Question

Find the value of cos12°cos24°cos36°cos48°cos72°cos96°

Answer 1

Step-by-step explanation:

We have,

$\tt{ cos(12\degree) \cdot cos(24\degree) \cdot \: cos(36\degree) \cdot \: cos(48\degree) \cdot \: cos(72\degree) \cdot \:cos(96\degree) }$

Let 12°=θ

So,

$\tt{ cos( \theta) \cdot cos(2 \theta) \cdot \: cos(3 \theta) \cdot \: cos(4 \theta) \cdot \: cos(6\theta) \cdot \:cos(8 \theta) }$

$\tt{ = cos( \theta) \: cos(2 \theta) \: cos(4 \theta) \: cos(8 \theta) \cdot \: cos(3 \theta) \: cos(6\theta)}$

$\tt{ = cos( \theta) \: cos(2 \theta) \: cos( {2}^{2} \theta) \: cos( {2}^{3} \theta) \cdot \: cos(3 \theta) \: cos(6\theta)}$

$\displaystyle \boxed{ \bf{We \: \: know, \: \: \: \prod^{n - 1}_{k = 0} cos \left(2^{k} \phi\right) = \dfrac{ sin( {2}^{n} \: \phi ) }{ {2}^{n} \cdot \: sin( \phi) }}}$

So,

$\tt{ = \dfrac{ sin( {2}^{4} \: \theta) }{ {2}^{4} \: sin( \theta) } \cdot \: cos(3 \theta) \: cos(6\theta)}$

$\tt{ = \dfrac{ sin( 16 \times 12 \degree) }{ 16 \: sin( 12 \degree) } \cdot \: cos(36 \degree) \: cos(7 2\degree)}$

$\tt{ = \dfrac{ sin( 192\degree) }{ 16 \: sin( 12 \degree) } \cdot \: cos(36 \degree) \: cos(7 2\degree)}$

$\tt{ = \dfrac{ sin( 180\degree + 12 \degree) }{ 16 \: sin( 12 \degree) } \cdot \: cos(36 \degree) \: cos(7 2\degree)}$

$\tt{ = - \dfrac{ sin( 12 \degree) }{ 16 \: sin( 12 \degree) } \cdot \: cos(36 \degree) \: cos(7 2\degree)}$

$\tt{ = - \dfrac{ 1}{ 16 } \cdot \: cos(36 \degree) \: cos(7 2\degree)}$

$\tt{ = - \dfrac{ 1}{ 16 } \cdot \: cos(36 \degree) \: cos(2 \times 36\degree)}$

$\tt{ = - \dfrac{ 1}{ 16 } \cdot \: \dfrac{ sin( {2}^{2} \times 36 \degree) }{ {2}^{2} \cdot sin(36 \degree) }}$

$\tt{ = - \dfrac{ 1}{ 16 } \cdot \: \dfrac{ sin( 4 \times 36 \degree) }{ 4 \times sin(36 \degree) }}$

$\tt{ = - \dfrac{ 1}{ 16 } \cdot \: \dfrac{ sin( 144 \degree) }{ 4 \times sin(36 \degree) }}$

$\tt{ = - \dfrac{ 1}{ 16 } \cdot \: \dfrac{ sin( 180\degree - 36 \degree) }{ 4 \times sin(36 \degree) }}$

$\tt{ = - \dfrac{ 1}{ 16 } \cdot \: \dfrac{ sin( 36 \degree) }{ 4 \times sin(36 \degree) }}$

$\tt{ = - \dfrac{ 1}{ 16 } \cdot \: \dfrac{ 1 }{ 4 }}$

$\tt{ = - \dfrac{ 1}{64 }}$

Answer 2

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

Given Trigonometric expression is

$\rm :\longmapsto\:cos12\degree cos24\degree cos36\degree cos48\degree cos72\degree cos96\degree$

can be re-arranged as

$\rm =\bigg[ cos12\degree cos24\degree cos48\degree cos96\degree \bigg]\bigg[cos36\degree cos72\degree \bigg]$

We know

$\red{\rm :\longmapsto\:\boxed{\tt{ cos36\degree = \frac{ \sqrt{5} + 1}{4}}}}$

and

$\red{\rm :\longmapsto\:\boxed{\tt{ cos72\degree = \frac{ \sqrt{5} - 1}{4}}}}$

and Let assume that, x = 12°

So, above expression can be rewritten as

$\rm = cosx \: cos2x \: cos4x \: cos8x \times \bigg[\dfrac{ \sqrt{5} + 1}{4} \times \dfrac{ \sqrt{5} - 1}{4} \bigg]$

$\rm = cosx \: cos2x \: cos {2}^{2} x \: cos {2}^{3} x \times \bigg[\dfrac{ (\sqrt{5})^{2} - {1}^{2} }{16} \bigg]$

We know,

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

So, using this, we get

$\rm \:  =  \: \dfrac{sin {2}^{3 + 1} x}{ {2}^{3 + 1}sinx} \times \dfrac{5 - 1}{16}$

$\rm \:  =  \: \dfrac{sin {2}^{4} x}{ {2}^{4}sinx} \times \dfrac{4}{16}$

$\rm \:  =  \: \dfrac{sin16 x}{16sinx} \times \dfrac{1}{4}$

$\rm \:  =  \: \dfrac{sin16(12\degree )}{64 \: sin12\degree }$

$\rm \:  =  \: \dfrac{sin192\degree }{64 \: sin12\degree }$

$\rm \:  =  \: \dfrac{sin(180\degree + 12\degree ) }{64 \: sin12\degree }$

$\rm \:  =  \: \dfrac{ - sin(12\degree ) }{64 \: sin12\degree }$

$\rm \:  =  \: - \: \dfrac{1}{64}$

Hence,

$\:\boxed{\tt{ cos12\degree cos24\degree cos36\degree cos48\degree cos72\degree cos96\degree = - \: \frac{1}{64}}}$

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Additional Information

$\boxed{\tt{ sin54\degree = cos36\degree = \frac{ \sqrt{5} + 1 }{4} \: }}$

$\boxed{\tt{ sin18\degree = cos72\degree = \frac{ \sqrt{5} - 1 }{4} \: }}$

$\boxed{\tt{ sin36\degree = cos54\degree = \frac{ \sqrt{10 - 2 \sqrt{5} } }{4} \: }}$

$\boxed{\tt{ sin72\degree = cos18\degree = \frac{ \sqrt{10 + 2 \sqrt{5} } }{4} \: }}$