Question

4
If COS A + cos B + cos C = 0 and
COS 3A + cos 3B + cos 3C = K cos A cos B
then K=​

Answer 1

Appropriate Question :-

If COS A + cos B + cos C = 0 and

COS 3A + cos 3B + cos 3C = K cos A cos B cos C

then K=

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

Given that,

$\rm :\longmapsto\:cosA + cosB + cosC = 0$

$\bf\implies \:\boxed{ \tt{ \: cosA + cosB = - cosC}} - - - - (1)$

Now, Further given that

$\rm :\longmapsto\:cos3A + cos3B + cos3C = KcosAcosBcosC$

Consider,

$\rm :\longmapsto\:cos3A + cos3B + cos3C$

$\rm=(4{cos}^{3}A -3cosA) + (4{cos}^{3}B -3cosB) + (4{cos}^{3}C -3cosC)$

can be re-arranged as

$\rm = 4( {cos}^{3}A + {cos}^{3}B + {cos}^{3}C) - 3(cosA + cosB + cosC)$

$\rm = 4\bigg[({cos}^{3}A + {cos}^{3}B) + {cos}^{3}C\bigg] - 3 \times 0$

$\red{\bigg \{ \because \:using \: equation \: (1) \bigg \}}$

We know,

$\boxed{ \tt{ \: {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y)}}$

So, using this identity, we get

$\rm =4\bigg[ {(cosA + cosB)}^{3} - 3cosAcosB(cosA + cosB) + {cos}^{3}C\bigg]$

$\rm =4\bigg[ {( - cosC)}^{3} - 3cosAcosB( -cosC) + {cos}^{3}C\bigg]$

$\rm =4\bigg[ {- cos}^{3}C + 3cosAcosBcosC + {cos}^{3}C\bigg]$

$\rm \: = \:12cosAcosBcosC$

Hence,

$\red{\rm :\longmapsto\:cos3A + cos3B + cos3C = 12cosAcosBcosC}$

So, on comparing with

$\pink{\rm :\longmapsto\:cos3A + cos3B + cos3C = KcosAcosBcosC}$

We get

$\bf\implies \:\boxed{ \tt{ \: K \: = \: 12 \: \: }}$

More to Know:-

$\boxed{ \tt{ \: sin2x = 2sinxcosx = \frac{2tanx}{1 + {tan}^{2}x } }}$

$\boxed{ \tt{ \: tan2x = \frac{2tanx}{1 - {tan}^{2}x } }}$

$\boxed{ \tt{ \: cos2x = 1 - {2sin}^{2}x = {2cos}^{2}x - 1}}$

$\boxed{ \tt{ \: cos2x = {cos}^{2}x - {sin}^{2}x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2} x}}}$

$\boxed{ \tt{ \: sin3x = 3sinx - {4sin}^{3}x}}$

$\boxed{ \tt{ \: tan3x = \frac{3tanx - {tan}^{3}x }{1 - {3tan}^{2}x }}}$

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$