In a ABC- prove that cos 2 A -cos 2 A -3-cos 2 A -3-3-2

We have LHS = cos^2 A + cos^2 ( A + π/3 )+cos^2 ( A π/3 ) =1/2(1+cos2A) + 1/2{ 1 + cos ( 2A + 2π/3 ) }+1/2{ 1 + cos ( 2A 2π/3 ) } =3/2+1/2cos2A+1/2{cos ( 2A ...

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Mathematics

Collinearity Condition

In a ABC, pro...

Question

In a $△$ ABC, prove that ${cos}^{2} A + {cos}^{2} (A + \frac{π}{3}) + {cos}^{2} (A - \frac{π}{3}) = \frac{3}{2}$

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Solution

We have

LHS = ${cos}^{2} A + {cos}^{2} (A + \frac{π}{3}) + {cos}^{2} (A - \frac{π}{3})$

= $\frac{1}{2} (1 + cos 2 A) + \frac{1}{2} {1 + cos (2 A + \frac{2 π}{3})} + \frac{1}{2} {1 + cos (2 A - \frac{2 π}{3})}$

= $\frac{3}{2} + \frac{1}{2} cos 2 A + \frac{1}{2} {cos (2 A + \frac{2 π}{3}) + cos (2 A - \frac{2 π}{3})}$

= $\frac{3}{2} + \frac{1}{2} cos 2 A + cos 2 A cos \frac{2 π}{3} [∵ cos (A + B) + cos (A - B) = 2 cos A cos B]$

= $\frac{3}{2} + \frac{1}{2} cos 2 A - \frac{1}{2} cos 2 A [∵ cos \frac{2 π}{3} = - \frac{1}{2}]$

= $\frac{3}{2} = RHS$

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In a △ABC, prove that cos2A+cos2(A+π3)+cos2(A−π3)=32

We have LHS = cos2A+cos2(A+π3)+cos2(A−π3) =12(1+cos2A)+12{1+cos(2A+2π3)}+12{1+cos(2A−2π3)} =32+12cos2A+12{cos(2A+2π3)+cos(2A−2π3)} =32+12cos2A+cos2A cos2π3 [∵cos(A+B)+cos(A−B)=2cosA cosB] =32+12cos2A−12cos2A[∵cos2π3=−12] =32=RHS

In a $△$ ABC, prove that ${cos}^{2} A + {cos}^{2} (A + \frac{π}{3}) + {cos}^{2} (A - \frac{π}{3}) = \frac{3}{2}$