# Prove the following identity:

\[{\cosh ^2}x{\cos ^2}x - {\sinh ^2}x{\sin ^2}x = \dfrac{1}{2}(1 + \cosh 2x\cos 2x)\]

Last updated date: 17th Mar 2023

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Answer

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Hint: Substitute the formula for hyperbolic sine and cosine, that is, \[\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}\] and \[\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}\] in the left-hand side of the equation and simplify to complete the proof.

Complete step by step answer:

Let us start proving by assigning the left-hand side of the equation to the LHS.

\[LHS = {\cosh ^2}x{\cos ^2}x - {\sinh ^2}x{\sin ^2}x.........(1)\]

Hyperbolic functions are similar to trigonometric functions but they are defined in terms of the exponential function. Like, sin x and cos x is defined on a circle, sinh x and cosh x are defined on a hyperbola, thus giving its name. The point (cos t, sin t) lies on the circle.

Similarly, the point (cosh t, sinh t) lies on a hyperbola.

We know the formula for hyperbolic cosine and sine in terms of exponential function.

\[\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}..........(2)\]

\[\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}..........(3)\]

Substituting equation (2) and equation (3) in equation (1), we get:

\[LHS = {\left( {\dfrac{{{e^x} + {e^{ - x}}}}{2}} \right)^2}{\cos ^2}x - {\left( {\dfrac{{{e^x} - {e^{ - x}}}}{2}} \right)^2}{\sin ^2}x\]

Evaluating the squares, we get:

\[LHS = \left( {\dfrac{{{e^{2x}} + {e^{ - 2x}} + 2}}{4}} \right){\cos ^2}x - \left( {\dfrac{{{e^{2x}} + {e^{ - 2x}} - 2}}{4}} \right){\sin ^2}x\]

Grouping together the common exponential terms, we get:

\[LHS = \dfrac{{{e^{2x}}}}{4}({\cos ^2}x - {\sin ^2}x) + \dfrac{{{e^{ - 2x}}}}{4}({\cos ^2}x - {\sin ^2}x) + \dfrac{2}{4}({\cos ^2}x + {\sin ^2}x)\]

We know that \[{\cos ^2}x + {\sin ^2}x = 1\], hence, we have:

\[LHS = \dfrac{{{e^{2x}}}}{4}({\cos ^2}x - {\sin ^2}x) + \dfrac{{{e^{ - 2x}}}}{4}({\cos ^2}x - {\sin ^2}x) + \dfrac{1}{2}\]

We know that \[{\cos ^2}x - {\sin ^2}x = \cos 2x\], hence, we have:

\[LHS = \dfrac{{{e^{2x}}}}{4}(\cos 2x) + \dfrac{{{e^{ - 2x}}}}{4}(\cos 2x) + \dfrac{1}{2}\]

Now, taking cos2x as a common term, we have:

\[LHS = \cos 2x\left( {\dfrac{{{e^{2x}} + {e^{ - 2x}}}}{4}} \right) + \dfrac{1}{2}\]

Now using equation (2), the term inside the bracket can be expressed in terms of cosh2x.

Hence, we get:

\[LHS = \cos 2x\left( {\dfrac{{\cosh 2x}}{2}} \right) + \dfrac{1}{2}\]

Taking the factor of half as a common term, we have:

\[LHS = \dfrac{1}{2}\left( {1 + \cos 2x\cosh 2x} \right)\]

The right hand side of the equation is nothing but the right hand side of the proof, hence, we get:

\[LHS = RHS\]

Hence, we proved.

Note: You can also use the relation between the hyperbolic sine and cosine, \[{\cosh ^2}x - {\sinh ^2}x = 1\] and \[\cosh 2x = {\cosh ^2}x + {\sinh ^2}x\] to complete the proof.

Complete step by step answer:

Let us start proving by assigning the left-hand side of the equation to the LHS.

\[LHS = {\cosh ^2}x{\cos ^2}x - {\sinh ^2}x{\sin ^2}x.........(1)\]

Hyperbolic functions are similar to trigonometric functions but they are defined in terms of the exponential function. Like, sin x and cos x is defined on a circle, sinh x and cosh x are defined on a hyperbola, thus giving its name. The point (cos t, sin t) lies on the circle.

Similarly, the point (cosh t, sinh t) lies on a hyperbola.

We know the formula for hyperbolic cosine and sine in terms of exponential function.

\[\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}..........(2)\]

\[\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}..........(3)\]

Substituting equation (2) and equation (3) in equation (1), we get:

\[LHS = {\left( {\dfrac{{{e^x} + {e^{ - x}}}}{2}} \right)^2}{\cos ^2}x - {\left( {\dfrac{{{e^x} - {e^{ - x}}}}{2}} \right)^2}{\sin ^2}x\]

Evaluating the squares, we get:

\[LHS = \left( {\dfrac{{{e^{2x}} + {e^{ - 2x}} + 2}}{4}} \right){\cos ^2}x - \left( {\dfrac{{{e^{2x}} + {e^{ - 2x}} - 2}}{4}} \right){\sin ^2}x\]

Grouping together the common exponential terms, we get:

\[LHS = \dfrac{{{e^{2x}}}}{4}({\cos ^2}x - {\sin ^2}x) + \dfrac{{{e^{ - 2x}}}}{4}({\cos ^2}x - {\sin ^2}x) + \dfrac{2}{4}({\cos ^2}x + {\sin ^2}x)\]

We know that \[{\cos ^2}x + {\sin ^2}x = 1\], hence, we have:

\[LHS = \dfrac{{{e^{2x}}}}{4}({\cos ^2}x - {\sin ^2}x) + \dfrac{{{e^{ - 2x}}}}{4}({\cos ^2}x - {\sin ^2}x) + \dfrac{1}{2}\]

We know that \[{\cos ^2}x - {\sin ^2}x = \cos 2x\], hence, we have:

\[LHS = \dfrac{{{e^{2x}}}}{4}(\cos 2x) + \dfrac{{{e^{ - 2x}}}}{4}(\cos 2x) + \dfrac{1}{2}\]

Now, taking cos2x as a common term, we have:

\[LHS = \cos 2x\left( {\dfrac{{{e^{2x}} + {e^{ - 2x}}}}{4}} \right) + \dfrac{1}{2}\]

Now using equation (2), the term inside the bracket can be expressed in terms of cosh2x.

Hence, we get:

\[LHS = \cos 2x\left( {\dfrac{{\cosh 2x}}{2}} \right) + \dfrac{1}{2}\]

Taking the factor of half as a common term, we have:

\[LHS = \dfrac{1}{2}\left( {1 + \cos 2x\cosh 2x} \right)\]

The right hand side of the equation is nothing but the right hand side of the proof, hence, we get:

\[LHS = RHS\]

Hence, we proved.

Note: You can also use the relation between the hyperbolic sine and cosine, \[{\cosh ^2}x - {\sinh ^2}x = 1\] and \[\cosh 2x = {\cosh ^2}x + {\sinh ^2}x\] to complete the proof.

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