Question

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Prove that,
$\cos^{2} x + cos {}^{2} (x + \frac{\pi}{3} ) + cos {}^{2} (x - \frac{\pi}{3} ) = \frac{3}{2}$
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Answer 1

Step-by-step explanation:

$cos²x+cos²( x+\frac{\pi}{3}) + cos²(x- \frac{\pi}{3} )\\ = \frac{1 + cos2x}{2} + \frac{1}{2} [2cos²(x+ \frac{\pi}{2} )+2cos²(x- \frac{\pi}{3} )] \\ = \frac{1 + cos2x}{2} + \frac{1}{2} [1 + cos(2x+ \frac{2\pi}{3} )+1+cos(2x - \frac{2\pi}{3} )] \\ =\frac{1+cos2x}{2} + \frac{1}{2} [2+cos(2x+ \frac{2\pi}{3} )+cos(2x- \frac{2\pi}{3} )] \\ = \frac{1 + cos2x}{2} + \frac{1}{2} [2+2cos( \frac{2x+ \frac{2\pi}{3} +2x- \frac{2\pi}{3} }{2} ) \times cos( \frac{2x+ \frac{2\pi}{3}-2x+ \frac{2\pi}{3} }{2} ] \\ = \frac{1 + cos2x}{2} + \frac{1}{2} [2+2cos2xcos \frac{2\pi}{3} ] \\ = \frac{1+cos2x}{2} + 1 + cos2x \times cos \frac{2\pi}{3} \\ = \frac{1}{2} + \frac{cos2x}{2} + 1 + cos2xcos(\pi - \frac{\pi}{3} ) \\ = \frac{1}{2} + \frac{cos2x}{2} +1 + cos2x(- \frac{1}{2} ) \\ = \frac{1}{2} + 1 \\ = \frac{3}{2}$

$what\:i\:have\:used\:\:\\cos2x = 2cos²x-1 \\ cos²x= \frac{1 + cos2x}{2} \\ 2cos²x=1+cos2x\\cosA+cosB=2cos \frac{A}{2} cos \frac{B}{2}$

Answer 2

Answer:

LHS = RHS

‎

Step-by-step explanation:

Taking the LHS we get,‎

$\sf \Longrightarrow cos^{2}x + cos^{2}\Bigg\{x + \dfrac{\pi}{3}\Bigg\} + cos^2\Bigg\{x - \dfrac{\pi}{3}\Bigg\}$

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We know that,

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$\sf \dashrightarrow cos2\theta = 2cos^{2}\theta - 1$

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$\sf \dashrightarrow cos2\theta + 1 = 2cos^{2}\theta$

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$\sf \dashrightarrow \ \dfrac{\textsf{\textbf{cos2}}\theta \textsf{\textbf{ + 1}}}{\textsf{\textbf{2}}} = \textsf{\textbf{cos}}^{\textsf{\textbf{2}}}\theta$

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On substituting this relation in the LHS we get, where,

• For the first addend, θ = x

• For the second addend,  θ = {x + (π/3)}

• For the final addend,  θ = {x - (π/3)}

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$\sf \Longrightarrow \dfrac{cos2x + 1}{2} + \dfrac{cos2\bigg\{x + \dfrac{\pi}{3}\bigg\} + 1}{2} + \dfrac{cos2\bigg\{x - \dfrac{\pi}{3}\bigg\} + 1}{2}$

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$\sf \Longrightarrow \dfrac{cos2x + 1 + cos2\bigg\{x + \dfrac{\pi}{3}\bigg\} + 1 + cos2\bigg\{x - \dfrac{\pi}{3} \bigg\} + 1}{2}$

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On re-arranging on the basis of like-terms we get,

‎

$\sf \Longrightarrow \dfrac{1 + 1 + 1 + cos2x + cos2\bigg\{x + \dfrac{\pi}{3}\bigg\} + cos2\bigg\{x - \dfrac{\pi}{3} \bigg\}}{2}$

