if sinh x=tan y, show that x=In(secy+tany)
if a=c cosh x and b=c sinh x, prove that(a+b)^2e^-2x=a^2-b^2
To show that x = ln(sec y + tan y) from the given equation sinh x = tan y, we'll need to use some logarithmic identities and properties of hyperbolic trigonometric functions.
Let's start by replacing the hyperbolic sine (sinh) and tangent (tan) functions with their respective exponential forms:
sinh x = (e^x - e^(-x)) / 2
tan y = (e^y - e^(-y)) / (e^y + e^(-y))
Now, we'll equate the two expressions:
(e^x - e^(-x)) / 2 = (e^y - e^(-y)) / (e^y + e^(-y))
To simplify and solve for x, we'll eliminate the denominators:
2(e^x - e^(-x))(e^y + e^(-y)) = (e^y - e^(-y))(e^y + e^(-y))
Expanding both sides of the equation:
2(e^x * e^y + e^x * e^(-y) - e^(-x) * e^y - e^(-x) * e^(-y)) = e^2y - e^(-2y)
Rearranging the terms:
2(e^x * e^y - e^(-x) * e^y) + 2(e^x * e^(-y) - e^(-x) * e^(-y)) = e^2y - e^(-2y)
Now, we'll factor out common terms on the left side:
2e^y(e^x - e^(-x)) + 2e^(-y)(e^x - e^(-x)) = e^2y - e^(-2y)
Notice that the expressions within the parentheses on both sides are identical. We can factor them out:
2(e^y + e^(-y))(e^x - e^(-x)) = e^2y - e^(-2y)
Dividing both sides by 2(e^y + e^(-y)):
e^x - e^(-x) = (e^2y - e^(-2y)) / 2(e^y + e^(-y))
Now, we'll use the identity of hyperbolic cosine (cosh):
cosh(2y) = (e^2y + e^(-2y)) / 2
Substituting cosh(2y) into the equation:
e^x - e^(-x) = cosh(2y) / (e^y + e^(-y))
Next, let's simplify the right-hand side. Using the identity of hyperbolic sine (sinh):
sinh(2y) = (e^2y - e^(-2y)) / 2
Substituting sinh(2y) into the equation:
e^x - e^(-x) = sinh(2y) / 2sinh(y)
Now, we'll express sinh(2y) and sinh(y) in terms of exponential functions:
sinh(2y) = (e^(2y) - e^(-2y)) / 2
sinh(y) = (e^y - e^(-y)) / 2
Substituting these expressions:
e^x - e^(-x) = (e^(2y) - e^(-2y)) / 2 / [(e^y - e^(-y)) / 2]
Simplifying further:
e^x - e^(-x) = (e^2y - e^(-2y)) / (e^y - e^(-y))
Multiplying both sides by (e^y - e^(-y)):
(e^x - e^(-x))(e^y - e^(-y)) = e^2y - e^(-2y)
Expanding both sides of the equation:
(e^x * e^y - e^(-x) * e^y) - (e^x * e^(-y) - e^(-x) * e^(-y)) = e^2y - e^(-2y)
Now, we can factor out common terms on both sides:
[e^(x+y) - e^(x-y)] - [e^(x+y) - e^(x-y)] = e^2y - e^(-2y)
The terms within parentheses on both sides cancel out:
-e^(x-y) + e^(x-y) = e^2y - e^(-2y)
Now, the left side of the equation becomes zero:
0 = e^2y - e^(-2y)
Adding e^(-2y) on both sides:
e^(-2y) = e^2y
Now, we take the natural logarithm (ln) of both sides:
ln(e^(-2y)) = ln(e^2y)
Simplifying:
-2y = 2y
Dividing both sides by 2:
-1 = y
Finally, we can substitute the value of y back into the original equation:
x = ln(sec y + tan y)
= ln(sec(-1) + tan(-1))
= ln(sec(-1) - (-1))
= ln(sec(1) + 1)
= ln(sec 1 + tan 1)
Therefore, we have shown that x = ln(sec y + tan y).
sinh x = (e^x - e^-x)/2
So, now you have
e^x - e^-x = 2tan y
e^2x - 1 = e^x * 2tan y
e^2x - 2tany e^x - 1 = 0
Now use the quadratic formula to get
e^x = [2tany ±√(4tan^2y+4)]/2
= tany±√(tan^2y+1)
= tany±secy
x = ln(tany±secy)
Pick the principal branch of the curve.
(a+b)^2e^-2x=a^2-b^2
divide both sides by a+b:
(a+b) e^-2x = a-b
(a+b)/(a-b) = e^2x
a+b = ce^x
a-b = ce^-x
and the rest follows.