We are asked to find the exact value of $\cos\left(\frac{13\pi}{12}\right)$ by using a sum or difference formula. The angle $\frac{13\pi}{12}$ is not a standard angle, so we need to express it as a sum or difference of two angles that are standard angles for which we know the cosine values.

We can break down $\frac{13\pi}{12}$ as:

$\frac{13\pi}{12} = \pi - \frac{\pi}{4}$

This works because $\pi = \frac{12\pi}{12}$, and $\frac{13\pi}{12} = \frac{12\pi}{12} + \frac{\pi}{12}$, making the expression equivalent to $\pi - \frac{\pi}{4}$.

Now, using the cosine difference identity: $\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)$

Let $A = \pi$ and $B = \frac{\pi}{4}$. Substituting these into the formula:

$\cos\left(\frac{13\pi}{12}\right) = \cos(\pi)\cos\left(\frac{\pi}{4}\right) + \sin(\pi)\sin\left(\frac{\pi}{4}\right)$

From the unit circle, we know:

• $\cos(\pi) = -1$
• $\sin(\pi) = 0$
• $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
• $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Now substitute these values:

$\cos\left(\frac{13\pi}{12}\right) = (-1) \cdot \frac{\sqrt{2}}{2} + 0 \cdot \frac{\sqrt{2}}{2}$

Simplifying this gives:

$\cos\left(\frac{13\pi}{12}\right) = -\frac{\sqrt{2}}{2}$

Thus, the exact value of $\cos\left(\frac{13\pi}{12}\right)$ is:

$\boxed{-\frac{\sqrt{2}}{2}}$

Would you like a detailed explanation on how to use this in similar problems?

Related questions:

1. How do you apply the sum and difference identities for sine?
2. What is the unit circle, and how does it help in trigonometry?
3. Can $\cos\left(\frac{13\pi}{12}\right)$ be calculated without a sum or difference identity?
4. How do you derive the cosine and sine values for standard angles like $\frac{\pi}{4}$?
5. What other angle sum or difference combinations could work for $\frac{13\pi}{12}$?

Tip:

Always ensure you are working with angles that are easily recognizable from the unit circle to simplify trigonometric calculations!