\(\operatorname{ctg}t-\sin t=2\sin^2\frac{t}{2}\)
[\(\sin^2\frac{t}{2}=\frac{1-\cos t}{2}\quad|\cdot2\)
\(2\cdot\sin^2\frac{t}{2}=2\cdot\frac{1-\cos t}{2}\)
\(2\sin^2\frac{t}{2}=1-\cos t\)]
\(\operatorname{ctg}t-\sin t=1-\cos t\)
\(\frac{\cos t}{\sin t}-\sin t=1-\cos t\quad|\cdot\sin t\)
\(\cos t-\sin^2 t=\sin t-\sin t\cos t\)
\(\cos t+\sin t\cos t=\sin t+\sin^2 t\)
\(\cos t\cdot(1+\sin t)=\sin t\cdot(1+\sin t)\)
\(\cos t(1+\sin t)-\sin t(1+\sin t)=0\)
\((\cos t-\sin t)\cdot(1+\sin t)=0\)
Якум ҳолатро дида мебароем:
\(\cos t-\sin t=0\)
\(\cos t=\sin t\)
\(\sin t=\cos t\quad|\cdot\frac{1}{\cos t}\)
\(\sin t\cdot\frac{1}{\cos t}=\cos t\cdot\frac{1}{\cos t}\)
\(\frac{\sin t}{\cos t}=1\)
\(\operatorname{tg}t=1\)
\(t=\operatorname{arctg}1+\pi n\)
\(t=\frac{\pi}{4}+\pi n\)
\(t=\pi n+\frac{\pi}{4}\)
\(t=\frac{\pi}{4}(4n+1)\).
Ҳолати дуюмро дида мебароем:
\(1+\sin t=0\)
\(\sin t=-1\)
\(t=-\frac{\pi}{2}+2\pi n\)
\(t=2\pi n-\frac{\pi}{2}\)
\(t=\frac{\pi}{2}(4n-1)\).
Ҷавоб: \(t_1=\frac{\pi}{4}(4n+1);t_2=\frac{\pi}{2}(4n-1)\).