what is the value of cos(3π/4)?\n√2/2\n-√2/2\n-1/2\n-√3/2

Explanation:

Step1: Recall cosine angle - relation

We know that $\cos(A + B)=\cos A\cos B-\sin A\sin B$. Also, $\frac{3\pi}{4}=\frac{\pi}{2}+\frac{\pi}{4}$. So, $\cos(\frac{3\pi}{4})=\cos(\frac{\pi}{2}+\frac{\pi}{4})$. By the formula $\cos(A + B)$ with $A=\frac{\pi}{2}$ and $B = \frac{\pi}{4}$, we have $\cos(\frac{\pi}{2}+\frac{\pi}{4})=\cos\frac{\pi}{2}\cos\frac{\pi}{4}-\sin\frac{\pi}{2}\sin\frac{\pi}{4}$. Since $\cos\frac{\pi}{2} = 0$ and $\sin\frac{\pi}{2}=1$, $\cos(\frac{\pi}{2}+\frac{\pi}{4})=0\times\cos\frac{\pi}{4}-1\times\sin\frac{\pi}{4}=-\sin\frac{\pi}{4}$.

Step2: Evaluate $\sin\frac{\pi}{4}$

We know that $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$. So, $\cos(\frac{3\pi}{4})=-\frac{\sqrt{2}}{2}$.

Answer:

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