Question

The value of \(cos^210^{\circ}-cos10^{\circ}cos50^{\circ}+cos^250^{\circ}\) is equal to....

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{3}{4}\)
• D. \(1\)

Hint:

Convert squares and the product into sums of cosines using product-to-sum identities.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We use \(\cos^2\theta=\dfrac{1+\cos2\theta}{2}\) and \(\cos A\cos B=\dfrac12[\cos(A+B)+\cos(A-B)]\).

Step 2: Rewrite the squares:
\[ \cos^210^{\circ}+\cos^250^{\circ}=1+\dfrac12(\cos20^{\circ}+\cos100^{\circ}) \]
\(\cos20^{\circ}+\cos100^{\circ}=2\cos60^{\circ}\cos40^{\circ}=\cos40^{\circ}\), so the sum is \(1+\dfrac12\cos40^{\circ}\).

Step 3: Rewrite the product:
\[ \cos10^{\circ}\cos50^{\circ}=\dfrac12(\cos60^{\circ}+\cos40^{\circ})=\dfrac14+\dfrac12\cos40^{\circ} \]

Step 4: Subtract:
\[ 1+\dfrac12\cos40^{\circ}-\dfrac14-\dfrac12\cos40^{\circ}=\dfrac34 \]

Step 5: Why the other options are wrong.
1/4, 1/2 and 1 would follow from dropping the "1" or the 1/4 term. The cosine of 40 degrees cancels, which leaves exactly 3/4.

Final Answer:
The value is 3/4. \[ \boxed{\text{(C) }\dfrac34} \]