Hyperbolic Functions (VSAQs)
Maths-1A | 9. Hyperbolic Functions – VSAQs:
Welcome to VSAQs in Chapter 9: Hyperbolic Functions. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.
VSAQ-1 : If sinhx = 3 then show that x = loge (3+√10).
Step 1: Definition of sinh x
The hyperbolic sine function, denoted as sinh x, is defined as:
$$\sinh x = \frac{e^x – e^{-x}}{2}$$
Given that sinhx=3, we can write:
$$\frac{e^x – e^{-x}}{2} = 3$$
Step 2: Definition of sinh−1x
The inverse hyperbolic sine function, denoted as sinh−1x or arsinhx, can be defined in terms of natural logarithms as follows:
$$\sinh^{-1} x = \log_e(x + \sqrt{x^2 + 1})$$
Step 3: Calculate x using sinhx=3
Given sinhx=3, we want to find x. According to the definition of sinh−1x, we can express x as:
$$x = \sinh^{-1}(3)$$
Using the formula for sinh−1x, we substitute 33 for x:
$$x = \log_e(3 + \sqrt{3^2 + 1})$$
Simplify the expression under the square root:
$$x = \log_e(3 + \sqrt{9 + 1})$$
$$x = \log_e(3 + \sqrt{10})$$
VSAQ-2 : If sinhx = 5 then show that x = log(5+√26).
Step 1: Definition of sinhx
The hyperbolic sine function, denoted as sinhx, is defined as:
$$\sinh x = \frac{e^x – e^{-x}}{2}$$
Step 2: Definition of sinh−1x
The inverse hyperbolic sine function, denoted as sinh−1x or arsinhx, can be expressed with the natural logarithm function (loge or ln) as follows:
$$\sinh^{-1} x = \log_e(x + \sqrt{x^2 + 1})$$
Step 3: Calculate x using sinhx=5
Given sinhx=5, to find x, we write:
$$x = \sinh^{-1}(5)$$
Using the formula for sinh−1x, we substitute 55 for x:
$$x = \log_e(5 + \sqrt{5^2 + 1})$$
Simplify the expression under the square root:
$$x = \log_e(5 + \sqrt{25 + 1})$$
$$x = \log_e(5 + \sqrt{26})$$
VSAQ-3 : Show that Tanh-11/2 = 1/2 loge3.
Step 1: Definition of tanh−1x
The inverse hyperbolic tangent function, denoted as tanh−1x, is defined as:
$$\tanh^{-1} x = \frac{1}{2}\log_e\left(\frac{1 + x}{1 – x}\right)$$
Step 2: Calculate
$$\tanh^{-1}\left(\frac{1}{2}\right)$$
$$\tanh^{-1}\left(\frac{1}{2}\right) = \frac{1}{2}\log_e\left(\frac{1 + \frac{1}{2}}{1 – \frac{1}{2}}\right)$$
Step 3: Simplify the expression
Simplify the fraction inside the logarithm:
$$\tanh^{-1}\left(\frac{1}{2}\right) = \frac{1}{2}\log_e\left(\frac{1 + \frac{1}{2}}{1 – \frac{1}{2}}\right) = \frac{1}{2}\log_e\left(\frac{\frac{3}{2}}{\frac{1}{2}}\right)$$
Simplify the complex fraction:
$$\tanh^{-1}\left(\frac{1}{2}\right) = \frac{1}{2}\log_e\left(3\right)$$
VSAQ-4 : Show that Tanh-11/4 = 1/2 loge5/3.
