Question

If
$f(x) = \frac{1}{2} ( {3}^{x} + {3}^{ - x} )and \\ g(x) = \frac{1}{2} ( {3}^{x} - {3}^{ - x} ) then \: prove \: that \: f(x + y) = f(x)f(y) + g(x)g(y).$
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Answer 1

Answer:

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\pink{ \boxed{ \boxed{ \begin{array}{cc} \leadsto \bf \: Given \: : \\ \\ \rm \to \: f(x) = \frac{1}{2} ( {3}^{x} + {3}^{ - x}) \\ \\ \rm \to \: g(x) = \frac{1}{2} ( {3}^{x} - {3}^{ - x}) \\ \\ \\ \blue{ \underline{ \sf \: we \: have \: to \: prove \: :}} \\ \\ \bigstar \: \rm \:f(x + y) = f(x)f(y) + g(x)g(y) \end{array}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \red{ \bf \: Solution \:: }}$

$\orange{ \boxed{ \boxed{ \begin{array}{cc} L.H.S\: = \rm \: f(x + y) \: \: \: \: \: \: \: \: \: \: \: \\ \\ \: \: = \rm\:\frac{1}{2}( {3}^{x + y} + {3}^{ - x - y} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{array}}}}$

${ \boxed{ \boxed{ \begin{array}{cc} R.H.S\: = \rm \: f(x)f(y) + g(x) g(y) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = \small{ \rm\frac{1}{2} ( {3}^{x} + {3}^{ - x}) \frac{1}{2} ( {3}^{y} + {3}^{ - y} ) + \frac{1}{2} ( {3}^{x} - {3}^{ - x} ) \frac{1}{2} ( {3}^{y} - {3}^{ - y} )} \\ \\ \small{ \rm = \frac{1}{4} ( {3}^{x + y} + \cancel{ {3}^{x - y}} + \cancel{ {3}^{ - x + y}} + {3}^{ - x - y} + {3}^{x + y} - \cancel{{3}^{x - y} }- \cancel{ {3}^{ - x + y} } + {3}^{ - x - y} )} \\ \\ \rm = \frac{1}{4} \times 2( {3}^{x + y} + {3}^{ - x - y} ) \\ \\ \rm = \frac{1}{2} ( {3}^{x + y} + {3}^{ - x - y} ) \\ \\ = L.H.S \end{array}}}}$

Hence proved.

Assalamu A'laikum,

kemon acho?

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onek din kono kotha hoy ni. most of time, tomar question a jokhon visit kori tokhon kono space paina.

However,

hope your study is also going well...

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