Question

$\large ✯ \underline{\large\sf \pink{Question :- }}$


Prove that

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


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Answer 1

EXPLANATION.

$\sf \implies cos^{2} x + cos^{2} \bigg(x + \dfrac{\pi}{3} \bigg) + cos^{2} \bigg(x - \dfrac{\pi}{3} \bigg) = \dfrac{3}{2}$

As we know that,

Formula of :

⇒ cos2x = 2cos²x - 1.

⇒ cos2x + 1/2 = cos²x.

⇒ cos²(x + π/3).

⇒ cos2(x + π/3) + 1/2.

⇒ cos(2x + 2π/3) + 1/2.

⇒ cos²(x - π/3).

⇒ cos2(x - π/3) + 1/2.

⇒ cos(2x - 2π/3) + 1/2.

Put this formula in equation, we get.

$\sf \implies \dfrac{1 + cos2x}{2} \ + \dfrac{cos\bigg(2x + \dfrac{2\pi}{3} \bigg) + 1}{2} \ + \dfrac{cos \bigg(2x - \dfrac{2\pi}{3}\bigg) + 1}{2}$

$\sf \implies \dfrac{1}{2} \bigg[ 3 + cos2x + cos\bigg(2x + \dfrac{2\pi}{3} \bigg) + cos \bigg(2x - \dfrac{2\pi}{3} \bigg)\bigg]$

As we know that,

Formula of :

$\sf \implies cos(x) + cos(y) = 2 \ cos \bigg(\dfrac{x + y}{2} \bigg) . cos \bigg(\dfrac{x - y}{2} \bigg)$

Using this formula in equation, we get.

$\sf \implies \dfrac{1}{2} \bigg[ 3 + cos 2x + 2 cos \bigg( \dfrac{2x + \dfrac{2\pi}{3} + 2x - \dfrac{2\pi}{3} }{2} \bigg) . cos \bigg( \dfrac{2x + \dfrac{2\pi}{3} - \bigg(2x - \dfrac{2\pi}{3} \bigg)}{2} \bigg)\bigg]$

$\sf \implies \dfrac{1}{2} \bigg[ 3 + cos 2x + 2cos \bigg(\dfrac{4x}{2} \bigg) . cos \bigg( \dfrac{\dfrac{4\pi}{3} }{2} \bigg) \bigg]$

$\sf \implies \dfrac{1}{2} \bigg[ 3 + cos 2x + 2cos (2x) . cos \bigg( \dfrac{2\pi}{3} \bigg) \bigg]$

$\sf \implies \dfrac{1}{2} \bigg[ 3 + cos 2x + 2cos (2x) . cos \bigg( \pi - \dfrac{\pi}{3} \bigg) \bigg]$

$\sf \implies \dfrac{1}{2} \bigg[ 3 + cos 2x + 2cos (2x) . \bigg( \dfrac{-1}{2} \bigg) \bigg]$

$\sf \implies \dfrac{1}{2} \bigg[ 3 + cos2x - cos2x \bigg] = \dfrac{3}{2}$

Answer 2

Step-by-step explanation:

$\huge{ \underline{ \underline{ \textsf{ \textbf{ \purple{Answer :-}}}}}}$

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

LHS :

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

$\bf \longmapsto \: \bigg( \frac{1 + cos2x}{2} \bigg) + \left[ \: \frac{1 + cos2 \bigg(x + \frac{\pi}{3} \bigg)} {2} \right] + \left[ \: \frac{1 + cos2 \bigg(x - \frac{\pi}{3} \bigg)} {2} \right]$

$\bf \left[ \: cos \: 2x = 2 {cos}^{2} x - 1 \: \: (or) {cos}^{2} x = \frac{1 + cos2x}{2} \right]$

$\bf \longmapsto \: \frac{1}{2} \left [ 1 + cos \: 2x + 1 + cos2 \bigg(x + \frac{\pi}{3} \bigg) \right] + 1 + cos2 \bigg(x - \frac{\pi}{3} \bigg)$

$\bf \longmapsto \: \frac{1}{2} \left[3 + cos \: 2x + cos2 \bigg(x - \frac{\pi}{3} \bigg) + cos2 \bigg(x - \frac{\pi}{3} \bigg) \right]$

$\tt \: using \: cos \: x \: + \: cos \: y \: = 2 \: cos \: \bigg( \frac{x + y}{2} \bigg)cos \bigg( \frac{x - y}{2} \bigg)$

$\tt \: replace \: \: x \: \: by \: \: 2x + \frac{2\pi}{3} \: \: and \: \: y \: \: by2x - \frac{2\pi}{3}$

$\bf \longmapsto \: \frac{1}{2} \left [ 3 + cos2x + 2cos \bigg( \frac{2x + \frac{2\pi}{3} + 2x - \frac{2\pi}{3} }{2} \bigg)cos \bigg( \frac{2x + \frac{2\pi}{3} - (2x - \frac{2\pi}{3}) }{2} \bigg)\right]$

$\bf \longmapsto \: \frac{1}{2} \left[ 3 + cos \: 2x + cos \bigg( \frac{4x}{2} \bigg)cos \bigg( \frac{4\pi}{ \frac{3}{2} } \bigg) \right]$

$\bf \longmapsto \: \frac{1}{2} \left [ 3 + cos2 + 2cos2x \: \: cos \bigg( \frac{2\pi}{3} \bigg) \right]$

$\bf \longmapsto \: \frac{1}{2} \left [ 3 + cos2 + 2cos2x \: \: cos \bigg(\pi - \frac{\pi}{3} \bigg) \right]$

$\bf \longmapsto \: \frac{1}{2} \left [ 3 + cos2 + 2cos2x \: \: cos \bigg( - cos \: \bigg( \frac{\pi}{3} \bigg) \bigg)\right]$

$\bf \: Using \: \: cos (π- \theta)=-cos \theta$

$\bf \longmapsto \: \frac{1}{2} \left[ 3 + cos2x + 2cos2x \bigg( \frac{ - 1}{2} \bigg) \right]$

$\bf \longmapsto \: \frac{1}{2} \left[ 3 + cos2x - cos2x\right]$

$\bf \longmapsto \purple{ \frac{3}{2} }$

LHS = RHS

HENCE PROVED !!