Question:

To evaluate the expression $\cos 12^\circ + \cos 84^\circ + \cos 156^\circ + \cos 132^\circ$, we proceed as follows: **Step 1:** Rewrite angles using supplementary angle identities. Using the identity $\cos(180^\circ - x) = -\cos x$: $$ \cos 156^\circ = \cos(180^\circ - 24^\circ) = -\cos 24^\circ $$ $$ \cos 132^\circ = \cos(180^\circ - 48^\circ) = -\cos 48^\circ $$ Substituting these into the original expression, we get: $$ \cos 12^\circ + \cos 84^\circ - \cos 24^\circ - \cos 48^\circ $$ **Step 2:** Rearrange and group terms strategically. Group the terms to facilitate the application of sum-to-product formulas: $$ (\cos 24^\circ + \cos 48^\circ) - (\cos 12^\circ + \cos 84^\circ) $$ **Step 3:** Apply the sum-to-product formula: $\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$. For the first group, $(\cos 24^\circ + \cos 48^\circ)$: $$ 2 \cos\left(\frac{24^\circ+48^\circ}{2}\right) \cos\left(\frac{24^\circ-48^\circ}{2}\right) = 2 \cos(36^\circ) \cos(-12^\circ) = 2 \cos(36^\circ) \cos(12^\circ) $$ For the second group, $(\cos 12^\circ + \cos 84^\circ)$: $$ 2 \cos\left(\frac{12^\circ+84^\circ}{2}\right) \cos\left(\frac{12^\circ-84^\circ}{2}\right) = 2 \cos(48^\circ) \cos(-36^\circ) = 2 \cos(48^\circ) \cos(36^\circ) $$ **Step 4:** Substitute these results back into the grouped expression. The expression becomes: $$ 2 \cos(36^\circ) \cos(12^\circ) - 2 \cos(48^\circ) \cos(36^\circ) $$ Factor out $2 \cos(36^\circ)$: $$ 2 \cos(36^\circ) [\cos 12^\circ - \cos 48^\circ] $$ **Step 5:** Apply the difference-to-product formula: $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$. For $[\cos 12^\circ - \cos 48^\circ]$: $$ -2 \sin\left(\frac{12^\circ+48^\circ}{2}\right) \sin\left(\frac{12^\circ-48^\circ}{2}\right) $$ $$ = -2 \sin(30^\circ) \sin(-18^\circ) $$ Since $\sin(-x) = -\sin x$: $$ = -2 \sin(30^\circ) (-\sin 18^\circ) = 2 \sin(30^\circ) \sin(18^\circ) $$ Substitute the value $\sin(30^\circ) = \frac{1}{2}$: $$ = 2 \cdot \frac{1}{2} \cdot \sin(18^\circ) = \sin(18^\circ) $$ **Step 6:** Substitute this result back into the factored expression. The expression is: $$ 2 \cos(36^\circ) [\sin(18^\circ)] = 2 \cos(36^\circ) \sin(18^\circ) $$ **Step 7:** Use the known exact values for $\sin(18^\circ)$ and $\cos(36^\circ)$. $$ \sin(18^\circ) = \frac{\sqrt{5}-1}{4} $$ $$ \cos(36^\circ) = \frac{\sqrt{5}+1}{4} $$ Substitute these values into the expression: $$ 2 \left(\frac{\sqrt{5}+1}{4}\right) \left(\frac{\sqrt{5}-1}{4}\right) $$ $$ = 2 \frac{(\sqrt{5})^2 - 1^2}{16} $$ $$ = 2 \frac{5 - 1}{16} $$ $$ = 2 \frac{4}{16} $$ $$ = 2 \cdot \frac{1}{4} $$ $$ = \frac{1}{2} $$ The final value of the expression is $\frac{1}{2}$.