Cos 42° + sin 402° find the value of

The sine function has a period of $360^\circ$, which means $\sin(\theta) = \sin(\theta + n \cdot 360^\circ)$ for any integer $n$.
We can write $402^\circ$ as $360^\circ + 42^\circ$.

$$\sin(402^\circ) = \sin(360^\circ + 42^\circ)$$

Using the identity $\sin(\theta + 360^\circ) = \sin(\theta)$:

$$\sin(402^\circ) = \sin(42^\circ)$$

The original expression is $\cos(42^\circ) + \sin(402^\circ)$.
Substitute $\sin(402^\circ) = \sin(42^\circ)$:

$$\cos(42^\circ) + \sin(42^\circ)$$

Since $42^\circ$ is not a standard angle for which exact trigonometric values are typically known without a calculator, this is the most simplified form of the expression.