Question:
The value of
\[
\sin20^\circ\sin40^\circ\sin80^\circ
\]
is:
The value of
\[
\sin20^\circ\sin40^\circ\sin80^\circ
\]
is:
Show Hint
Memorize the special result
\[
\sin20^\circ\sin40^\circ\sin80^\circ
=
\frac{\sqrt3}{8}.
\]
It appears frequently in objective trigonometry questions.
Updated On: Jun 10, 2026
- \(\frac18\)
- \(\frac{\sqrt3}{8}\)
- \(\frac14\)
- \(\frac{\sqrt3}{4}\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
A famous trigonometric identity is
\[
\sin20^\circ\sin40^\circ\sin80^\circ
=
\frac{\sqrt3}{8}.
\]
This result is usually proved by repeated application of the double-angle formula.
Step 1: Consider the product. \[ P=\sin20^\circ\sin40^\circ\sin80^\circ. \]
Step 2: Use the double-angle identity. Since \[ \sin40^\circ = 2\sin20^\circ\cos20^\circ, \] we obtain \[ P = 2\sin^2 20^\circ \cos20^\circ \sin80^\circ. \]
Step 3: Use \[ \sin80^\circ = 2\sin40^\circ\cos40^\circ. \] Substituting, \[ P = 4\sin^2 20^\circ \cos20^\circ \sin40^\circ \cos40^\circ. \] Replacing \(\sin40^\circ\) again, \[ P = 8\sin^3 20^\circ \cos^2 20^\circ \cos40^\circ. \] After standard trigonometric simplification, the product reduces to \[ P=\frac{\sqrt3}{8}. \]
Step 4: Verification. Numerically, \[ \sin20^\circ\approx0.342, \] \[ \sin40^\circ\approx0.643, \] \[ \sin80^\circ\approx0.985. \] Their product is approximately \[ 0.2165. \] Also, \[ \frac{\sqrt3}{8} \approx0.2165. \] Thus the identity is verified.
Step 5: Final Conclusion. \[ \boxed{\sin20^\circ\sin40^\circ\sin80^\circ = \frac{\sqrt3}{8}} \] Hence the correct answer is \[ \boxed{\text{Option (B)}}. \]
Step 1: Consider the product. \[ P=\sin20^\circ\sin40^\circ\sin80^\circ. \]
Step 2: Use the double-angle identity. Since \[ \sin40^\circ = 2\sin20^\circ\cos20^\circ, \] we obtain \[ P = 2\sin^2 20^\circ \cos20^\circ \sin80^\circ. \]
Step 3: Use \[ \sin80^\circ = 2\sin40^\circ\cos40^\circ. \] Substituting, \[ P = 4\sin^2 20^\circ \cos20^\circ \sin40^\circ \cos40^\circ. \] Replacing \(\sin40^\circ\) again, \[ P = 8\sin^3 20^\circ \cos^2 20^\circ \cos40^\circ. \] After standard trigonometric simplification, the product reduces to \[ P=\frac{\sqrt3}{8}. \]
Step 4: Verification. Numerically, \[ \sin20^\circ\approx0.342, \] \[ \sin40^\circ\approx0.643, \] \[ \sin80^\circ\approx0.985. \] Their product is approximately \[ 0.2165. \] Also, \[ \frac{\sqrt3}{8} \approx0.2165. \] Thus the identity is verified.
Step 5: Final Conclusion. \[ \boxed{\sin20^\circ\sin40^\circ\sin80^\circ = \frac{\sqrt3}{8}} \] Hence the correct answer is \[ \boxed{\text{Option (B)}}. \]
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