Find-the-value-of-4-sin-pi-24-cos-pi-12-cos-pi-6-

Question Number 17435 by 786786AM last updated on 06/Jul/17

Find the value of 4 \sin \frac{π}{24} \cos \frac{π}{12} \cos \frac{π}{6} .

Answered by alex041103 last updated on 07/Jul/17

Let A = 4 s i n \frac{π}{24} c o s \frac{π}{12} c o s \frac{π}{6} .

Then we use s i n 2 θ = 2 s i n θ c o s θ :

A = 2 (2 s i n \frac{π}{24} c o s \frac{π}{24}) c o s \frac{π}{12} c o s \frac{π}{6} \frac{1}{c o s \frac{π}{24}}

= (2 s i n \frac{π}{12} c o s \frac{π}{12}) c o s \frac{π}{6} \frac{1}{c o s \frac{π}{24}}

= \frac{1}{2 c o s \frac{π}{24}} (2 s i n \frac{π}{6} c o s \frac{π}{6})

= \frac{s i n \frac{π}{3}}{2 c o s \frac{π}{24}}

We know that c o s 2 θ = 2 {c o s}^{2} θ - 1

\Rightarrow {c o s}^{2} \frac{θ}{2} = \frac{1 + c o s θ}{2}

When θ \in [0, \frac{π}{2}], \frac{θ}{2} \in [0, \frac{π}{2}] \Rightarrow c o s \frac{θ}{2} ≧ 0

\Rightarrow c o s \frac{θ}{2} = \sqrt{\frac{1 + c o s θ}{2}}

From this we can evaluate c o s \frac{π}{24}

c o s \frac{π}{12} = \sqrt{\frac{1 + c o s \frac{π}{6}}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}

= \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}

c o s \frac{π}{24} = \sqrt{\frac{1 + c o s \frac{π}{12}}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2 + \sqrt{3}}}{2}}{2}}

= \sqrt{\frac{2 + \sqrt{2 + \sqrt{3}}}{4}} = \frac{\sqrt{2 + \sqrt{2 + \sqrt{3}}}}{2}

And now we solve for A :

A = \frac{\frac{\sqrt{3}}{2}}{2 \frac{\sqrt{2 + \sqrt{2 + \sqrt{3}}}}{2}} = \frac{\sqrt{3}}{2 \sqrt{2 + \sqrt{2 + \sqrt{3}}}}

A = \frac{1}{2} \sqrt{\frac{3}{2 + \sqrt{2 + \sqrt{3}}}}

A = 4 s i n \frac{π}{24} c o s \frac{π}{12} c o s \frac{π}{6} = \frac{1}{2} \sqrt{\frac{3}{2 + \sqrt{2 + \sqrt{3}}}} \approx 0.436749 \dots

Commented by mrW1 last updated on 07/Jul/17

very good sir!

Commented by alex041103 last updated on 07/Jul/17

Can you see my answer to Q : 17514 . I^{'} ll

post the hole prove at some point .