I am new to trigonometry. I need to evaluate the below expression:

$$ 96\sqrt{3} \ \sin\left(\frac \pi {48}\right) \cos\left(\frac \pi {48}\right) \cos\left(\frac \pi{24}\right)\cos\left(\frac \pi {12}\right) \cos\left(\frac \pi 6\right) $$

I need to use basic identities, such as double angle formulae, Trigonometric ratios


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  • $\begingroup$ Do you know abou the half angle formulas? $\endgroup$ – Ovi Jul 13 '17 at 18:36
  • $\begingroup$ we get $$\sin(\pi/48)=\frac{1}{4} \left(-\sqrt{2-\sqrt{2+\sqrt{2}}}-\sqrt{2 \left(2-\sqrt{2+\sqrt{2}}\right)}-\sqrt{\left(2+\sqrt{2}\right) \left(2-\sqrt{2+\sqrt{2}}\right)}\right)+\frac{1}{4} \sqrt{3} \left(\sqrt{2+\sqrt{2+\sqrt{2}}}+\sqrt{2 \left(2+\sqrt{2+\sqrt{2}}\right)}-\sqrt{\left(2+\sqrt{2}\right) \left(2+\sqrt{2+\sqrt{2}}\right)}\right)$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 13 '17 at 18:38
  • $\begingroup$ @ovi Yes, how to use it here $\endgroup$ – Ravi Prakash Jul 13 '17 at 18:38
  • 2
    $\begingroup$ Borrow a 2 from the 96 and put it with the first $\sin$ and $\cos$. Use the identity $2\sin x \cos x = \sin 2x$. Repeat. $\endgroup$ – B. Goddard Jul 13 '17 at 18:39
  • 1
    $\begingroup$ @Dr.SonnhardGraubner : To say the least, that's definitely the hard way! See my answer below. $\endgroup$ – Michael Hardy Jul 13 '17 at 18:39

$$ 96\sin(\pi/48)\cos(\pi/48) = 48 \times \underbrace{2\sin\left(\frac \pi {48} \right) \cos\left( \frac \pi{48}\right)} = 48 \underbrace{\sin\left( 2\times\frac \pi{48} \right)} $$ by the usual double-angle formula, and then $$ = 48\sin \left( \frac \pi {24}\right). $$ Next, do the same thing with $24$ that we just did with $48,$ then with $12,$ then with $6.$


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