Prove that: cos 7π/12 + cos π/12 = sin 5π/12 − sin π/12

Question

Prove that:

\[ \cos \frac{7\pi}{12}+\cos \frac{\pi}{12} = \sin \frac{5\pi}{12}-\sin \frac{\pi}{12} \]

Proof

Consider the left-hand side:

\[ \cos \frac{7\pi}{12}+\cos \frac{\pi}{12} \]

Using the identity:

\[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \]

Let

\[ C=\frac{7\pi}{12}, \qquad D=\frac{\pi}{12} \]

Then,

\[ \cos \frac{7\pi}{12}+\cos \frac{\pi}{12} \]

\[ = 2\cos\left(\frac{\frac{7\pi}{12}+\frac{\pi}{12}}{2}\right) \cos\left(\frac{\frac{7\pi}{12}-\frac{\pi}{12}}{2}\right) \]

\[ = 2\cos\left(\frac{8\pi}{24}\right) \cos\left(\frac{6\pi}{24}\right) \]

\[ = 2\cos\frac{\pi}{3}\cos\frac{\pi}{4} \]

Now,

\[ \cos\frac{\pi}{3}=\frac{1}{2}, \qquad \cos\frac{\pi}{4}=\frac{1}{\sqrt{2}} \]

Therefore,

\[ = 2\times\frac{1}{2}\times\frac{1}{\sqrt{2}} \]

\[ = \frac{1}{\sqrt{2}} \]

Now consider the right-hand side:

\[ \sin \frac{5\pi}{12}-\sin \frac{\pi}{12} \]

Using the identity:

\[ \sin C-\sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2} \]

Let

\[ C=\frac{5\pi}{12}, \qquad D=\frac{\pi}{12} \]

Then,

\[ \sin \frac{5\pi}{12}-\sin \frac{\pi}{12} \]

\[ = 2\cos\left(\frac{\frac{5\pi}{12}+\frac{\pi}{12}}{2}\right) \sin\left(\frac{\frac{5\pi}{12}-\frac{\pi}{12}}{2}\right) \]

\[ = 2\cos\left(\frac{6\pi}{24}\right) \sin\left(\frac{4\pi}{24}\right) \]

\[ = 2\cos\frac{\pi}{4}\sin\frac{\pi}{6} \]

Now,

\[ \cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}, \qquad \sin\frac{\pi}{6}=\frac{1}{2} \]

Therefore,

\[ = 2\times\frac{1}{\sqrt{2}}\times\frac{1}{2} \]

\[ = \frac{1}{\sqrt{2}} \]

Hence,

\[ \cos \frac{7\pi}{12}+\cos \frac{\pi}{12} = \sin \frac{5\pi}{12}-\sin \frac{\pi}{12} \]

Hence proved.