2
$\begingroup$

Prove that $$\sin(12^\circ)\sin(48^\circ)\sin(54^\circ)=\frac18.$$ Without using a calculator. I tried all identities I know but I have no idea how to proceed. I always get stuck on finding $\sin36^\circ$.

$\endgroup$

closed as off-topic by heropup, David H, Cookie, Adam Hughes, Gina Jul 26 '14 at 6:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, David H, Cookie, Adam Hughes, Gina
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ This is false as stated. Please, do mention that you measure angles in degrees, not in radians. The overwhelming majority of mathematicians will by default treat the argument of $\sin$ as being in radians. $\endgroup$ – Dan Shved Jul 26 '14 at 6:06
  • $\begingroup$ Thankyou for the suggestion $\endgroup$ – Kanishk Jul 26 '14 at 6:08
  • $\begingroup$ We have $$ \sin12^\circ\sin48^\circ\sin54^\circ=\frac18\tag1 $$ Multiplying $(1)$ by $2$ yields $$ 2\sin12^\circ\sin48^\circ\sin54^\circ=\frac14\tag2 $$ Using identities: $2\sin a\sin b=\cos(a-b)-\cos(a+b)$ and $\sin(90^\circ-x)=\cos x$ yields $$ \left(\cos36^\circ-\frac12\right)\cos36^\circ=\frac14\tag3 $$ Now, setting $x=\cos36^\circ$ and multiplying $(3)$ by $4$ yields $$ 4x^2-2x-1=0\tag4 $$ Solving $(4)$ then use identity $\sin\theta=\sqrt{1-\cos^2\theta}$. $\endgroup$ – Tunk-Fey Jul 26 '14 at 7:23
  • $\begingroup$ @Tunk-Fey, $$(3)\implies\cos36^\circ-\cos72^\circ=\frac12$$ (math.stackexchange.com/questions/827540/…) $\endgroup$ – lab bhattacharjee Jul 26 '14 at 8:08
6
$\begingroup$

Use the following: $$\sin\theta \sin\left(\frac{\pi}{3}-\theta\right)\sin\left(\frac{\pi}{3}+\theta\right)=\frac{1}{4}\sin(3\theta)$$ Then you have: $$\sin(12^{\circ})\sin(60^{\circ}-12^{\circ})\sin(60^{\circ}+12^{\circ})=\frac{1}{4}\sin(36^{\circ})$$ $$\Rightarrow \sin(12^{\circ})\sin(48^{\circ})=\frac{1}{4}\frac{\sin(36^{\circ})}{\sin(72^{\circ})}$$ $$\Rightarrow \sin(12^{\circ})\sin(48^{\circ})\sin(54^{\circ})=\frac{1}{4}\frac{\sin(36^{\circ})}{\sin(72^{\circ})}\sin(54^{\circ})=\frac{1}{4}\frac{\sin(36^{\circ})}{\sin(72^{\circ})}\cos(36^{\circ})$$ $$\Rightarrow \sin(12^{\circ})\sin(48^{\circ})\sin(54^{\circ})=\boxed{\dfrac{1}{8}}$$

$\endgroup$
1
$\begingroup$

$2\sin 12\cdot \sin 48 = \cos (48 - 12) - \cos (48 + 12) = \cos 36 - \dfrac{1}{2}$. So the problem is to find $\cos 36$. Use $1 - 2x^2 = 3x - 4x^3$ to solve for $\sin 18$ ( not hard ), then find $\cos 36$, and $\sin 54$. In fact, the equation is: $4x^3 - 2x^2 - 3x + 1 = 0 \to (x-1)(4x^2 + 2x - 1) = 0$, so $\sin 18 = x = \dfrac{\sqrt{5} - 1}{4}$. You can take it from here.

$\endgroup$
  • $\begingroup$ Can you please elaborate a little bit? $\endgroup$ – Kanishk Jul 26 '14 at 6:11
  • $\begingroup$ the problem is equivalent to finding the diagonal of a pentagon (its easy with similar triangles). $\endgroup$ – John Joy Jul 26 '14 at 14:27

Not the answer you're looking for? Browse other questions tagged or ask your own question.