Answer
Verified
414.9k+ views
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Explain zero factorial class 11 maths CBSE
What is the past participle of wear Is it worn or class 10 english CBSE