sinh 0 = tanh 0 = 0 , cosh 0 = 1 .
\sinh 0 =\tanh 0 =0 \;,\qquad \cosh 0 =1 \;.
sinh 0 = tanh 0 = 0 , cosh 0 = 1 .
для аргумента z = π k / 2 z=\pi k/2 z = π k / 2 cos 0 = 1 , sin 0 = 0 , tan 0 = 0 ,
\cos 0 = 1 \;,\quad \sin 0 = 0 \;,\quad \tan 0 = 0 \;,
cos 0 = 1 , sin 0 = 0 , tan 0 = 0 ,
cos ( π / 2 ) = 0 , sin ( π / 2 ) = 1 , tan ( π / 2 ) = ∞ ,
\cos(\pi/2) = 0 \;,\quad \sin(\pi/2) = 1 \;,\quad \tan(\pi/2) = \infty \;,
cos ( π / 2 ) = 0 , sin ( π / 2 ) = 1 , tan ( π / 2 ) = ∞ ,
cos ( π ) = − 1 , sin ( π ) = 0 , tan ( π ) = 0 ,
\cos(\pi) =-1 \;,\quad \sin(\pi) = 0 \;,\quad \tan(\pi) = 0 \;,
cos ( π ) = − 1 , sin ( π ) = 0 , tan ( π ) = 0 ,
cos ( 3 π / 2 ) = 0 , sin ( 3 π / 2 ) = − 1 , tan ( 3 π / 2 ) = ∞ .
\cos(3\pi/2) = 0 \;,\quad \sin(3\pi/2) =-1 \;,\quad \tan(3\pi/2) = \infty \;.
cos ( 3 π / 2 ) = 0 , sin ( 3 π / 2 ) = − 1 , tan ( 3 π / 2 ) = ∞ . cos φ \cos \varphi cos φ sin φ \sin \varphi sin φ ξ = e i φ \xi =e^{i \,\varphi} ξ = e i φ φ = π k / 2 \varphi =\pi k/2 φ = π k / 2 ξ \xi ξ
В таблице 1 представлены значения основных тригонометрических функций для некоторых значений аргумента z z z ( 0 , π / 2 ) (0,\pi/2) ( 0 , π / 2 ) z = π / 6 z=\pi/6 z = π / 6 z = π / 4 z=\pi/4 z = π / 4 z = π / 3 z=\pi/3 z = π / 3
Таблица 1. Некоторые значения тригонометрических функций
z z z cos z \cos z cos z sin z \sin z sin z tan z \tan z tan z
π / 12 \pi/12 π / 1 2 ( 1 / 4 ) ( 6 + 2 ) (1/4) \,\bigl(\sqrt{6}+\sqrt{2}\bigr) ( 1 / 4 ) ( 6 + 2 ) ( 1 / 4 ) ( 6 − 2 ) (1/4) \,\bigl(\sqrt{6}-\sqrt{2}\bigr) ( 1 / 4 ) ( 6 − 2 ) 2 − 3 2-\sqrt{3} 2 − 3
π / 10 \pi/10 π / 1 0 ( 1 / 4 ) 10 + 2 5 (1/4) \,\sqrt{ 10 +2 \,\sqrt{5} } ( 1 / 4 ) 1 0 + 2 5 ( 1 / 4 ) ( 5 − 1 ) (1/4) \,\bigl(\sqrt{5}-1\bigr) ( 1 / 4 ) ( 5 − 1 ) ( 1 / 5 ) 25 − 10 5 (1/5) \,\sqrt{ 25 -10 \,\sqrt{5} } ( 1 / 5 ) 2 5 − 1 0 5
π / 8 \pi/8 π / 8 ( 1 / 2 ) 2 + 2 (1/2) \,\sqrt{ 2 +\sqrt{2} } ( 1 / 2 ) 2 + 2 ( 1 / 2 ) 2 − 2 (1/2) \,\sqrt{ 2 -\sqrt{2} } ( 1 / 2 ) 2 − 2 2 − 1 \sqrt{2}-1 2 − 1
