Question

The value of \(cos(\frac{π}{5})\) is ...

• A. \(\frac{-1+\sqrt{5}}{4}\)
• B. \(\frac{-1+\sqrt{5}}{2}\)
• C. \(\frac{1+\sqrt{5}}{4}\)
• D. \(\frac{1-\sqrt{5}}{4}\)

Hint:

Let A = cos 36 and B = cos 72; their product is 1/4 and their difference is 1/2.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We want \(\cos 36^{\circ}\). Let \(A=\cos36^{\circ}\) and \(B=\cos72^{\circ}\). Two simple facts about \(A\) and \(B\) will fix \(A\).

Step 2: Find the Product:
Multiply and divide by \(\sin36^{\circ}\) and use the double-angle formula twice:
\[ AB=\frac{\sin36^{\circ}\cos36^{\circ}\cos72^{\circ}}{\sin36^{\circ}}=\frac{\tfrac12\sin72^{\circ}\cos72^{\circ}}{\sin36^{\circ}}=\frac{\tfrac14\sin144^{\circ}}{\sin36^{\circ}}=\frac14 \]
because \(\sin144^{\circ}=\sin36^{\circ}\).

Step 3: Find the Difference:
\(A-B=\cos36^{\circ}-\cos72^{\circ}=2\sin54^{\circ}\sin18^{\circ}\). Now \(\sin54^{\circ}=\cos36^{\circ}=A\) and \(\sin18^{\circ}=\cos72^{\circ}=B\). So
\[ A-B=2AB=\frac12 \]

Step 4: Solve for A:
Put \(B=A-\tfrac12\) in \(AB=\tfrac14\):
\[ A\left(A-\frac12\right)=\frac14 \Rightarrow 4A^2-2A-1=0 \Rightarrow A=\frac{2\pm\sqrt{20}}{8}=\frac{1\pm\sqrt5}{4} \]
\(\cos36^{\circ}\) is positive, so \(A=\dfrac{1+\sqrt5}{4}\approx0.809\).

Step 5: Check the Other Options:
Option (A) \(\frac{-1+\sqrt5}{4}\approx0.309\) is \(\cos72^{\circ}\). Option (D) is its negative. Option (B) is about 0.618, which is too small for \(\cos36^{\circ}\).

Final Answer:
\(\cos(\pi/5)=\dfrac{1+\sqrt5}{4}\), option (C). \[ \boxed{\text{(C) } \frac{1+\sqrt{5}}{4}} \]