Find $$\frac {\cos 81^{\circ}}{\sin3^{\circ}\sin57^{\circ}\sin63^{\circ}} $$
This expression seemed quite easy to solve, but now the $63$ and $57$ in the equation are posing me with difficulties. I tried multiplying by $\cos9$ but I was unable to solve it further.
Would someone please help me to solve this question? Thanks in advance.
Let $\alpha = 60^\circ$ and $\theta = 3^\circ$. The numerator is $$\begin{align} \cos(81^\circ) = \sin(9^\circ) &= \sin(3\theta)\\ &= \sin\theta\cos(2\theta) + \cos\theta\sin(2\theta)\\ &= \sin\theta(\cos(2\theta) + 2\cos^2\theta)\\ &= \sin\theta(2\cos(2\theta) + 1)\end{align}$$
while the denominator equals to $$\begin{align} \sin 3^\circ\sin 57^\circ \sin 63^\circ &= \sin\theta\sin(\alpha-\theta)\sin(\alpha+\theta)\\ &= \frac12\sin\theta(\cos(2\theta) - \cos(2\alpha))\\ &= \frac14\sin\theta(2\cos(2\theta) + 1)\end{align}$$ This means the expression at hand equals to $4$.
Note that $$\frac{\sin 9}{\sin 3} =\frac{3\sin 3-4\sin^3 3 }{\sin 3} =1+2\cos 6= 4\sin 57 \sin 63 $$
Thus $$\frac {\cos 81}{\sin3 \sin57 \sin63 } =\frac{\frac{\sin 9}{\sin 3 }}{ \sin57 \sin63 } =4$$
Let x=3°, y=60°, rewrite expression as $\{\frac{\sin(3x)}{\sin(x)}\} \div \{\sin(y-x)\sin(y+x)\} $.
$$\frac{\sin(3x)}{\sin(x)} = \frac{3\sin(x)-4(\sin(x))^3}{\sin(x)} = 3 - 4(\sin(x))^2 = 1 + 2\cos(2x)$$
$$\sin(y-x)\sin(y+x) = \frac{1}{2} \{-\cos(2y) + \cos(2x)\} = \frac{1}{4}(1 + 2 \cos(2x)) $$
Thus, expression simplified to 4
