Question

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$ \Large \mathtt \red{Question :}$
Prove That :
$\rm{ {cos}^{2}x + {cos}^{2}( x + \frac{\pi}{3} ) + {cos}^{2}(x - \frac{\pi}{3}) = \frac{3}{2} }$

Help ASAP!!!
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Answer 1

Given :-

Cos²X + Cos²(x + π/3) + Cos²(x - π/3) = 3/2

To Show :-

• L.H.S = R.H.S

Solution :-

• By applying trigonometry ratio :-

Cos²X + Cos²(x + π/3) + Cos²(x - π/3)

• Applying formula here :-
• Cos2x = 2cos²x - 1
• 2cos²x = Cos2x + 1

⇒ 1 + Cos2x/2 + 1 + cos(2x + 2π/3)/2 + 1 + cos(2x - 2π/3)/2

⇒ 1/2[ 1 + Cos2x + 1 + cos(2x + 2π/3) + 1 + cos(2x - 2π/3)]

⇒ 1/2[3 + Cos2x + Cos(2x + 2π/3) + Cos(2x - 2π/3)]

• Using Cos(x + y) = Cosx + cosy
• Cosx + cosy = 2cos(x + y/2) × Cos(x - y/2)
• X = 2x + 2π/3 ,. Y = 2x - 2π/3

⇒ 1/2[3 + cos2x + 2Cos(2x + 2π/3 + 2x - 2π/3/2) * Cos(2x + 2π/3 - 2x + 2π/3/2)]

⇒ 1/2[3 + Cos2x + 2cos(4x/2) * Cos({4π/3}/2)]

⇒ 1/2[3 + Cos2x + 2cos2x * cos(2π/3)]

• Using Cos 2π/3 = Cos(π - π/3)
• Cos( π - x) = - Cosx here x = theta
• - Cosx = -1/2

⇒ 1/2[3 + cos2x + 2cos2x *(-1/2)]

⇒ 1/2[3 + Cos2x - 2co2x * 1/2]

⇒ 1/2[3 + Co2x - Cos2x]

⇒ 1/2[3]

⇒ 3/2

Hence, proved L.H.S = R.H.S

Answer 2

Answer:

Given:-

• ${cos}^{2} x + {cos}^{2} (x + \frac{\pi}{3} ) + {cos}^{2} (x - \frac{\pi}{3} )= \frac{3}{2}$ here we can do this method easily

To prove :-

• Here we should find LHS=RHS.

Explanation :-

• Here we can apply trigonometric ratio property we get value so,
• $cos2x = 2 {cos}^{2} x - 1$
• $2 {cos}^{2} x = cos2x + 1$
• $\frac{1}{2} (1 + cos2x + 1 + cos(2x + \frac{2\pi}{3} ) + 1 + cos(2x - \frac{2\pi }{3} )$
• $\frac{1}{2} (1 + cos2x + 1 + cos(2x + \frac{2\pi}{3} ) + 1 + cos(2x - \frac{2\pi}{3} )$
• $\frac{1}{2} (3 + cos2x + cos(2x + \frac{2\pi}{3} ) + cos(2x - \frac{2\pi}{3} )$
• Here by using ,
• $cos(x + y) = cos \: x + cos \: y$
• $cos \: x + cos \: y = 2cos(x + \frac{y}{2} ) \times cos(x - \frac{y}{2} )$
• here i will answer it direct ok.
• Using
• Cos 2pi/3=cos (pi-pi/3)

• $\frac{1}{2} (3 + cos2x + 2cos( \frac{4x}{2} )$
• $\frac{1}{2} (3 + cos2x + 2cos( \frac{4x}{2} ) \times cos( \frac{ \frac{4x}{3} }{2} )$
• $\frac{1}{2} (3 + cos \: 2x + 2cos2x \times cos \frac{2\pi}{3}$
• $\frac{1}{2} (3 + cos2x + 2cos2x \times \frac{ - 1}{2}$
• $\frac{1}{2} (3 + cos2x - 2cos2x \times \frac{1}{2}$
• $\frac{1}{2} (3 + cos2x - cos2x)$
• 1/2(3)
• $\frac{3}{2}$

Therefore ,

LHS=RHS.

Hope it helps u mate .

Thank you .