If sinh x = frac{sqrt{21}}{2},then cosh 2x + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $\sinh x = \frac{\sqrt{21}}{2}$.We know that $\cosh^2 x - \sinh^2 x = 1$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + \frac{21}{4}} = \sqr

Vedclass · Accepted Answer

$\frac{23 + 5\sqrt{21}}{2}$

Vedclass · Answer

(C) Given $\sinh x = \frac{\sqrt{21}}{2}$.We know that $\cosh^2 x - \sinh^2 x = 1$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + \frac{21}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.Using the identities $\sinh 2x = 2 \sinh x \cosh x$ and $\cosh 2x = \cosh^2 x + \sinh^2 x$:$\sinh 2x = 2 	imes \frac{\sqrt{21}}{2} 	imes \frac{5}{2} = \frac{5\sqrt{21}}{2}$.$\cosh 2x = (\frac{5}{2})^2 + (\frac{\sqrt{21}}{2})^2 = \frac{25}{4} + \frac{21}{4} = \frac{46}{4} = \frac{23}{2}$.Therefore,$\cosh 2x + \sinh 2x = \frac{23}{2} + \frac{5\sqrt{21}}{2} = \frac{23 + 5\sqrt{21}}{2}$.

If $\sinh x = \frac{\sqrt{21}}{2}$,then $\cosh 2x + \sinh 2x = $

Similar Questions

$\text{If } \sin(\alpha+\beta)=1, \sin(\alpha-\beta)=\frac{1}{2}, \alpha, \beta \in [0, \frac{\pi}{2}], \text{ then } \tan(\alpha+2\beta) \cdot \tan(2\alpha+\beta) = ?$

Evaluate: $\sin ^4 \frac{\pi}{8} + \sin ^4 \frac{3\pi}{8} + \sin ^4 \frac{5\pi}{8} + \sin ^4 \frac{7\pi}{8} = $

If $x = 3 \sin \theta$,$y = 3 \cos \theta \cos \phi$,and $z = 3 \cos \theta \sin \phi$,then $x^{2} + y^{2} + z^{2} =$

If $p = \frac{2\sin \theta}{1 + \cos \theta + \sin \theta}$ and $q = \frac{\cos \theta}{1 + \sin \theta}$,then

For $0 \leq x \leq \pi$,if $81^{\sin ^2 x}+81^{\cos ^2 x}=30$,then $x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module