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Question: Show that \(\cos hx + \sin hx = {e^x}\) and simplify \(\cos hx - \sin hx = ?\) . By considering \({(...

Show that cos⁡hx+sin⁡hx=ex\cos hx + \sin hx = {e^x} and simplify cos⁡hx−sin⁡hx=?\cos hx - \sin hx = ? . By considering (cos⁡hx+sin⁡hx)2+(cos⁡hx−sin⁡hx)2{(\cos hx + \sin hx)^2} + {(\cos hx - \sin hx)^2} show that cos⁡2hx−sin⁡2hx=cos⁡h2x.{\cos ^2}hx - {\sin ^2}hx = \cos h2x.

Explanation

Solution

Hint:-In this problem firstly you have to define cos⁡hx\cos hx and sin⁡hx\sin hx in terms of exponential function ex{e^x}. Then you have to apply various operations addition, subtraction, squaring of obtained expressions to get the required result or to simplify the given expression.

Complete step by step answer:
Now we know that
The hyperbolic functions have similar names to the trigonometric functions, but they are defined in terms of the exponential function. The hyperbolic functions cos⁡hx\cos hx and sin⁡hx\sin hx are defined using the exponential function ex{e^x}as:
cos⁡hx=ex+e−x2 eq.1 and sinhx=ex−e−x2 eq.2 adding eq.1 and eq.2 we get cos⁡hx+sin⁡hx=ex+e−x2 + ex−e−x2 ⇒cos⁡hx+sin⁡hx=ex eq.3 subtracting eq.2 from eq.1, we get cos⁡hx+sin⁡hx=ex+e−x2 − ex−e−x2 ⇒cos⁡hx−sin⁡hx=e−x eq.4 On squaring the eq.3 and eq.4,we get ⇒ (cos⁡hx+sin⁡hx)2=e2x eq.5 ⇒ (cos⁡hx−sin⁡hx)2=e−2x eq.6  add eq.5 and eq.6, we get ⇒ (cos⁡hx+sin⁡hx)2+(cos⁡hx−sin⁡hx)2=e2x+e−2x  On further solving the above equation, we get ⇒2(cos⁡2hx+sin⁡2hx)=e2x+e−2x  ∴ cos2hx + sin2hx = 1 ⇒(cos⁡2hx+sin⁡2hx)=e2x+e−2x2 ∴ cos⁡h2x = e2x+e−2x2 ⇒(cos⁡2hx+sin⁡2hx)=cos⁡h2x eq.7  \cos hx = \dfrac{{{e^x} + {e^{ - x}}}}{2}{\text{ eq}}{\text{.1}} \\\ {\text{and sin}}hx = \dfrac{{{e^x} - {e^{ - x}}}}{2}{\text{ eq}}{\text{.2}} \\\ {\text{adding eq}}{\text{.1 and eq}}{\text{.2 we get}} \\\ \cos hx + \sin hx = \dfrac{{{e^x} + {e^{ - x}}}}{2}{\text{ }} + {\text{ }}\dfrac{{{e^x} - {e^{ - x}}}}{2} \\\ \Rightarrow \cos hx + \sin hx = {e^x}{\text{ eq}}{\text{.3}} \\\ {\text{subtracting eq}}{\text{.2 from eq}}{\text{.1, we get}} \\\ \cos hx + \sin hx = \dfrac{{{e^x} + {e^{ - x}}}}{2}{\text{ }} - {\text{ }}\dfrac{{{e^x} - {e^{ - x}}}}{2} \\\ \Rightarrow \cos hx - \sin hx = {e^{ - x}}{\text{ eq}}{\text{.4}} \\\ {\text{On squaring the eq}}{\text{.3 and eq}}{\text{.4,we get}} \\\ \Rightarrow {\text{ (}}\cos hx + \sin hx{)^2} = {e^{2x}}{\text{ eq}}{\text{.5}} \\\ \Rightarrow {\text{ }}{(\cos hx - \sin hx)^2} = {e^{ - 2x}}{\text{ eq}}{\text{.6 }} \\\ {\text{add eq}}{\text{.5 and eq}}{\text{.6, we get}} \\\ \Rightarrow {\text{ (}}\cos hx + \sin hx{)^2} + {(\cos hx - \sin hx)^2} = {e^{2x}} + {e^{ - 2x}}{\text{ }} \\\ {\text{On further solving the above equation, we get}} \\\ \Rightarrow 2({\cos ^2}hx + {\sin ^2}hx) = {e^{2x}} + {e^{ - 2x}}{\text{ \\{ }}\therefore {\text{ co}}{{\text{s}}^2}{\text{hx + si}}{{\text{n}}^2}{\text{hx = 1\\} }} \\\ \Rightarrow ({\cos ^2}hx + {\sin ^2}hx) = \dfrac{{{e^{2x}} + {e^{ - 2x}}}}{2} \\\ \therefore {\text{ }}\cos h2x{\text{ = }}\dfrac{{{e^{2x}} + {e^{ - 2x}}}}{2} \\\ \Rightarrow ({\cos ^2}hx + {\sin ^2}hx) = \cos h2x{\text{ eq}}{\text{.7}} \\\
Hence proved,
cos⁡hx+sin⁡hx=ex\cos hx + \sin hx = {e^x} {∴\therefore from eq.3}
cos⁡2hx−sin⁡2hx=cos⁡h2x.{\cos ^2}hx - {\sin ^2}hx = \cos h2x. {∴\therefore from eq.7}
And simplification of cos⁡hx−sin⁡hx=e−x\cos hx - \sin hx = {e^{ - x}} {∴\therefore from eq.4}

Note:- Whenever you get this type of problem the key concept of solving is that you have knowledge about hyperbolic function. There are two base equation from which all other results will be derived are cos⁡hx=ex+e−x2 and sinhx=ex−e−x2 \cos hx = \dfrac{{{e^x} + {e^{ - x}}}}{2}{\text{ and sin}}hx = \dfrac{{{e^x} - {e^{ - x}}}}{2}{\text{ }}.Then by simple operation like squaring , addition , subtraction you can obtained the desired result. And one more thing to be remembered that hyperbolic functions are different from trigonometric functions.