Question

Find the value of $\cos 24^\circ+\cos 55^\circ+\cos 125^\circ+\cos 156^\circ$.

• A. $0$
• B. $2$
• C. $3$
• D. $1$

Hint:

Whenever you see angles that are supplements ($\theta$ and $180^\circ-\theta$), pair them using $\cos(180^\circ-\theta)=-\cos\theta$ to simplify quickly.

Answer 1

The Correct Option is A

Step 1: Use $\cos(180^\circ-\theta)=-\cos\theta$. 
$\cos 156^\circ=\cos(180^\circ-24^\circ)=-\cos 24^\circ$, \quad $\cos 125^\circ=\cos(180^\circ-55^\circ)=-\cos 55^\circ$. 

Step 2: Pair and sum. 
\[ (\cos24^\circ+\cos156^\circ)+(\cos55^\circ+\cos125^\circ) =(\cos24^\circ-\cos24^\circ)+(\cos55^\circ-\cos55^\circ)=0. \] \[ \boxed{0} \]

Answer 2

Instead of pairing supplementary angles directly, this value can be found by grouping adjacent terms with the sum-to-product formula.

Grouping and converting to products. Using \( \cos X+\cos Y = 2\cos\!\left(\dfrac{X+Y}{2}\right)\cos\!\left(\dfrac{X-Y}{2}\right) \):
\[ \cos24^\circ+\cos55^\circ = 2\cos(39.5^\circ)\cos(15.5^\circ), \] \[ \cos125^\circ+\cos156^\circ = 2\cos(140.5^\circ)\cos(15.5^\circ). \] Since \( 140.5^\circ = 180^\circ-39.5^\circ \), we have \( \cos(140.5^\circ) = -\cos(39.5^\circ) \). Adding the two products: \[ 2\cos(15.5^\circ)\big[\cos(39.5^\circ)-\cos(39.5^\circ)\big] = 2\cos(15.5^\circ)\times 0 = 0. \]

1. Option \(0\): Matches the value obtained above exactly — the two bracketed cosine terms cancel completely.
2. Option \(2\): Would only arise if all four cosines were equal and added constructively, which they do not; ruled out.
3. Option \(3\): Far too large — the terms partially cancel by construction, so a sum this size is impossible.
4. Option \(1\): Would require a leftover non-zero cosine term after cancellation, which the algebra above shows does not occur.

The grouping shows the whole expression collapses to zero regardless of the exact value of \(\cos(15.5^\circ)\), since it is multiplied by a bracket that vanishes.

So the correct answer is \(0\).