Question

Cos²A + Cos²B - Cos²C = 1-2 sinA×sinB×cosc

Answer 1

Question:-

In triangle ABC, prove that cos²A + cos²B - cos²C = 1-2 sinA sinB cosC

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

Given, A triangle ABC

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

Consider LHS

$\rm \: {cos}^{2}A + {cos}^{2}B - {cos}^{2}C$

can be rewritten as

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

can be further rearrange as

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

We know,

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

So, using this result, we get

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

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

$\: \: \: \: \: \: \: \: \: \: \: \: \{ \: \rm \: \because \: A + B + C = \pi \: \}$

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

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

$\rm \: = \: 1 - cos C \:[cos(A - B) + {cos}C]$

$\rm \: = \: 1 - cos C \:[cos(A - B) + {cos}(\pi - (A + B))]$

$\: \: \: \: \: \: \: \: \: \: \: \: \{ \: \rm \: \because \: A + B + C = \pi \: \}$

$\rm \: = \: 1 - cos C \:[cos(A - B) - {cos}((A + B)]$

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

We know,

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

So, using this result, we get

$\rm \: = \: 1 - cosC \: (2sinA \: sinB)$

$\rm \: = \: 1 - 2sinA \: sinB \: cosC$

Hence,

$\boxed{\tt{ \rm \: {cos}^{2}A + {cos}^{2}B - {cos}^{2}C=\: 1 - 2sinA \: sinB \: cosC}}$

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ADDITIONAL INFORMATION

$\rm \: \: {sin}^{2}x - {sin}^{2}y = sin(x + y) \: sin(x - y) \:$

$\rm \: \: 2sinA \: cosB = sin(A + B) + sin(A - B) \:$

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

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

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

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

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

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