10 Most VSAQ’s of Hyperbolic Functions Chapter in Inter 1st Year Maths-1A (TS/AP)
2 Marks
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)
$$\cosh 2x = \cosh^2 x + \sinh^2 x$$
$$\cosh 2x = 2\cosh^2 x – 1$$