Question

Evaluate \[ \cos\frac{6\pi}{17}\cos\frac{10\pi}{17}\cos\frac{12\pi}{17}\cos\frac{14\pi}{17}. \]

• A. $-\dfrac{1}{16}$
• B. $\dfrac{1}{16}$
• C. $-16$
• D. $\dfrac{1}{4}$

Hint:

Use $\cos(\pi-\theta)=-\cos\theta$ first to convert angles greater than $\frac{\pi}{2}$ into acute angles before applying product identities.

Answer 1

The Correct Option is A

Step 1: Make the angles smaller using a rule.
Use $\cos(\pi-\theta)=-\cos\theta$. Since $\dfrac{10\pi}{17}=\pi-\dfrac{7\pi}{17}$, we get $\cos\dfrac{10\pi}{17}=-\cos\dfrac{7\pi}{17}.$
Step 2: Do the same for the others.
$\cos\dfrac{12\pi}{17}=-\cos\dfrac{5\pi}{17}$ and $\cos\dfrac{14\pi}{17}=-\cos\dfrac{3\pi}{17}.$ Three minus signs appear.
Step 3: Collect the signs.
The product becomes \[ P=\cos\tfrac{6\pi}{17}\big(-\cos\tfrac{7\pi}{17}\big)\big(-\cos\tfrac{5\pi}{17}\big)\big(-\cos\tfrac{3\pi}{17}\big)=-\cos\tfrac{3\pi}{17}\cos\tfrac{5\pi}{17}\cos\tfrac{6\pi}{17}\cos\tfrac{7\pi}{17}. \]
Step 4: Recognise a standard product.
The angles $\dfrac{3\pi}{17},\dfrac{5\pi}{17},\dfrac{6\pi}{17},\dfrac{7\pi}{17}$ form a known product whose value is $\dfrac1{16}.$ This comes from the cosine product identity for the $17$th roots.
Step 5: Substitute the known value.
So $P=-\dfrac1{16}.$
Step 6: State the answer.
A numerical check also gives $-0.0625$. \[ \boxed{-\dfrac1{16}} \]