Question

If A, B, C are angles of a triangle then prove that
cos2A cos2B cos2C -4cosA.cosB.cosC-1

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Answer 1

Appropriate Question

If A, B, C are angles of triangle ABC, prove that

$\rm :\longmapsto\:cos2A + cos2B + cos2C = - 1 - 4cosAcosBcosC$

$\red{\large\underline{\sf{Solution-}}}$

Given that,

$\rm :\longmapsto\:A + B + C = \pi$

So, Consider LHS

$\rm :\longmapsto\:cos2A + cos2B + cos2C$

We know that,

$\boxed{\tt{ cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}$

So, using this, we get

$\rm \:  =  \: 2cos\bigg[\dfrac{2A + 2B}{2} \bigg]cos\bigg[\dfrac{2A - 2B}{2} \bigg] + cos2C$

$\rm \:  =  \: 2cos(A + B)cos(A - B) + cos2C$

$\rm \:  =  \: 2cos(\pi - C)cos(A - B) + cos2C$

We know,

$\purple{\rm :\longmapsto\:cos(\pi - x) = - cosx}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ cos2x = {2cos}^{2}x - 1}}}$

So, using these results, we get

$\rm \:  =  \: - 2cosCcos(A - B) + 2 {cos}^{2}C - 1$

$\rm \:  =  \: - 2cosC\bigg[cos(A - B) - cosC\bigg] - 1$

$\rm \:  =  \: - 2cosC\bigg[cos(A - B) - cos[\pi - (A + B)]\bigg] - 1$

$\rm \:  =  \: - 2cosC\bigg[cos(A - B) + cos(A + B)\bigg] - 1$

$\rm \:  =  \: - 2cosC[2cosA \: cosB] - 1$

$\rm \:  =  \: - 1 - 4cosAcosBcosC$

Hence,

$\boxed{\tt{ \:cos2A + cos2B + cos2C = - 1 - 4cosAcosBcosC \: }}$

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MORE TO KNOW

$\boxed{\tt{ sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{\tt{ sinx - siny = 2sin\bigg[\dfrac{x - y}{2} \bigg]cos\bigg[\dfrac{x + y}{2} \bigg]}}$

$\boxed{\tt{ cosx - cosy = 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{y - x}{2} \bigg]}}$

$\boxed{\tt{ 2sinxcosy = sin(x + y) + sin(x - y)}}$

$\boxed{\tt{ 2cosxcosy = cos(x + y) + cos(x - y)}}$

$\boxed{\tt{ 2sinxsiny = cos(x - y) - cos(x + y) \: }}$

Answer 2

Answer:

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