Step 1: Definition of tanh−1x
The inverse hyperbolic tangent function, denoted as tanh−1x, is given by the formula:
$$\tanh^{-1} x = \frac{1}{2}\log_e\left(\frac{1 + x}{1 – x}\right)$$
Step 2: Apply the formula to 1/4
$$\tanh^{-1}\left(\frac{1}{4}\right) = \frac{1}{2}\log_e\left(\frac{1 + \frac{1}{4}}{1 – \frac{1}{4}}\right)$$
Step 3: Simplify the expression
$$\tanh^{-1}\left(\frac{1}{4}\right) = \frac{1}{2}\log_e\left(\frac{\frac{5}{4}}{\frac{3}{4}}\right)$$
$$\tanh^{-1}\left(\frac{1}{4}\right) = \frac{1}{2}\log_e\left(\frac{5}{3}\right)$$
VSAQ-5 : If sinhx = ¾ find cosh2x and sinh2x.
Step 1: Calculate cosh(x)
Given that $$\cosh(x) = \sqrt{\sinh^2(x) + 1}$$
we first compute sinh2(x): $$\sinh^2(x) = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$$
Next, add 1 to sinh2(x) and take the square root:
$$\cosh(x) = \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{9}{16} + \frac{16}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$$
Step 2: Calculate cosh(2x)
Using the formula $$\cosh(2x) = \cosh^2(x) + \sinh^2(x)$$
$$\cosh^2(x) = \left(\frac{5}{4}\right)^2 = \frac{25}{16}$$
$$\sinh^2(x) = \frac{9}{16}$$
Adding these results gives us:
$$\cosh(2x) = \frac{25}{16} + \frac{9}{16} = \frac{34}{16} = \frac{17}{8}$$
Step 3: Calculate sinh(2x)
The formula for sinh(2x) is 2sinh(x)cosh(x):
$$\sinh(2x) = 2 \times \frac{3}{4} \times \frac{5}{4} = \frac{15}{8}$$
VSAQ-6 : If coshx = 5/2, find cosh(2x), sinh(2x).
Step 1: Calculate sinh(x)
From the identity $$\cosh^2(x) – \sinh^2(x) = 1$$
we can solve for sinh(x): $$\sinh^2(x) = \cosh^2(x) – 1$$
Given $$\cosh(x) = \frac{5}{2}$$
$$\sinh^2(x) = \left(\frac{5}{2}\right)^2 – 1 = \frac{25}{4} – 1 = \frac{25}{4} – \frac{4}{4} = \frac{21}{4}$$
Taking the square root (and considering the positive root because cosh(x) is positive):
$$\sinh(x) = \sqrt{\frac{21}{4}} = \frac{\sqrt{21}}{2}$$
Step 2: Calculate cosh(2x)
Using the double angle identity $$\cosh(2x) = 2\cosh^2(x) – 1$$
$$\cosh(2x) = 2\left(\frac{5}{2}\right)^2 – 1 = 2 \times \frac{25}{4} – 1 = \frac{50}{4} – 1 = \frac{50}{4} – \frac{4}{4} = \frac{46}{4} = \frac{23}{2}$$
Step 3: Calculate sinh(2x)
Using the identity $$\sinh(2x) = 2\sinh(x)\cosh(x)$$
$$\sinh(2x) = 2 \times \frac{\sqrt{21}}{2} \times \frac{5}{2} = \sqrt{21} \times \frac{5}{2} = \frac{5\sqrt{21}}{2}$$
VSAQ-7 : Prove that (coshx+sinhx)n=cosh(nx)+sinh(nx).
L.H.S $$= (\cosh(x) + \sinh(x))^n$$
$$\cosh(x) = \frac{e^x + e^{-x}}{2}, \quad \sinh(x) = \frac{e^x – e^{-x}}{2}$$
$$= \left( \frac{e^x + e^{-x}}{2} + \frac{e^x – e^{-x}}{2} \right)^n$$
Simplify the expression: $$= \left( \frac{2e^x}{2} \right)^n = (e^x)^n$$
Simplify further: $$= e^{nx} \quad \dots \text{(1)}$$
R.H.S $$= \cosh(nx) + \sinh(nx)$$
Again, using the definitions of hyperbolic cosine and hyperbolic sine, but this time with nx instead of x:
$$\cosh(nx) = \frac{e^{nx} + e^{-nx}}{2}, \quad \sinh(nx) = \frac{e^{nx} – e^{-nx}}{2}$$
Substitute these definitions into the R.H.S: $$= \frac{e^{nx} + e^{-nx}}{2} + \frac{e^{nx} – e^{-nx}}{2}$$
Simplify the expression: $$= \frac{2e^{nx}}{2}$$
Simplify further: $$= e^{nx} \quad \dots \text{(2)}$$
$$L.H.S = R.H.S$$
VSAQ-8 : Prove that (coshx-sinhx)n=cosh(nx)-sinh(nx).
