1provethefollowingidentities:a.cosh(2x)=cosh^2(x)+sinh^2(x)b.cosh(x+y)=cosh(x)cosh(y)+sinh(x)sinh(y)2.showthattheinversehyperboliccosinefunctioniscosh^-1(x)=ln(x+根号下x^2-1)byadaptingthemethodusedinclasstoderivetheinvers

Question

1 prove the following identities:
a.cosh(2x)=cosh^2(x)+sinh^2(x)
b.cosh(x+y)=cosh(x)cosh(y)+sinh(x)sinh(y)
2.show that the inverse hyperbolic cosine function is cosh^-1(x)=ln( x+根号下x^2-1 ) by adapting the method used in class to derive the inverse of the hyperbolic sine function.
3.verify the differentiation formula d/dx[sech^-1(x)]=-1/x根号下1-x^2
3
4.let x>0.show that ∫ e^xt dt=2sinh(3x)/x
-3

Answer 1

1.
a.
cosh(2x)
= [e^(2x) + e^(-2x)]/2
= [(e^x)²+(e^-x)²]/2
= [(e^x + e^-x)² + (e^x - e^-x)²]/4
= (e^x + e^-x)²/4 + (e^x - e^-x)²/4
= cosh²(x) + sinh²(x)
b.
cosh(x+y)
= [e^(x+y) + e^(-x-y)]/2
= [e^x * e^y + e^-x * e^-y]/2
= [e^x * e^y + e^x * e^-y +e^-x * e^y + e^-x * e^-y]/4 + [e^x * e^y - e^x * e^-y -e^-x * e^y + e^-x * e^-y]/4
= (e^x + e^-x)(e^y + e^-y)/4 + (e^x - e^-x)(e^y - e^-y)/4
= cosh(x)cosh(y)+sinh(x)sinh(y)
2.
cosh(x) = (e^x + e^-x)/2
令y = cosh(x),即x = cosh^-1(y),则2y = e^x + e^-x
(e^x)² - 2ye^x + 1 = 0
e^x = y ± √(y²-1)
x = ln[y ± √(y²-1)] = cosh^-1(y)
即cosh^-1(x) = ln[x ± √(x²-1)]
因为x - √(x²-1) = 1/[x + √(x²-1)] 恒不大于1（小于或等于1）
则ln[x - √(x²-1)] ≤ 0
一般取该函数的正支,即cosh^-1(x) = ln[x + √(x²-1)]
3.
令y = sech^-1(x),则x = sech(y),dx = [sech(y)]'dy
而d[sech^-1(x)]/dx = dy/dx = 1/[sech(y)]'
sech(y) = 2/(e^y + e^-y)
[sech(y)]' = [2/(e^y + e^-y)]' = -2(e^y - e^-y)/(e^y + e^-y)² = -sinh(y)/cosh²(y)
dy/dx = 1/[sech(y)]' = -cosh²(y)/sinh(y)
因为x = sech(y),所以cosh(y) = 1/x,sinh(y) = √(1 - 1/x²)
代入上式得到dy/dx = -1/x² * 1/√(1 - 1/x²) = -1/[x√(x²-1)]
即d[sech^-1(x)]/dx = -1/[x√(x²-1)]
4.
∫(-3,3)e^xt dt
= 1/x * ∫(-3,3)e^xt dxt
= 1/x * e^xt|(-3,3)
= 1/x * (e^3x - e^-3x)
= 2sinh(3x)/x