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$\sf \Longrightarrow \dfrac{3+ cos2x + cos2\bigg\{x + \dfrac{\pi}{3}\bigg\} + cos2\bigg\{x - \dfrac{\pi}{3} \bigg\}}{2}$

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On taking out 1/2 outside as it's a common factor, we get,

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$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3+ cos2x + cos2\bigg\{x + \dfrac{\pi}{3}\bigg\} + cos2\bigg\{x - \dfrac{\pi}{3} \bigg\}\Bigg\}$

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$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3+ cos2x + cos\bigg\{2x + \dfrac{2\pi}{3}\bigg\} + cos\bigg\{2x - \dfrac{2\pi}{3} \bigg\}\Bigg\}$

‎

We know that,

‎

$\sf \dashrightarrow \textsf{\textbf{cos}}\alpha + \textsf{\textbf{cos}}\beta \textsf{\textbf{= 2cos}}\bigg\{\dfrac{\alpha + \beta}{\textsf{\textbf{2}}}\bigg\} \ \textsf{\textbf{cos}}\bigg\{\dfrac{\alpha - \beta}{\textsf{\textbf{2}}}\bigg\}$

‎

Considering cos{2x + (2π/3)} + cos{2x - (2π/3)}, Let

• α = {2x + (2π/3)}

• β = {2x - (2π/3)}

On using this formula in the simplified form of the LHS we've derived we get,

‎

[Expression has been broken into two lines]

‎

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

$... \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sf \times \ cos\bigg\{\dfrac{2x + ((2\pi)/3) - \big(2x - ((2\pi)/3)\big)}{2}\bigg\}$

‎

$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3 + cos2x + 2cos\bigg\{\dfrac{2x + 2x}{2}\bigg\} \times cos\bigg\{\dfrac{((2\pi)/3) + ((2\pi)/3)}{2}\bigg\}$

‎

$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3 + cos2x + 2cos\bigg\{\dfrac{4x}{2}\bigg\} \times cos\bigg\{\dfrac{((4\pi)/3)}{2}\bigg\}$

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$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3 + cos2x + 2cos2x \times cos\bigg\{\dfrac{4\pi}{3 \times 2}\bigg\}\Bigg\}$

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$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3 + cos2x + 2cos2x \times cos\bigg\{\dfrac{4\pi}{6}\bigg\}\Bigg\}$

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$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3 + cos2x + 2cos2x \times cos\bigg\{\dfrac{2\pi}{3}\bigg\}\Bigg\}$

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2π can be written as 3π - π.

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$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3 + cos2x + 2cos2x \times cos\bigg\{\dfrac{3\pi - \pi}{3}\bigg\}\Bigg\}$

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$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3 + cos2x + 2cos2x \times cos\bigg\{\dfrac{\not{3}\pi}{\not{3}} - \dfrac{\pi}{3}\bigg\}\Bigg\}$

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$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3 + cos2x + 2cos2x \times cos\bigg\{\pi - \dfrac{\pi}{3}\bigg\}\Bigg\}$

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We know that cos(π - θ) = -cosθ, in our case, θ = π/3

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$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3 + cos2x + 2cos2x \times \bigg(-cos\bigg\{ \dfrac{\pi}{3}\bigg\}\bigg)\Bigg\}$

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We know that -cos(π/3) = -1/2, on substituting the value we get,

‎

$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3 + cos2x + \not{2}cos2x\bigg(-\dfrac{1}{\not{2}}\bigg)\Bigg\}$

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$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3 + cos2x + cos2x\big(-1\big)\Bigg\}$

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$\sf \Longrightarrow \dfrac{1}{2}\Bigg\{3 + cos2x - cos2x\Bigg\}$

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$\sf \Longrightarrow \dfrac{1}{2}\Big\{3\Big\}$

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$\sf \Longrightarrow \dfrac{3}{2}$

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LHS = RHS

Hence proved.