Question

QUESTION :

$\sf{If\:\dfrac{\cos\:x}{a_1}=\dfrac{\cos\:2x}{a_2}=\dfrac{\cos\:3x}{a_3}}$
$\sf{Prove\: that\:\:\sin^2\dfrac{x}{2}=\dfrac{2a_2-a_1-a_3}{4a_2}}$

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Answer 1

${\underline{\sf{Given}}}$

$\sf\dfrac{\cos\:x}{a_1}=\dfrac{\cos\:2x}{a_2}=\dfrac{\cos\:3x}{a_3}$

${\underline{\sf{To\:Prove}}}$

$\sf\sin^2\dfrac{x}{2}=\dfrac{2a_2-a_1-a_3}{4a_2}$

${\underline{\sf{Solution}}}$

We have to Prove

$\sf\sin^2\dfrac{x}{2}=\dfrac{2a_2-a_1-a_3}{4a_2}$

LHS :

$\sf=\sin^2(\dfrac{x}{2})$

RHS

$\sf=\dfrac{2a_2-a_1-a_3}{4a_2}$

$\sf=\dfrac{1}{4}\times\dfrac{2a_2-a_1-a_3}{ a_2}$

$\sf=\frac{1}{4}\times(\dfrac{2a_2}{a_2}-\dfrac{a_1}{a_2}-\dfrac{a_3}{a_2})$

$\sf=\dfrac{1}{4}\times(2-\dfrac{a_1}{a_2}-\dfrac{a_3}{a_2})$

$\sf=\dfrac{1}{4}\times(2-\dfrac{\cos\:x}{\cos2x}-\dfrac{\cos3x}{\cos2x})$

$\sf=\dfrac{1}{4}\times(2-\dfrac{\cos\:x+\cos3x}{\cos2x})$

We know that

$\displaystyle{\cos\:x+\cos\:y=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})}$

$\sf=\dfrac{1}{4}\times(2-\dfrac{2\cos(\frac{x+3x}{2})\cos(\frac{x-3x}{2})}{\cos2x})$

$\sf=\dfrac{1}{4}\times(2-\dfrac{2\cos2x\cos(-x)}{\cos2x})$

$\sf=\dfrac{1}{4}\times(2-\dfrac{2\cos2x\cos\:x}{\cos2x})$

$\sf=\dfrac{1}{4}\times(2-\dfrac{2\cancel{cos2x}\cos(x)}{\cancel{\cos2x}})$

$\sf=\dfrac{2}{4}\times(1-\cos\:x)$

$\sf=\dfrac{1}{2}\times[2\sin^2(\dfrac{x}{2})]$

$\sf=\sin^2(\dfrac{x}{2})$

Now , LHS = RHS

Hence Proved

$\rule{200}2$

${\underline{\sf{Formula's\:Used}}}$

$1)\sf1-\cos\theta =2\sin^2(\dfrac{\theta}{2})$

$2)\displaystyle{\cos\:x+\cos\:y=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})}$

Answer 2

Answer:

$\sf Let\:\:\:\dfrac{\cos\:x}{a_1}=\dfrac{\cos\:2x}{a_2}=\dfrac{\cos\:3x}{a_3} = 1\\\\\\:\implies\sf \dfrac{\cos\:x}{a_1} = 1\\\\:\implies \sf\cos \:x =a_1 \\\\\\\\:\implies\sf \dfrac{\cos\:2x}{a_2} = 1 \\\\:\implies \sf\cos \:2x =a_2 \\\\\\\\:\implies\sf \dfrac{\cos\:3x}{a_3} = 1 \\ \\:\implies \sf\cos \:3x =a_3$

⠀⠀⠀$\rule{160}{1.2}$

☯ $\underline{\boldsymbol{According\: to \:the\: Question :}}$

$\begin{aligned}:\implies\sf\:\:\sin^2 \left(\dfrac{x}{2} \right)=\dfrac{2a_2-a_1-a_3}{4a_2}\\\\\\\sf = \dfrac{1}{4} \times \bigg(\dfrac{2a_2-a_1-a_3}{a_2}\bigg)\\\\\\\sf = \dfrac{1}{4} \times \bigg(\dfrac{2a_2}{a_2} - \dfrac{a_1}{a_2} - \frac{a_3}{a_2} \bigg)\\\\\\\sf = \dfrac{1}{4} \times \bigg(2 - \dfrac{a_1}{a_2} - \frac{a_3}{a_2} \bigg)\\\\\\\sf = \dfrac{1}{4} \times \bigg(2 - \dfrac{(a_1 + a_3)}{a_2} \bigg)\\\\\\\sf = \dfrac{1}{4} \times \bigg(2 - \dfrac{(\cos\:x + \cos\:3x)}{\cos\:2x} \bigg)\\\\\\\sf = \dfrac{1}{4} \times \bigg(2 - \dfrac{2\cos\:(\frac{3x + x}{2})\cos \:( \frac{3x - x}{2})}{\cos\:2x} \bigg)\\\\\\\sf = \dfrac{1}{4} \times \bigg(2 - \dfrac{2\cos\:2x\cos \:x}{\cos\:2x} \bigg)\\\\\\\sf = \dfrac{1}{4} \times\bigg(2 - 2\cos \:x\bigg)\\\\\\\sf = \dfrac{1}{4} \times 2\bigg(1- \cos \:x\bigg)\\\\\\\sf = \dfrac{1}{2} \times 2\sin^2 \left(\dfrac{x}{2} \right)\\\\\\\sf =\sin^2 \left(\dfrac{x}{2}\right)\end{aligned}$