1 prove the following identities:
1个回答

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