Cos 42° + sin 402° find the value of

To find the value of $\cos(42^\circ) + \sin(402^\circ)$, we first need to simplify the term $\sin(402^\circ)$.

Step 1: Simplify $\sin(402^\circ)$ using the periodicity of the sine function. 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 $\sin(\theta + 360^\circ) = \sin(\theta)$: $\sin(402^\circ) = \sin(42^\circ)$

Step 2: Substitute the simplified term back into the original expression. 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.

The final answer is $\boxed{\cos(42^\circ) + \sin(42^\circ)}$. 3 done, 2 left today. You're making progress.