How do I solve [math]2 \cosh 2 x + 10 \sinh 2 x = 5[/math]?
The hyperbolic functions sinh x and cosh x are defined asSinh x = (e^x - e^-x)/2 and cosh x = (e^x + e^-x)/2Hence, the given equation is2 {(e^2x + e^-2x)/2} + 10 { (e^2x-e^-2x)/2} = 5e^2x + e^-2x +5 (e^2x - e^-2x) = 5Simplifying we get 6e^2x - 4 e^-2x -5 =0Dividing by e^-2x we get6e^4x -5e^2x -4 = 0(3e^2x -4)(2e^2x +1) = 0e^2x =4/3 or e^2x = -1/2e^x cannot be negative.So e^2x =4/3Taking log , 2x = log 4/3 = log 4 - log 3x = log 2 - 1/2(log 3)
Solve the equation 2 cosh 2x - sinh 2x = 2?
2 cosh 2x - sinh 2x = 2 2 cosh²x + 2 sinh²x - 2 sinhx coshx = 2 2 sin²x + 2 sinh²x - 2 sinhx coshx = 0 sinx(2sinhx - cosx) = 0 Answer: x1 = 0 x2 = artanh(1/2) = 1/2 ln(3)
Solve for real values of x: 3 cosh 2x = 3 + sinh 2x?
cosh(t) = (1/2) * (e^(t) + e^(-t)) sinh(t) = (1/2) * (e^(t) - e^(-t)) 3 * cosh(2x) = 3 + sinh(2x) 3 * (1/2) * (e^(2x) + e^(-2x)) = 3 + (1/2) * (e^(2x) - e^(-2x)) 3 * (e^(2x) + e^(-2x)) = 6 + e^(2x) - e^(-2x) 3 * e^(2x) - e^(2x) + 3 * e^(-2x) + e^(-2x) - 6 = 0 2 * e^(2x) - 6 + 4 * e^(-2x) = 0 e^(2x) - 3 + 2 * e^(-2x) = 0 e^(4x) - 3 * e^(2x) + 2 = 0 e^(2x) = (3 +/- sqrt(9 - 8)) / 2 e^(2x) = (3 +/- 1) / 2 e^(2x) = 4/2 , 2/2 e^(2x) = 2 , 1 2x = ln(2) , 0 x = ln(2) / 2 , 0
If a = c cosh x and b = c sinh x, prove that (a + b)^2 (e^-2x) = a^2 - b^2?
(a + b)^2 * (e^(-2x)) => (c * cosh(x) + c * sinh(x))^2 * e^(-2x) => c^2 * (cosh(x) + sinh(x))^2 * e^(-2x) => c^2 * ((1/2) * (e^(x) + e^(-x)) + (1/2) * (e^(x) - e^(-x)))^2 * e^(-2x) => c^2 * (1/4) * (e^(x) + e^(x) + e^(-x) - e^(-x))^2 * e^(-2x) => (1/4) * c^2 * (2e^(x))^2 * e^(-2x) => (1/4) * c^2 * 4 * e^(2x) * e^(-2x) => c^2 a^2 - b^2 => c^2 * cosh(x)^2 - c^2 * sinh(x)^2 => c^2 * (cosh(x)^2 - sinh(x)^2) => c^2 * 1 => c^2 c^2 = c^2
If tanh(x) = 2/3, what is cosh(x) and cosh(2x)?
Note first that sech^2(x) = 1 - tanh^2(x), so in this case sech^2(x) = 1 - (2/3)^2 = 1 - (4/9) = 5/9, and thus cosh^2(x) = 1/sech^2(x) = 9/5. As cosh(x) > 0 for all x this means that cosh(x) = 3/sqrt(5). Next, cosh(2x) = cosh^2(x) + sinh^2(x) = cosh^2(x) + (cosh^2(x) - 1) = 2*cosh^2(x) - 1. So in this case we have cosh(2x) = 2*(9/5) - 1 = (18/5) - 1 = 13/5.
How to solve these hyperbolic functions?
1) When k = -2, then k^3 - 3k + 4 = 2. When k = -3, then k^3 - 3k + 4 = -14. Since the expression changes sign here, it has a zero between -3 and -2. Substituting y cosh x for k gives y^3 (cosh x)^3 = 3y cosh x - 4. The first two terms of thisequation will give us the form 4 (cosh x)^3 - 3 cosh x if (y^3)/3y = 4/3, so y^2 = 2 or -2. Since cosh x cannot be negative and we are looking for a negative value of k, choose y = -2. So the equation now is -8 (cosh x)^3 = -6 cosh x - 4, which becomes, using the given formula, 2 cosh 3x = 4. Solve this for x and then use k = 2 cosh x. 2) Square both sides of each equation and then add, using the property (cosh x)^2 + (sinh x)^2 = cosh 2x, and so get (x^2 + y^2) cosh 2w = 2 (i) Subracting instead of adding, and using (cosh x)^2 - (sinh x)^2 = 1, also 2 sinh x cosh x = sinh 2x, we get x^2 - y^2 = 2xy sinh 2w (ii) Hence (x^2 - y^2)^2 = 4 x^2 y^2 (sinh 2w)^2 = 4 x^2 y^2 ((cosh 2w)^2 -1) Add 4 x^2 y^2 to both sides: (x^2 + y^2)^2 = 4x^2 y^2 (cosh 2w)^2 Now multiply both sides by (x^2 + y^2)^2 and use equation (i) to replace (x^2 + y^2)^2 (cosh 2w)^2 by 4, giving (x^2 + y^2)^4 = 16 x^2 y^2 Taking the square root of both sides gives (x^2 + y^2)^2 = 4xy (3) From the given formula, 2(sinh A)^2 = cosh 2A - 1, and a related formula gives 2(cosh A)^2 = cosh 2A + 1. Square each of these and add, remembering that 2 (cosh 2A)^2 -1 = cosh 4A, and you have the result. Subtract instead of adding, and the second result is immediate.
Limits with hyperbolic functions?
Use the definitions of the hyperbolic functions as exponentials. lim(x→ ∞) sinh x = lim(x→ ∞) (1/2)(e^x - e^(-x)) = ∞, since lim(x→ ∞) e^(-x) = 0. lim(x→ -∞) sinh x = lim(x→ -∞) (1/2)(e^x - e^(-x)) = 0 - ∞ = -∞. -------------- Similarly, using cosh x = (1/2)(e^x + e^(-x), we have lim(x→ ±∞) cosh x = ∞. -------------- lim(x→ ∞) tanh x = lim(x→ ∞) (e^x - e^(-x))/(e^x + e^(-x)) = lim(x→ ∞) (1 - e^(-2x))/(1 + e^(-2x)), dividing each term by e^x = (1 - 0)/(1 + 0) = 1. lim(x→ -∞) tanh x = lim(x→ -∞) (e^x - e^(-x))/(e^x + e^(-x)) = lim(x→ -∞) (e^(2x) - 1)/(e^(2x) + 1), dividing each term by e^(-x) = (0 - 1)/(0 + 1) = -1. ------------------- For the other three, use the Reciprocal identities. lim(x→ ±∞) sech x = lim(x→ ±∞) 1/cosh x = 0 lim(x→ ±∞) csch x = lim(x→ ±∞) 1/sinh x = 0 lim(x→ ±∞) coth x = lim(x→ ±∞) 1/tanh x = ±1. I hope this helps!
Is there a difference between tan (x) and tanh (x)?
tan(x) = sin(x)/cos(x) tanh(x) = (e^x - e^-x)/(e^x +e^-x)