Question

Prove it :-

$\\\bullet\quad\sf \cos\theta+i\sin\theta=e^{i\theta}$

Answer 1

$\large\underline{\sf{Solution-}}$

We know

$\rm :\longmapsto\:cosx = 1 - \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{4} }{4!} - \dfrac{ {x}^{6} }{6!} + - -$

$\rm :\longmapsto\:sinx = x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} - \dfrac{ {x}^{7} }{7!} + - - -$

Also, we know

$\rm :\longmapsto\: {e}^{x} = 1 + x + \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{3}}{3!} + \dfrac{ {x}^{4} }{4!} + \dfrac{ {x}^{5} }{5!} - - -$

Now, Replace x by ix

So, we get

$\rm :\longmapsto\: {e}^{ix} = 1 + ix + \dfrac{ {(ix)}^{2} }{2!} + \dfrac{ {(ix)}^{3}}{3!} + \dfrac{ {(ix)}^{4} }{4!} + \dfrac{ {(ix)}^{5} }{5!} - - -$

$\rm :\longmapsto\: {e}^{ix} = 1 + ix + \dfrac{ { {i}^{2} x}^{2} }{2!} + \dfrac{{ {i}^{3} x}^{3}}{3!} + \dfrac{ { {i}^{4} x}^{4} }{4!} + \dfrac{ { {i}^{5} x}^{5} }{5!} - - -$

We know,

$\boxed{ \tt{ \: {i}^{2} = - 1}} \: \: \boxed{ \tt{ \: {i}^{3} = - i}} \: \: \boxed{ \tt{ \: {i}^{4} = 1 \: }} \: \: \boxed{ \tt{ \: {i}^{5} = i \: }}$

So, using this, we get

$\rm :\longmapsto\: {e}^{ix} = 1 + ix - \dfrac{{x}^{2} }{2!} - \dfrac{{ix}^{3}}{3!} + \dfrac{{x}^{4} }{4!} + i\dfrac{{x}^{5} }{5!} - - -$

$\rm :\longmapsto\: {e}^{ix}=\bigg[1 - \dfrac{ {x}^{2} }{2!}+\dfrac{ {x}^{4} }{4!} + - - \bigg] + i\bigg[x - \dfrac{ {x}^{3} }{3!}\dfrac{ {x}^{5}}{5!} + - -\bigg]$

So, using expansion of sinx and cosx, we get

$\rm :\longmapsto\: {e}^{ix} = cosx \: + \: i \: sinx$

Now, Replace

$\red{\rm :\longmapsto\:x \: with \: \theta \: \: we \: get \: }$

$\rm :\longmapsto\: \boxed{ \tt{ \: {e}^{i\theta} = cos\theta \: + \: i \: sin\theta \: }}$

Hence, Proved

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Learn More :-

Argument of complex number

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Complex \: number & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x + iy & \sf {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg| \\ \\ \sf - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf - x - iy & \sf - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}} \\ \end{gathered}$

Answer 2

Given Expression :-

$\\ \quad \bullet \quad \sf \bigg(\dfrac{x}{y} \bigg)^{a-b} \times \bigg( \dfrac{y}{z} \bigg)^{b-c} \times \bigg( \dfrac{z}{x} \bigg)^{c-a}$

We know that :-

$\\ \quad \mapsto \quad \boxed{ \frak{ \bigg(\dfrac{a}{b} \bigg) ^{c - d} = \dfrac{ {a}^{c} }{ {b}^{d} } } } \bigstar$

Calculation :-

$\\ \quad \longrightarrow \quad \sf \bigg(\dfrac{x}{y} \bigg)^{a-b} \times \bigg( \dfrac{y}{z} \bigg)^{b-c} \times \bigg( \dfrac{z}{x} \bigg)^{c-a}$

$\\ \quad \longrightarrow \quad \sf \dfrac{ {x}^{a} }{ {y}^{b} } \times \dfrac{ {y}^{b} }{ {z}^{c} } \times \dfrac{ {z}^{c} }{ {x}^{a} }$

$\\ \quad \therefore \quad \boxed{ \frak{1}} \bigstar$

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