题目
Problem
1.
(a) Use the definition of sinh x \sinh x sinh x
sinh 3 x ≡ 4 sinh 3 x + 3 sinh x . \sinh 3x \equiv 4\sinh^3 x + 3\sinh x. sinh 3 x ≡ 4 sinh 3 x + 3 sinh x . (2)
(b) Hence determine the exact coordinates of the points of intersection of the curve with equation y = sinh 3 x y = \sinh 3x y = sinh 3 x y = 19 sinh x y = 19\sinh x y = 19 sinh x
(5)
题目中文翻译
(a) 利用 sinh x \sinh x sinh x
sinh 3 x ≡ 4 sinh 3 x + 3 sinh x \sinh 3x \equiv 4\sinh^3 x + 3\sinh x sinh 3 x ≡ 4 sinh 3 x + 3 sinh x (2)
(b) 因此求方程 y = sinh 3 x y = \sinh 3x y = sinh 3 x y = 19 sinh x y = 19\sinh x y = 19 sinh x
(5)
解答
(a)
解法一
思路
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严格从 sinh x = ( e x − e − x ) / 2 \sinh x=(e^x-e^{-x})/2 sinh x = ( e x − e − x ) /2 4 sinh 3 x + 3 sinh x 4\sinh^3x+3\sinh x 4 sinh 3 x + 3 sinh x e 3 x e^{3x} e 3 x e − 3 x e^{-3x} e − 3 x
答题过程
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Using
sinh x = e x − e − x 2 , \sinh x=\frac{e^x-e^{-x}}{2}, sinh x = 2 e x − e − x , we have
4 sinh 3 x + 3 sinh x = 4 ( e x − e − x 2 ) 3 + 3 ( e x − e − x 2 ) = 1 2 ( e 3 x − 3 e x + 3 e − x − e − 3 x ) + 3 2 ( e x − e − x ) = e 3 x − e − 3 x 2 = sinh 3 x . \begin{align*}
4\sinh^3x+3\sinh x
=&\,4\big(\frac{e^x-e^{-x}}{2}\big)^3
+3\big(\frac{e^x-e^{-x}}{2}\big)\\[4mm]
=&\,\frac12
\bigl(e^{3x}-3e^x+3e^{-x}-e^{-3x}\bigr)\\[4mm]
&\,\hspace{2pt}+\frac32(e^x-e^{-x})\\[4mm]
=&\,\frac{e^{3x}-e^{-3x}}{2}\\[4mm]
=&\,\sinh3x.
\end{align*} 4 sinh 3 x + 3 sinh x = = = = 4 ( 2 e x − e − x ) 3 + 3 ( 2 e x − e − x ) 2 1 ( e 3 x − 3 e x + 3 e − x − e − 3 x ) + 2 3 ( e x − e − x ) 2 e 3 x − e − 3 x sinh 3 x . Therefore
sinh 3 x ≡ 4 sinh 3 x + 3 sinh x . \boxed{\sinh3x\equiv4\sinh^3x+3\sinh x}. sinh 3 x ≡ 4 sinh 3 x + 3 sinh x .
