Question

Prove that :-
$\tt\cos12 \degree \cos24 \degree \cos36 \degree \cos48 \degree \cos72 \degree \cos84 \degree = \dfrac{1}{ {2}^{6} }$
Topic : Compound and Multiple Angles of Trigonometry.
( 11th ISC )
Relevant Answers Needed !!​

Answer 1

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

Consider LHS

$\rm \: \cos12 \degree \cos24 \degree \cos36 \degree \cos48 \degree \cos72 \degree \cos84 \degree$

can be rewritten as

$\rm \: = \: (\cos12 \degree \cos24 \degree cos48 \degree \cos84 \degree)cos36 \degree \: cos72 \degree$

We know,

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

and

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

Now, using these values, we get

$\rm \: = [cos12 \degree cos24 \degree cos48 \degree cos(180 \degree - 96 \degree )] \times \dfrac{ \sqrt{5} + 1}{4} \times \dfrac{ \sqrt{5} - 1}{4}$

We know,

$\boxed{\tt{ \: cos(180 \degree - x) = - \: cosx \: }}$

So, using this result, we get

$\rm \: = - [cos12 \degree cos24 \degree cos48 \degree cos96 \degree )] \times \dfrac{( \sqrt{5})^{2} - 1}{16}$

Let assume that 12° = x

So, above expression can be rewritten as

$\rm \: = - ( cosx \: cos2x \: cos4x \: cos8x) \times \dfrac{5 - 1}{16}$

can be rewritten as

$\rm \: = - ( cosx \: cos2x \: cos {2}^{2} x \: cos {2}^{3} x) \times \dfrac{4}{16}$

We know,

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

So, using this result, we get

$\rm \: = \: - \: \dfrac{sin {2}^{3 + 1} x}{ {2}^{3 + 1} sinx} \times \dfrac{1}{4}$

$\rm \: = \: - \: \dfrac{sin {2}^{4} x}{ {2}^{4} sinx} \times \dfrac{1}{4}$

$\rm \: = \: - \: \dfrac{sin16x}{ {2}^{6} \: sinx}$

On substituting the value of x = 12°, we get

$\rm \: = \: - \: \dfrac{sin(16 \times 12 \degree )}{ {2}^{6} \: sin12 \degree }$

$\rm \: = \: - \: \dfrac{sin192 \degree }{ {2}^{6} \: sin12 \degree }$

$\rm \: = \: - \: \dfrac{sin(180 \degree + 12 \degree )}{ {2}^{6} \: sin12 \degree }$

We know,

$\boxed{\tt{ \: sin(180 \degree + x) = \: - \: sinx \: }}$

So, using this result, we get

$\rm \: = \: \dfrac{sin12 \degree }{ {2}^{6} \: sin12 \degree }$

$\rm \: = \: \dfrac{1}{ {2}^{6} }$

Hence,

$\boxed{\sf{ \cos12 \degree \cos24 \degree \cos36 \degree \cos48 \degree \cos72 \degree \cos84 \degree \: = \: \frac{1}{ {2}^{6} } \: }}$

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ADDITIONAL INFORMATION

$\rm \: sin18 \degree = cos72 \degree = \dfrac{ \sqrt{5} - 1}{4}$

$\rm \: cos18 \degree = sin72 \degree = \dfrac{ \sqrt{10 + 2 \sqrt{5} }}{4}$

$\rm \: cos36 \degree = sin54 \degree = \dfrac{ \sqrt{5} + 1}{4}$

$\rm \: cos54 \degree = sin18 \degree = \dfrac{ \sqrt{10 - 2 \sqrt{5} }}{4}$

Answer 2

Answer:

Given ✿

cos12.cos24.cos36.cos48.cos72.cos96

$= \frac{2sin \: 12.cos12. \: cos24 \: .cos48. \: cos96. \: cos36 \: .cos72</p><p>}{ \: 2sin12} \\ \\ = \frac{2sin24.cos24. \: cos48. \: cos96 \: .cos36. \: cos72</p><p>}{4sin12} \\ \\ = \frac{2sin48 \: cos48. \: cos96. \: cos36. \: cos72</p><p>}{8sin12} \\ \\ = \frac{2sin96.cos96. \: cos36. \: cos72</p><p>}{16sin12} \\ \\ = \frac{sin192. \: cos36. \: cos72}{16sin12} = \\ \\ = \frac{sin(180+12).cos36. \: cos72</p><p>}{16sin12} \\ \\ =− \frac{1}{16} cos36 \: .cos72 \\ \\ = - \frac{1}{16} ( \frac{ \sqrt{5 - 1} }{4} ) (\frac{ \sqrt{5 - 1} }{4} ) \\ \\ = - \frac{1}{64} \\ \\ = \frac{1}{ {2}^{6} }$

Step-by-step explanation:

Hope it helps ✿