Question

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Answer 1

Hey there!

Taking the left hand side or manipulating the left hand side to proceed further;

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

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Using the following identity rule for cos that is,

$\bf{cos(s - t) = cos(s)cos(t) + sin(s)sin(t)}$

AND, Using the following identity rule for cos that is,

$\bf{cos(s + t) = cos(s)cos(t) - sin(s)sin(t)}$

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

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Apply the following trivial identities;

$\bf{First \: trivial \: identity: \: cos(\dfrac{\pi}{3}) = \dfrac{1}{2}}$

$\bf{Second \: trivial \: identity: \: sin(\dfrac{\pi}{3}) = \dfrac{\sqrt{3}}{2}}$

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$\bf{\therefore \quad cos^2(x) + (\dfrac{1}{2} cos(x) - \dfrac{\sqrt{3}}{2} sin(x))^2 + (\dfrac{1}{2} cos(x) + \dfrac{\sqrt{3}}{2} sin(x))^2}$

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$\bf{cos^2(x) + (\dfrac{1}{2} cos(x))^2 - 2 \times \dfrac{1}{2} cos(x) \dfrac{\sqrt{3}}{2} sin(x) + (\dfrac{\sqrt{3}}{2} sin(x))^2 + (\dfrac{1}{2} cos(x))^2}$ $\bf{+ 2 \times \dfrac{1}{2} cos(x) \dfrac{\sqrt{3}}{2} sin(x) + (\dfrac{\sqrt{3}}{2} sin(x))^2}$

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$\bf{cos^2(x) + \dfrac{1}{4} cos^2(x) - \dfrac{\sqrt{3}}{2} cos(x)sin(x) + \dfrac{3}{4} sin^2(x)}$ $\bf{+ \dfrac{1}{4} cos^2(x) + \dfrac{\sqrt{3}}{2} cos(x)sin(x) + \dfrac{3}{4} sin^2(x)}$

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$\bf{\dfrac{3}{2} cos^2(x) + \dfrac{3}{4} sin^2(x) \dfrac{3}{4} sin^2(x)}$

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$\bf{\dfrac{3}{2} cos^2(x) + (\dfrac{3}{4} + \dfrac{3}{4}) sin^2(x)}$

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$\bf{\dfrac{3}{2} cos^2(x) + \dfrac{6}{4} sin^2(x)}$

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$\bf{\dfrac{3}{2} cos^2(x) + \dfrac{3}{2} sin^2(x)}$

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$\bf{\dfrac{3 cos^2(x)}{2} + \dfrac{3 sin^2(x)}{2}}$

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$\bf{\dfrac{3 cos^2(x) + 3 sin^2(x)}{2}}$

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Since,  sin^2(x) = 1 - cos^2(x).

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$\bf{\therefore \quad \dfrac{(1 - cos^2(x)) \times 3 + 3 cos^2(x)}{2}}$

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$\bf{\therefore \quad \dfrac{3(1 - cos^2(x)) + 3 cos^2(x)}{2}}$

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$\bf{\therefore \quad \dfrac{3 - 3 cos^2(x) + 3 cos^2(x)}{2}}$

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$\boxed{\bf{\underline{Hence \: Proved, \: \dfrac{3}{2}}}}$

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Which is the required solution or the final proof for this type of query.

Hope it helps you and clears your doubt for proving the two trigonometric identities !!!!!

Answer 2

Given Equation is cos^2x + cos^2(x + π/3) + cos^2(x - π/3).

⇒ cos^2x + [cos(x + π/3)]^2 + [cos(x - π/3)]^2

We know that cos (a + b) =  cos a cos b - sin a sin b

We know that cos(a - b) = cos a cos b + sin a sin b.

⇒ cos^2x + [cos x * cos π/3 - sin x * sin π/3)^2 + [cos x * cos π/3 + sin x * sin π/3]^2

⇒ cos^2x + [cos x * (1/2) - sin x * (√3/2)]^2 + [cos x * (1/2) + sin x * (√3/2)]^2

We know that (a - b)^2 = a^2 + b^2 - 2ab.

We know that (a + b)^2 = a^2 + b^2 + 2ab.

⇒ cos^2x + (1/4)[cos^2x + 3 sin^2x - 2√3 sinx cosx] + (1/4)[cos^2x + 3 sin^2x + 2√3 sinx cosx ]

⇒ cos^2x + [cos^2(x/4) + 3 sin^2(x/4)] + (cos^2(x/4) + (3/4) sin^2x]

⇒ cos^2x + cos^2(x/4) + (3/4) sin^2x + cos^2(x/4) + (3/4) sin^2x]

Rearrange the terms, we get

⇒ cos^2x + cos^2(x/4) + cos^2(x/4) + (3/4)sin^2x + (3/4)sin^2x

⇒ cos^2x[1 + 1/4 + 1/4] + sin^2x[3/4 + 3/4]

⇒ cos^2x[3/2] + sin^2x[3/2]

⇒ (3/2)[sin^2x + cos^2x]

⇒ (3/2)[1]

⇒ (3/2).

Hope it helps!