Trigonometry problem: Evaluate sin 20° sin 40° sin 60° sin 80°. Interactive solution with detailed explanation

Trigonometric Problem: Evaluate \(\sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ\)

Find the value of \[ \sin20^\circ \cdot \sin40^\circ \cdot \sin60^\circ \cdot \sin80^\circ. \]

Select the correct option:

A. \(-\frac{3}{16}\)

B. \(\frac{5}{16}\)

C. \(\frac{3}{16}\)

D. \(-\frac{5}{16}\)

Detailed Solution

Step 1: Write the product as \[ P = \sin20^\circ \cdot \sin40^\circ \cdot \sin60^\circ \cdot \sin80^\circ. \] Notice that \(\sin60^\circ = \frac{\sqrt{3}}{2}\). Thus, we have: \[ P = \frac{\sqrt{3}}{2} \left( \sin20^\circ \cdot \sin40^\circ \cdot \sin80^\circ \right). \]

Step 2: Simplify the sub-product \(Q = \sin20^\circ \cdot \sin40^\circ\). Using the product-to-sum formula: \[ \sin A \sin B = \frac{1}{2}\Big[\cos(A-B) - \cos(A+B)\Big], \] with \(A=20^\circ\) and \(B=40^\circ\), we get: \[ \sin20^\circ \sin40^\circ = \frac{1}{2}\Big[\cos(20^\circ-40^\circ) - \cos(20^\circ+40^\circ)\Big] = \frac{1}{2}\Big[\cos(-20^\circ) - \cos60^\circ\Big]. \] Since \(\cos(-20^\circ) = \cos20^\circ\) and \(\cos60^\circ = \frac{1}{2}\), this becomes: \[ \sin20^\circ \sin40^\circ = \frac{1}{2}\left(\cos20^\circ - \frac{1}{2}\right) = \frac{1}{2}\cos20^\circ - \frac{1}{4}. \]

Step 3: Multiply the result by \(\sin80^\circ\) to get: \[ \sin20^\circ \cdot \sin40^\circ \cdot \sin80^\circ = \left(\frac{1}{2}\cos20^\circ - \frac{1}{4}\right) \sin80^\circ. \] Distribute \(\sin80^\circ\): \[ = \frac{1}{2} \sin80^\circ \cos20^\circ - \frac{1}{4} \sin80^\circ. \] Now, use the sine addition formula in reverse on the first term. Recall: \[ \sin80^\circ \cos20^\circ = \frac{1}{2}\left[\sin(80^\circ+20^\circ) + \sin(80^\circ-20^\circ)\right] = \frac{1}{2}\left[\sin100^\circ + \sin60^\circ\right]. \] Since \(\sin100^\circ = \sin(80^\circ)\) (because \(\sin100^\circ = \sin(180^\circ-100^\circ) = \sin80^\circ\)), we have: \[ \sin80^\circ \cos20^\circ = \frac{1}{2}\left[\sin80^\circ + \sin60^\circ\right]. \]

Substituting back: \[ \sin20^\circ \sin40^\circ \sin80^\circ = \frac{1}{2} \cdot \frac{1}{2}\left[\sin80^\circ + \sin60^\circ\right] - \frac{1}{4}\sin80^\circ = \frac{1}{4}\left[\sin80^\circ + \sin60^\circ\right] - \frac{1}{4}\sin80^\circ. \] Simplify by canceling the \(\sin80^\circ\) terms: \[ = \frac{1}{4}\sin60^\circ. \]

Step 4: Substitute \(\sin60^\circ = \frac{\sqrt{3}}{2}\) into the expression: \[ \sin20^\circ \sin40^\circ \sin80^\circ = \frac{1}{4} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}. \]

Step 5: Now return to the original product: \[ P = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{8} = \frac{3}{16}. \]

Thus, the value of \(\sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ\) is \(\frac{3}{16}\), which corresponds to option C.