Question

Алгебра нужна помощь

Answer 1

Ответ:

Пошаговое объяснение:

$1)f(\frac{\pi }{6} )=3ctg(\frac{\pi }{6} )+2sin(\frac{\pi }{6})=3*\sqrt{3}+2*\frac{1}{2}=3\sqrt{3}+1;\\f'(x)=\frac{-3}{sin^2x}+2cosx;\\f'(\frac{\pi }{6})=\frac{-3}{sin^2(\frac{\pi }{6})}+2cos(\frac{\pi }{6})=-12+\sqrt{3};\\y=f(\frac{\pi }{6})+f'(\frac{\pi }{6})(x-\frac{\pi }{6});\\y=3\sqrt{3}+1+(-12+\sqrt{3} )(x-\frac{\pi }{6}).$

2)

$f(\frac{\pi }{4})=\frac{\pi }{4}-2cos(\frac{\pi }{4})+3tg(\frac{\pi }{4})=\frac{\pi }{4}-\sqrt{2}+3;\\f'(x)=1+2sinx+\frac{3}{cos^2x};\\\\f'(\frac{\pi }{4} )=1+2sin(\frac{\pi }{4})+\frac{3}{cos^2(\frac{\pi }{4})}=1+\sqrt{2}+6=7+\sqrt{2};\\y=f(\frac{\pi }{4})+f'(\frac{\pi }{4})(x-\frac{\pi }{4});\\y=\frac{\pi }{4}-\sqrt{2}+3+(7+\sqrt{2} )(x-\frac{\pi }{4});\\y=\frac{\pi }{4}-\sqrt{2}+3+(7+\sqrt{2} )x-\frac{\pi (7+\sqrt{2}) }{4} ;\\y=\frac{\pi (\sqrt{2}-6 )}{4}-\sqrt{2}+3+(7+\sqrt{2} )x.$