Given:
$$LHS = (\cosh(x) – \sinh(x))^n$$
Start with the definition of cosh(x) and sinh(x):
$$\cosh(x) = \frac{e^x + e^{-x}}{2},$$
$$\sinh(x) = \frac{e^x – e^{-x}}{2}$$
Substitute these definitions into the LHS:
$$LHS = \left(\frac{e^x + e^{-x}}{2} – \frac{e^x – e^{-x}}{2}\right)^n.$$
Simplify the expression within the parenthesis:
$$LHS = \left(\frac{2e^{-x}}{2}\right)^n = (e^{-x})^n.$$
Further simplification gives:
$$LHS = e^{-nx}.$$
For the RHS:
Start with the definitions of cosh(nx) and sinh(nx):
$$\cosh(nx) = \frac{e^{nx} + e^{-nx}}{2},$$
$$\sinh(nx) = \frac{e^{nx} – e^{-nx}}{2}.$$
Substitute these definitions into the RHS:
$$RHS = \left(\frac{e^{nx} + e^{-nx}}{2}\right) – \left(\frac{e^{nx} – e^{-nx}}{2}\right).$$
Simplify the expression:
$$RHS = \frac{2e^{-nx}}{2} = e^{-nx}.$$
$$LHS = RHS$$
VSAQ-9 : Prove that cosh2x-sinh2x=1.
L.H.S: $$\cosh(2x) – \sinh(2x)$$
First, let’s recall the definitions of hyperbolic cosine and hyperbolic sine functions:
$$\cosh(x) = \frac{e^x + e^{-x}}{2}$$
$$\sinh(x) = \frac{e^x – e^{-x}}{2}$$
Substituting 2x in place of x, we have:
$$\cosh(2x) = \frac{e^{2x} + e^{-2x}}{2}$$
$$\sinh(2x) = \frac{e^{2x} – e^{-2x}}{2}$$
The expression becomes:
$$\cosh(2x) – \sinh(2x) = \left(\frac{e^{2x} + e^{-2x}}{2}\right) – \left(\frac{e^{2x} – e^{-2x}}{2}\right)$$
Simplifying:
$$= \frac{(e^{2x} + e^{-2x}) – (e^{2x} – e^{-2x})}{2}$$
$$= \frac{2e^{2x} – 2e^{2x}}{2} + \frac{2e^{-2x}}{2}$$
$$= \frac{2e^{-2x}}{2}$$
$$= e^{-2x} + e^{2x} – e^{2x}$$
$$= e^{-2x}$$
$$(a + b)^2 – (a – b)^2 = 4ab$$
VSAQ-10 : Prove that cosh2x = 2cosh2x-1.
Given:
$$\text{R.H.S} = 2 \cosh 2x – 1$$
First, we use the definition of the hyperbolic cosine function:
$$\cosh x = \frac{e^x + e^{-x}}{2}$$
Therefore,
$$\cosh 2x = \frac{e^{2x} + e^{-2x}}{2}$$
Substitute this into the given R.H.S:
$$\text{R.H.S} = 2\left( \frac{e^{2x} + e^{-2x}}{2} \right) – 1$$
Simplify the expression:
$$\text{R.H.S} = e^{2x} + e^{-2x} – 1$$
(OR)
$$\cosh 2x = \cosh^2 x + \sinh^2 x$$
$$\cosh 2x = 2\cosh^2 x – 1$$