Question

10
3
π
Prove that cos2 x + cos2x+
3
2.
+ COS
1
X
3​

Answer 1

$\large\underline\blue{\bold{Given \:  Question \: (correct \: statement) :-  }}$

☆ Prove that

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

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge \red{AηsωeR } ✍$

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☆Identity used :-

$\sf \:  ⟼1. \: cos(x + y) \: = \: cosx \: cosy \: - \: sinx \: siny$

$\sf \:  ⟼2. \: cos(x - y) \: = \: cosx \: cosy \: + \: sinx \: siny$

$\sf \:  ⟼ \: 3. \: {sin}^{2} x + {cos}^{2}x = 1$

$\sf \:  ⟼ \: 4. \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}$

$\sf \:  ⟼ \: 5. \: {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}$

$\sf \:  ⟼ \: 6. \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2} )$

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large\underline\purple{\bold{Solution :- }}$

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$\large\underline\red{\bold{❥︎Step :- 1 }}$

☆Consider

$\sf \:  ⟼{cos}^{2} (x + \dfrac{\pi}{3} )$

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

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

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

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

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$\large\underline\red{\bold{❥︎Step :- 2 }}$

☆Consider

$\sf \:  ⟼ \: {cos}^{2} (x - \dfrac{\pi}{3})$

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

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

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

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

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$\large\underline\red{\bold{❥︎Step :- 3 }}$

☆Cosider LHS

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

☆ On substituting, the values from equation (1) and (2), we get

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

$\sf \:  ⟼ \: {cos}^{2} x + 2\sf \: { \bigg( \dfrac{1}{4} {cos}^{2} x + \dfrac{3}{4} {sin}^{2}x \bigg)}$

$\sf \:  ⟼ {cos}^{2} x + \dfrac{1}{2} {cos}^{2} x \: + \dfrac{3}{2} \: {sin}^{2} x$

$\sf \:  ⟼\dfrac{3}{2} {cos}^{2} x + \dfrac{3}{2} {sin}^{2} x$

$\sf \:  ⟼ \: \dfrac{3}{2} ( {cos}^{2} x + {sin}^{2} x)$

$\sf \:  ⟼ \: \dfrac{3}{2} \: \times \: 1 \: = \dfrac{3}{2}$

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

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$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$