Question

Prove that

${cos}^{2} x + {cos}^{2} (x + \frac{\pi}{3}) + {cos}^{2}(x - \frac{\pi}{3}) = \frac{3}{2}$
without using formulae of multiple angles.

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Answer 1

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

Consider LHS

$\rm \: {cos}^{2} x + {cos}^{2}\bigg (x + \dfrac{\pi}{3}\bigg) + {cos}^{2}\bigg(x - \dfrac{\pi}{3}\bigg)$

Let consider

$\rm \: {cos}^{2}\bigg (x + \dfrac{\pi}{3}\bigg)$

$\rm \: =  \: {\bigg(cos\bigg(x + \dfrac{\pi}{3}\bigg)\bigg) }^{2}$

$\rm \: =  \: {\bigg(cosx \: cos\dfrac{\pi}{3} - \: sinx \: sin\dfrac{\pi}{3}\bigg) }^{2}$

$\rm \: =  \: {\bigg(\dfrac{1}{2}cosx - \dfrac{ \sqrt{3} }{2}sinx\bigg) }^{2}$

$\rm\implies \: {cos}^{2}\bigg(x + \dfrac{\pi}{3} \bigg) =  \: {\bigg(\dfrac{1}{2}cosx - \dfrac{ \sqrt{3} }{2}sinx\bigg) }^{2} - - (1)$

Now, Consider

$\rm \: {cos}^{2}\bigg (x - \dfrac{\pi}{3}\bigg)$

$\rm \: =  \: {\bigg(cos\bigg(x - \dfrac{\pi}{3}\bigg)\bigg) }^{2}$

$\rm \: =  \: {\bigg(cosx \: cos\dfrac{\pi}{3} + \: sinx \: sin\dfrac{\pi}{3}\bigg) }^{2}$

$\rm \: =  \: {\bigg(\dfrac{1}{2}cosx + \dfrac{ \sqrt{3} }{2}sinx\bigg) }^{2}$

$\rm\implies \: {cos}^{2}\bigg(x - \dfrac{\pi}{3} \bigg) =  \: {\bigg(\dfrac{1}{2}cosx + \dfrac{ \sqrt{3} }{2}sinx\bigg) }^{2} - - (2)$

Now, on substituting these in LHS,

$\rm \: {cos}^{2} x + {cos}^{2}\bigg (x + \dfrac{\pi}{3}\bigg) + {cos}^{2}\bigg(x - \dfrac{\pi}{3}\bigg)$

$\rm \: =  \: {cos}^{2}x + {\bigg(\dfrac{1}{2}cosx - \dfrac{ \sqrt{3} }{2}sinx\bigg) }^{2} + {\bigg(\dfrac{1}{2}cosx + \dfrac{ \sqrt{3} }{2}sinx\bigg) }^{2}$

We know,

$\boxed{\sf{  \:\rm \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2}) \: \: }}$

So, using this identity, we get

$\rm \: =  \: {cos}^{2}x + 2\bigg[\dfrac{1}{4} {cos}^{2}x + \dfrac{3}{4} {sin}^{2}x \bigg]$

$\rm \: =  \: {cos}^{2}x + \dfrac{1}{2} {cos}^{2}x + \dfrac{3}{2} {sin}^{2}x$

$\rm \: =  \: \dfrac{3}{2} {cos}^{2}x + \dfrac{3}{2} {sin}^{2}x$

$\rm \: =  \: \dfrac{3}{2} \bigg( {cos}^{2}x + {sin}^{2}x\bigg)$

$\rm \: =  \: \dfrac{3}{2}$

Hence,

$\boxed{{\:\rm \: {cos}^{2} x + {cos}^{2}\bigg (x + \dfrac{\pi}{3}\bigg) + {cos}^{2}\bigg(x - \dfrac{\pi}{3}\bigg) = \frac{3}{2} \: }}$

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Formulae Used :-

$\boxed{\sf{  \:\rm \: cos(x + y) = cosx \: cosy \: - \: sinx \: siny \: \: }}$

$\boxed{\sf{  \:\rm \: cos(x - y) = cosx \: cosy \: + \: sinx \: siny \: \: }}$

$\boxed{\sf{  \:\rm \: cos\dfrac{\pi}{3} = \frac{1}{2} \: \: }}$

$\boxed{\sf{  \:\rm \: sin\dfrac{\pi}{3} = \frac{ \sqrt{3} }{2} \: \: }}$

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Additional Information :-

$\boxed{\sf{  \:\rm \: sin(x + y) = sinxcosy + sinycosx \: }}$

$\boxed{\sf{  \:\rm \: sin(x - y) = sinxcosy - sinycosx \: }}$

$\boxed{\sf{  \:\rm \: tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany} \: \: }}$

$\boxed{\sf{  \:\rm \: tan(x - y) = \frac{tanx - tany}{1 + tanx \: tany} \: \: }}$

Answer 2

Step-by-step explanation:

$\pmb{ \text{LHS \( \tt =\cos ^{2} x+\cos ^{2}\left(x+\dfrac{\pi}{3}\right)+\cos ^{2}\left(x-\dfrac{\pi}{3}\right) \)}}$

$\pmb{ \tt=\left(\dfrac{1+\cos 2 x}{2}\right)+\left[\dfrac{1+\cos 2\left(x+\dfrac{\pi}{3}\right)}{2}\right]+\left[\dfrac{1+\cos 2\left(x-\dfrac{\pi}{3}\right)}{2}\right]}$

$\tt=\dfrac{1}{2}\left[1+\cos 2 x+1+\cos 2\left(x+\dfrac{\pi}{3}\right)\right]+1+\cos 2\left(x-\dfrac{\pi}{3}\right)$

$\tt=\dfrac{1}{2}\left[3+\cos 2 x+\cos 2\left(x+\dfrac{\pi}{3}\right)+\cos 2\left(x-\dfrac{\pi}{3}\right)\right]$

$\text{Using \( \tt \cos x+\cos y=2 \cos \left(\dfrac{x+y}{2}\right) \cos \left(\dfrac{x-y}{2}\right) \)}$

$\text{Replace \( \tt x \) by \( \tt \left(2 x+\dfrac{2 \pi}{3}\right) \) and \( \tt y \) by \( \tt 2 x-\dfrac{2 \pi}{3} \)}$

$\tt=\dfrac{1}{2}\left[3+\cos 2 x+2 \cos \left(\dfrac{2 x+\dfrac{2 \pi}{3}+2 x-\dfrac{2 \pi}{3}}{2}\right)\right. \left.\cos \left(\dfrac{2 x+\dfrac{2 \pi}{3}-\left(2 x-\dfrac{2 \pi}{3}\right)}{2}\right)\right]$

$\color{darkcyan} \pmb{ \begin{aligned}=& \tt \frac{1}{2}\left[3+\cos 2 x+\cos \left(\frac{4 x}{2}\right) \cos \left(\frac{4 \pi}{\frac{3}{2}}\right)\right] \\ \\ =& \tt\frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x \cos \left(\frac{2 \pi}{3}\right)\right] \\ \\ =& \tt \frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x \cos \left(\pi-\frac{\pi}{3}\right)\right] \\ \\ =& \tt \frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x\left(-\cos \left(\frac{\pi}{3}\right)\right)\right] \\ \\ &[\text { using } \tt \cos (\pi-\theta)=-\cos \theta] \\ \\ =& \tt\frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x\left(-\frac{1}{2}\right)\right] \\ \\ =& \tt \frac{1}{2}[3+\cos 2 x-\cos 2 x] \\ \\ =& \tt\frac{3}{2}=\text { RHS } \end{aligned} }$