To evaluate $\cos\left(\frac{7\pi}{12}\right)$ as $\cos\left(\frac{\pi}{4} + \frac{\pi}{3}\right)$, we will use the sum identity for cosine:

$\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$

In this case, $A = \frac{\pi}{4}$ and $B = \frac{\pi}{3}$. Now let's substitute these into the identity:

$\cos\left(\frac{\pi}{4} + \frac{\pi}{3}\right) = \cos\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{3}\right)$

We know the exact values of these trigonometric functions:

• $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
• $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$
• $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
• $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

Now substituting these values:

$\cos\left(\frac{7\pi}{12}\right) = \frac{\sqrt{2}}{2} \cdot \frac{1}{2} - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}$

Simplifying:

$\cos\left(\frac{7\pi}{12}\right) = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}$

Factoring out $\frac{1}{4}$:

$\cos\left(\frac{7\pi}{12}\right) = \frac{1}{4} \left(\sqrt{2} - \sqrt{6}\right)$

Thus, the value of $\cos\left(\frac{7\pi}{12}\right)$ is:

$\boxed{\frac{1}{4} \left(\sqrt{2} - \sqrt{6}\right)}$

Would you like further explanation or details? Here are 5 related questions:

1. How do you derive the sum and difference identities for cosine?
2. Can you evaluate $\sin\left(\frac{7\pi}{12}\right)$ using a similar approach?
3. How would you evaluate trigonometric expressions involving other angles like $\frac{5\pi}{12}$?
4. Can you prove the exact values of $\cos\left(\frac{\pi}{4}\right)$ and $\sin\left(\frac{\pi}{3}\right)$?
5. How can you use sum identities to solve trigonometric equations?

Tip: Familiarize yourself with the common angles and their exact trigonometric values; they are frequently useful in many problems!