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Q.

Choose whether the given statement is true or false.

By using the definitions of $sinhx$ and $coshx$ in terms of exponential functions we can prove that

$\cosh 2 x = 2 \cos h^{2} x - 1$ .

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a

True

b

False

answer is A.

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Detailed Solution

Here, we have been provided with the hyperbolic functions

sinhx

and

coshx

and we are asked to prove the relation

\cosh 2 x = 2 \cos h^{2} x - 1

.
We now know that hyperbolic functions in mathematics are analogues of ordinary trigonometric functions defined for the hyperbola rather than the circle. The points (cost,sint) form the right half of the equilateral hyperbola in the same way that the points (cost,sint) form a circle with a unit radius. These hyperbolic functions are expressed in exponential form as follows: -
(i) sinhx =

\frac{e^{x} - e^{- x}}{2}

(ii) coshx =

\frac{e^{x} + e^{- x}}{2}

Now, let us come to the question, we have to prove

\cosh 2 x = 2 \cos h^{2} x - 1

. So, considering relation (ii) from the above listed relations, we have,
⇒coshx =

\frac{e^{x} + e^{- x}}{2}

Replacing ‘x’ with ‘2x’ in the above relation, we have,
⇒cosh2x=

\frac{e^{2 x} + e^{- 2 x}}{2}

- (1)
Now, squaring both sides of relation (ii), we get,
⇒

\cos h^{2} 2 x = {\frac{(e^{2 x} + e^{- 2 x})}{2^{2}}}^{2}

⇒

\cos h^{2} 2 x = {\frac{(e^{2 x} + e^{- 2 x})}{4}}^{2}

Applying the algebraic identity,

{(a + b)}^{2} = a^{2} + b^{2} + 2 ab

, we get,
⇒

\cos h^{2} x = \frac{e^{2 x} + e^{- 2 x} + 2 \times e^{x} - e^{- x}}{4}

⇒

\cos h^{2} x = \frac{e^{2 x} + e^{- 2 x} + 2}{4}

Multiplying both sides with 2, we get,
⇒2

\cos h^{2} x = 2 \times \frac{e^{2 x} + e^{- 2 x} + 2}{4}

⇒2

\cos h^{2} x = \frac{e^{2 x} + e^{- 2 x} + 2}{2}

Subtracting 1 from both sides, we get,
⇒2

\cos h^{2} x - 1 = \frac{e^{2 x} + e^{- 2 x} + 2}{2} - 1

Taking L.C.M we get,
⇒2

\cos h^{2} x - 1 = \frac{e^{2 x} + e^{- 2 x} + 2 - 2}{2}

⇒2

\cos h^{2} x - 1 = \frac{e^{2 x} + e^{- 2 x}}{2}

- (2)
Clearly, we can see that the R.H.S of equation (1) and (2) are same, so comparing and equating their L.H.S, we get,
⇒cosh 2x=2

\cos h^{2} x - 1

Hence, proved
So, the given statement is true.

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