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Answered by sharanyalanka7
5

Step-by-step explanation:

To Prove :

32√3.sin(π/48).cos(π/48).cos(π/24).cos(π/12).cos(π/6) = 3

Solution :-

Taking L.H.S :-

= 32√3.sin(π/48).cos(π/48).cos(π/24).cos(π/12).cos(π/6)

= 16×2×√3.sin(π/48).cos(π/48).cos(π/24).cos(π/12).cos(π/6)

= 16√3.(2sin(π/48).cos(π/48))).cos(π/24).cos(π/12).cos(π/6)

= 16√3.(sin2(π/48)).cos(π/24).cos(π/12).cos(π/6)

= 16√3.sinπ/24.cos(π/24).cos(π/12).cos(π/6)

= 8√3.2.sinπ/24.cos(π/24).cos(π/12).cos(π/6)

= 8√3.sin2(π/24).cos(π/12).cos(π/6)

= 8√3.sin(π/12).cos(π/12).cos(π/6)

= 4√3.2.sin(π/12).cos(π/12).cos(π/6)

= 4√3.sin2(π/12).cos(π/6)

= 4√3.sin(π/6).cos(π/6)

= 2√3.2.sin(π/6).cos(π/6)

= 2√3.sin2(π/6)

= 2√3.sinπ/3

= 2√3 × sin60°

= 2√3 × √3/2

= √3 × √3

= 3

= R.H.S

Hence Proved.

Formula Required :-

1) sin2A = 2sinAcosA

2) π/3 = 60°

3) sin60° = √3/2

Answered by mathdude500
5

 \green{\large\underline{\sf{Given \:Question - }}}

Prove that,

\rm :\longmapsto\:32 \sqrt{3}sin\bigg[\dfrac{\pi}{48} \bigg]cos\bigg[\dfrac{\pi}{48} \bigg]cos\bigg[\dfrac{\pi}{24} \bigg]cos\bigg[\dfrac{\pi}{12} \bigg]cos\bigg[\dfrac{\pi}{6} \bigg] = 3

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

Consider, LHS

\rm :\longmapsto\:32 \sqrt{3}sin\bigg[\dfrac{\pi}{48} \bigg]cos\bigg[\dfrac{\pi}{48} \bigg]cos\bigg[\dfrac{\pi}{24} \bigg]cos\bigg[\dfrac{\pi}{12} \bigg]cos\bigg[\dfrac{\pi}{6} \bigg]

can be rewritten as

\rm\:=\:32 \sqrt{3}sin\bigg[\dfrac{\pi}{48} \bigg]cos\bigg[\dfrac{\pi}{48} \bigg]cos\bigg[\dfrac{2\pi}{48} \bigg]cos\bigg[\dfrac{4\pi}{48} \bigg]cos\bigg[\dfrac{8\pi}{48} \bigg]

Let assume that,

\red{ \boxed{ \sf{ \:  \:\dfrac{\pi}{48} \:  = \:  x \:  \: }}}

So, above expression can be rewritten as

\rm \:  =  \: 32 \sqrt{3}sinx \: cosx \: cos2x \: cos4x \: cos8x

\rm \:  =  \: 32 \sqrt{3}sinx \: cosx \: cos2x \: cos {2}^{2} x \: cos {2}^{3} x

We know,

\red{ \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 identity, we get

\rm \:  = 32 \sqrt{3}sinx \times \dfrac{sin {2}^{3 + 1} x}{ {2}^{3 + 1}sinx }

\rm \:  = 32 \sqrt{3}sinx \times \dfrac{sin {2}^{4} x}{ {2}^{4}sinx }

\rm \:  = 2 \:  \:  \:  \: \cancel{32} \sqrt{3} \: \cancel{sinx} \:  \times \dfrac{sin16x}{ \cancel{16} \: \cancel{sinx} }

\rm \:  =  \: 2 \sqrt{3} \: sin16x

\rm \:  =  \: 2 \sqrt{3} \: sin\bigg[16 \times \dfrac{\pi}{48} \bigg]

\rm \:  =  \: 2 \sqrt{3} \: sin\bigg[\dfrac{\pi}{3} \bigg]

\rm \:  =  \: 2 \sqrt{3} \:  \times \dfrac{ \sqrt{3} }{2}

\rm \:  =  \: 3

Hence,

\red{ \boxed{ \sf{ \:32 \sqrt{3}sin\bigg[\dfrac{\pi}{48} \bigg]cos\bigg[\dfrac{\pi}{48} \bigg]cos\bigg[\dfrac{\pi}{24} \bigg]cos\bigg[\dfrac{\pi}{12} \bigg]cos\bigg[\dfrac{\pi}{6} \bigg] = 3}}}

Additional Information :-

\red{ \boxed{ \sf{ \:cosx \: cos\bigg[\dfrac{\pi}{3}  - x\bigg] \: cos\bigg[\dfrac{\pi}{3} + x \bigg] =  \frac{1}{4}cos3x}}}

\red{ \boxed{ \sf{ \:sinx \: sin\bigg[\dfrac{\pi}{3}  - x\bigg] \: sin\bigg[\dfrac{\pi}{3} + x \bigg] =  \frac{1}{4}sin3x}}}

\red{ \boxed{ \sf{ \:tanx \: tan\bigg[\dfrac{\pi}{3}  - x\bigg] \: tan\bigg[\dfrac{\pi}{3} + x \bigg] =  tan3x}}}

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