π / 6 \pi/6 π / 6 ( 1 / 2 ) 3 (1/2) \,\sqrt{3} ( 1 / 2 ) 3 1 / 2 1/2 1 / 2 1 / 3 1/\sqrt{3} 1 / 3
π / 5 \pi/5 π / 5 ( 1 / 4 ) ( 5 + 1 ) (1/4) \,\bigl(\sqrt{5} +1\bigr) ( 1 / 4 ) ( 5 + 1 ) ( 1 / 4 ) 10 − 2 5 (1/4) \,\sqrt{ 10 -2 \,\sqrt{5} } ( 1 / 4 ) 1 0 − 2 5 5 − 2 5 \sqrt{ 5 -2 \,\sqrt{5} } 5 − 2 5
π / 4 \pi/4 π / 4 1 / 2 1/\sqrt{2} 1 / 2 1 / 2 1/\sqrt{2} 1 / 2 1 1 1
3 π / 10 3\pi/10 3 π / 1 0 ( 1 / 4 ) 10 − 2 5 (1/4) \,\sqrt{ 10 -2 \,\sqrt{5} } ( 1 / 4 ) 1 0 − 2 5 ( 1 / 4 ) ( 5 + 1 ) (1/4) \,\bigl(\sqrt{5} +1\bigr) ( 1 / 4 ) ( 5 + 1 ) 1 + ( 2 / 5 ) 5 \sqrt{ 1 +(2/5) \,\sqrt{5} } 1 + ( 2 / 5 ) 5
π / 3 \pi/3 π / 3 1 / 2 1/2 1 / 2 ( 1 / 2 ) 3 (1/2) \,\sqrt{3} ( 1 / 2 ) 3 3 \sqrt{3} 3
3 π / 8 3\pi/8 3 π / 8 ( 1 / 2 ) 2 − 2 (1/2) \,\sqrt{ 2 -\sqrt{2} } ( 1 / 2 ) 2 − 2 ( 1 / 2 ) 2 + 2 (1/2) \,\sqrt{ 2 +\sqrt{2} } ( 1 / 2 ) 2 + 2 2 + 1 \sqrt{2}+1 2 + 1
2 π / 5 2\pi/5 2 π / 5 ( 1 / 4 ) ( 5 − 1 ) (1/4) \,\bigl(\sqrt{5}-1\bigr) ( 1 / 4 ) ( 5 − 1 ) ( 1 / 4 ) 10 + 2 5 (1/4) \,\sqrt{ 10 +2 \,\sqrt{5} } ( 1 / 4 ) 1 0 + 2 5 5 + 2 5 \sqrt{ 5 +2 \,\sqrt{5} } 5 + 2 5
5 π / 12 5\pi/12 5 π / 1 2 ( 1 / 4 ) ( 6 − 2 ) (1/4) \,\bigl(\sqrt{6}-\sqrt{2}\bigr) ( 1 / 4 ) ( 6 − 2 ) ( 1 / 4 ) ( 6 + 2 ) (1/4) \,\bigl(\sqrt{6}+\sqrt{2}\bigr) ( 1 / 4 ) ( 6 + 2 ) 2 + 3 2+\sqrt{3} 2 + 3
lim z → 0 ( z − 1 ⋅ sinh ( β z ) ) = lim z → 0 ( z − 1 ⋅ tanh ( β z ) )
\lim_{z\to 0} \bigl(z^{-1}\cdot \sinh(\beta z)\bigr)
=\lim_{z\to 0} \bigl(z^{-1}\cdot \tanh(\beta z)\bigr)
lim z → 0 ( z − 1 ⋅ sinh ( β z ) ) = lim z → 0 ( z − 1 ⋅ tanh ( β z ) )
= lim z → 0 ( z − 1 ⋅ sin ( β z ) ) = lim z → 0 ( z − 1 ⋅ tan ( β z ) ) = β .
=\lim_{z\to 0} \bigl(z^{-1}\cdot \sin(\beta z)\bigr)
=\lim_{z\to 0} \bigl(z^{-1}\cdot \tan(\beta z)\bigr) =\beta \;.
= lim z → 0 ( z − 1 ⋅ sin ( β z ) ) = lim z → 0 ( z − 1 ⋅ tan ( β z ) ) = β .
Научная статья по математике на заказ от проверенных экспертов по низкой цене!
Комментарии