Animated Solution for Mathematics - Trigonometry: If ${\sin }^2(10^{\circ })\sin (20^{\circ })\sin (40^{\circ })\sin (50^{\circ })\sin (70^{\circ })=\alpha -\frac{1}{16}\sin (10^{\circ })$, then $16+{\alpha }^{-1}$ is equal to ______.

A Journey of Symmetry

Welcome, fellow traveler of the mathematical landscape. Today, we are going to tackle a problem that, at first glance, looks like a tangled mess of sine terms.

We are faced with the expression $E={\sin }^2(10^{\circ })\sin (20^{\circ })\sin (40^{\circ })\sin (50^{\circ })\sin (70^{\circ })$. It is easy to feel overwhelmed by such a product, but remember: in trigonometry, complexity is often just a mask for hidden symmetry.

Our goal is to find the value of $16+{\alpha }^{-1}$, where $E=\alpha -\frac{1}{16}\sin (10^{\circ })$. Let us peel back the layers together.

The Hidden Identity

Whenever you see angles like $10^{\circ }$, $50^{\circ }$, and $70^{\circ }$ in a product, your intuition should immediately scream, "Identity!" Specifically, look at the relationship between these angles.

Notice that $50^{\circ }=60^{\circ }-10^{\circ }$ and $70^{\circ }=60^{\circ }+10^{\circ }$. This is the hallmark of the beautiful identity:

If we set $\theta =10^{\circ }$, the expression inside our product perfectly matches this structure. This is the key that unlocks the entire problem.

Surgical Rearrangement

Now, let us look at our expression $E$ again. We have ${\sin }^2(10^{\circ })$, which is just $\sin (10^{\circ })\cdot \sin (10^{\circ })$. To make the identity work, we need to group the terms carefully.

Let us rewrite $E$ as:

By isolating the bracketed terms, we can apply our identity directly. The bracketed part becomes $\frac{1}{4}\sin (3\times 10^{\circ })$, which is $\frac{1}{4}\sin (30^{\circ })$.

Since $\sin (30^{\circ })=\frac{1}{2}$, the entire bracket collapses into $\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}$. Suddenly, our massive expression is reduced to $E=\frac{1}{8}\sin (10^{\circ })\sin (20^{\circ })\sin (40^{\circ })$.

The Product-to-Sum Transformation

We are left with a product of three sine terms. To simplify this, we need to convert products into sums using the identity $2\sin (A)\sin (B)=\cos (A-B)-\cos (A+B)$.

To use this, we multiply and divide by $2$ to get a factor of $2$ inside:

Applying the formula with $A=40^{\circ }$ and $B=20^{\circ }$, we get:

Now, substitute this back:

The Final Cancellation

Look at the first term: $\frac{1}{16}\sin (10^{\circ })\cos (20^{\circ })$. We use the product-to-sum formula again, $2\sin (A)\cos (B)=\sin (A+B)+\sin (A-B)$.

Multiplying and dividing by $2$ again:

Since $\sin (30^{\circ })=\frac{1}{2}$ and $\sin (-10^{\circ })=-\sin (10^{\circ })$, this becomes:

Combining this with the second term from Phase 3, we get:

Comparing this to $\alpha -\frac{1}{16}\sin (10^{\circ })$, we find $\alpha =\frac{1}{64}$. Finally, $16+{\alpha }^{-1}=16+64=80$. We have conquered the jungle!