(b)
解法一
思路
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承接 (a) 的恒等式，把交点条件化为关于 sinh x \sinh x sinh x 0 , 2 , − 2 0,2,-2 0 , 2 , − 2
答题过程
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At a point of intersection,
sinh 3 x = 19 sinh x . \sinh3x=19\sinh x. sinh 3 x = 19 sinh x . Using the identity from part (a),
4 sinh 3 x + 3 sinh x = 19 sinh x , 4\sinh^3x+3\sinh x=19\sinh x, 4 sinh 3 x + 3 sinh x = 19 sinh x , so
4 sinh x ( sinh 2 x − 4 ) = 0. 4\sinh x\bigl(\sinh^2x-4\bigr)=0. 4 sinh x ( sinh 2 x − 4 ) = 0. Hence
sinh x = 0 , sinh x = 2 , or sinh x = − 2. \sinh x=0,
\qquad \sinh x=2,
\qquad \text{or}\qquad \sinh x=-2. sinh x = 0 , sinh x = 2 , or sinh x = − 2. These give
x = 0 , x = ln ( 2 + 5 ) , x = − ln ( 2 + 5 ) , x=0,
\qquad x=\ln(2+\sqrt5),
\qquad x=-\ln(2+\sqrt5), x = 0 , x = ln ( 2 + 5 ) , x = − ln ( 2 + 5 ) , respectively. Since y = 19 sinh x y=19\sinh x y = 19 sinh x
( 0 , 0 ) , \boxed{(0,0)}, ( 0 , 0 ) , ( ln ( 2 + 5 ) , 38 ) , \boxed{\bigl(\ln(2+\sqrt5),38\bigr)}, ( ln ( 2 + 5 ) , 38 ) , and
( − ln ( 2 + 5 ) , − 38 ) . \boxed{\bigl(-\ln(2+\sqrt5),-38\bigr)}. ( − ln ( 2 + 5 ) , − 38 ) .
解法二
思路
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将交点方程完全写成指数形式。令 t = e 2 x > 0 t=e^{2x}>0 t = e 2 x > 0
答题过程
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Writing both hyperbolic sines in exponential form,
e 3 x − e − 3 x 2 = 19 e x − e − x 2 . \frac{e^{3x}-e^{-3x}}{2}
=19\frac{e^x-e^{-x}}{2}. 2 e 3 x − e − 3 x = 19 2 e x − e − x . Multiplying by 2 e 3 x 2e^{3x} 2 e 3 x
e 6 x − 19 e 4 x + 19 e 2 x − 1 = 0. e^{6x}-19e^{4x}+19e^{2x}-1=0. e 6 x − 19 e 4 x + 19 e 2 x − 1 = 0. Let t = e 2 x t=e^{2x} t = e 2 x t > 0 t>0 t > 0
t 3 − 19 t 2 + 19 t − 1 = ( t − 1 ) ( t 2 − 18 t + 1 ) = 0. \begin{align*}
t^3-19t^2+19t-1
=&\,(t-1)(t^2-18t+1)\\[4mm]
=&\,0.
\end{align*} t 3 − 19 t 2 + 19 t − 1 = = ( t − 1 ) ( t 2 − 18 t + 1 ) 0. Thus
t = 1 , t = 9 + 4 5 , t = 9 − 4 5 . t=1,
\qquad t=9+4\sqrt5,
\qquad t=9-4\sqrt5. t = 1 , t = 9 + 4 5 , t = 9 − 4 5 . Since
9 + 4 5 = ( 2 + 5 ) 2 9+4\sqrt5=(2+\sqrt5)^2 9 + 4 5 = ( 2 + 5 ) 2 and
9 − 4 5 = ( 5 − 2 ) 2 = 1 ( 2 + 5 ) 2 , 9-4\sqrt5=(\sqrt5-2)^2
=\frac1{(2+\sqrt5)^2}, 9 − 4 5 = ( 5 − 2 ) 2 = ( 2 + 5 ) 2 1 , the corresponding values of x x x
0 , ln ( 2 + 5 ) , − ln ( 2 + 5 ) . 0,
\qquad \ln(2+\sqrt5),
\qquad -\ln(2+\sqrt5). 0 , ln ( 2 + 5 ) , − ln ( 2 + 5 ) . Substitution into y = 19 sinh x y=19\sinh x y = 19 sinh x
( 0 , 0 ) , ( ln ( 2 + 5 ) , 38 ) , ( − ln ( 2 + 5 ) , − 38 ) . \boxed{(0,0),\quad
\bigl(\ln(2+\sqrt5),38\bigr),\quad
\bigl(-\ln(2+\sqrt5),-38\bigr)}. ( 0 , 0 ) , ( ln ( 2 + 5 ) , 38 ) , ( − ln ( 2 + 5 ) , − 38 ) . 