sin(20)sin(40)sin(60)sin(80)
Understand the Problem
The question consists of a trigonometric expression involving the product of sine functions with arguments 20, 40, 60, and 80 degrees. The task is likely to simplify this expression and find its exact value.
Answer
$\frac{3}{16}$
Answer for screen readers
$\frac{3}{16}$
Steps to Solve
- Substitute $\sin(60^\circ)$
We know $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, so we can rewrite the expression as: $$ \sin(20^\circ) \sin(40^\circ) \sin(60^\circ) \sin(80^\circ) = \sin(20^\circ) \sin(40^\circ) \cdot \frac{\sqrt{3}}{2} \cdot \sin(80^\circ) $$
- Rearrange the terms
Rearrange to group the sine terms together: $$ \frac{\sqrt{3}}{2} \sin(20^\circ) \sin(40^\circ) \sin(80^\circ) $$
- Use the identity $\sin(x) = \cos(90^\circ - x)$
Rewrite $\sin(80^\circ)$ as $\cos(10^\circ)$: $$ \frac{\sqrt{3}}{2} \sin(20^\circ) \sin(40^\circ) \cos(10^\circ) $$
- Use the product-to-sum identity $2\sin(A)\sin(B) = \cos(A-B) - \cos(A+B)$
Apply the product-to-sum identity to $\sin(20^\circ) \sin(40^\circ)$: $$ \sin(20^\circ) \sin(40^\circ) = \frac{1}{2} [\cos(40^\circ - 20^\circ) - \cos(40^\circ + 20^\circ)] = \frac{1}{2} [\cos(20^\circ) - \cos(60^\circ)] $$ Since $\cos(60^\circ) = \frac{1}{2}$, we have: $$ \sin(20^\circ) \sin(40^\circ) = \frac{1}{2} \cos(20^\circ) - \frac{1}{4} $$
- Substitute back into the expression
Substitute this back into the original expression: $$ \frac{\sqrt{3}}{2} (\frac{1}{2} \cos(20^\circ) - \frac{1}{4}) \cos(10^\circ) = \frac{\sqrt{3}}{2} (\frac{1}{2} \cos(20^\circ) \cos(10^\circ) - \frac{1}{4} \cos(10^\circ)) $$ $$ = \frac{\sqrt{3}}{4} \cos(20^\circ) \cos(10^\circ) - \frac{\sqrt{3}}{8} \cos(10^\circ) $$
- Use the product-to-sum identity $2\cos(A)\cos(B) = \cos(A-B) + \cos(A+B)$
Apply the product-to-sum identity to $\cos(20^\circ) \cos(10^\circ)$: $$ \cos(20^\circ) \cos(10^\circ) = \frac{1}{2} [\cos(20^\circ - 10^\circ) + \cos(20^\circ + 10^\circ)] = \frac{1}{2} [\cos(10^\circ) + \cos(30^\circ)] $$ Since $\cos(30^\circ) = \frac{\sqrt{3}}{2}$, we have: $$ \cos(20^\circ) \cos(10^\circ) = \frac{1}{2} \cos(10^\circ) + \frac{\sqrt{3}}{4} $$
- Substitute back into the expression
Substitute this back into the expression: $$ \frac{\sqrt{3}}{4} (\frac{1}{2} \cos(10^\circ) + \frac{\sqrt{3}}{4}) - \frac{\sqrt{3}}{8} \cos(10^\circ) = \frac{\sqrt{3}}{8} \cos(10^\circ) + \frac{3}{16} - \frac{\sqrt{3}}{8} \cos(10^\circ) $$
- Simplify
Simplify the expression: $$ \frac{\sqrt{3}}{8} \cos(10^\circ) + \frac{3}{16} - \frac{\sqrt{3}}{8} \cos(10^\circ) = \frac{3}{16} $$
$\frac{3}{16}$
More Information
The problem involves trigonometric identities and simplification. By using the product-to-sum identities and known trigonometric values, the expression simplifies to a constant value.
Tips
A common mistake is not remembering or misapplying the trigonometric identities correctly. Another mistake is making algebraic errors during the simplification process.
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