Question

$\mathbb{PROVE \: \: QUESTION} \\ \\ \longrightarrow \rm \cos^{2} x + \cos ^{2} \bigg\{ x + \dfrac{\pi}{3} \bigg\} + \cos^{2} \bigg\{ x - \dfrac{\pi}{3} \bigg \} = \dfrac{3}{2} \ \textless \ br /\ \textgreater \ \\ \\ \ \textless \ br /\ \textgreater \ \mathbb{NOTE \: IRRELEVANT \: ANSWER \: WILL \: BE \: REPORTED}$
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Answer 1

TO PROVE :–

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$\bf \longrightarrow \cos^{2} x + \cos ^{2} \left\{ x + \dfrac{\pi}{3} \right\} + \cos ^{2} \left\{ x - \dfrac{\pi}{3} \right\} = \dfrac{3}{2}$

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SOLUTION :–

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• Let's take L.H.S. –

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$\bf = \cos^{2} x + \cos ^{2} \left\{ x + \dfrac{\pi}{3} \right\} + \cos ^{2} \left\{ x - \dfrac{\pi}{3} \right\}$

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• Using identity –

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$\bf \implies \cos{2 \theta} = 2 { \cos}^{2}{ \theta} - 1$

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$\bf \implies { \cos}^{2}{ \theta} = \dfrac{1 +\cos{2 \theta}}{2}$

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• So that –

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

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

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

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

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• Using identity –

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$\bf \implies \cos(x) + \cos(y) = 2 \cos \left( \dfrac{x + y}{2} \right) \cos \left( \dfrac{x - y}{2} \right)$

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

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$\bf = \dfrac{1}{2} \left \{ 3+ \cos(2x) + 2\cos \left( 2x \right)\cos \left(\dfrac{2\pi}{3} \right) \right \}$

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• We know that –

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$\bf \implies \cos \left(\dfrac{2\pi}{3} \right) = - \dfrac{1}{2}$

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• So that –

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$\bf = \dfrac{1}{2} \left \{ 3+ \cos(2x) + 2\cos \left( 2x \right) \left( - \dfrac{1}{2} \right) \right \}$

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$\bf = \dfrac{1}{2} \left \{ 3+ \cos(2x) - \cos \left( 2x \right)\right \}$

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$\bf = \dfrac{1}{2} \left \{ 3\right \}$

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$\bf = \dfrac{3}{2}$

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$\bf = R.H.S.$

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$\bf \: \: \: \: \: \: \: { \underline{ \underline{ \bf Hence \: \: proved}}}$

Answer 2

EXPLANATION.

FORMULA USED.

$1) = \bold{\cos(2x) = 2 \cos {}^{2}x \: - 1}$

$\bold{2) = \cos {}^{2} (x) = \frac{ 1 + \cos(2x) }{2} }$

$\bold{3) = \cos(x) + \cos(y) = 2 \cos( \frac{x + y}{2}) \cos( \frac{x - y}{2} ) ) }$

$\bold{4 ) = \cos(\pi \: - \: \theta) = - \cos( \theta) }$

Note = For solution see the image

attachment.