8. (1 point) Given cos40^circ =a what is sin320^circ a. a b. sqrt (1-a^2) c. sqrt (1+a^2) d. -sqrt (1-a^2)

## Step 1: Rewrite $sin(320^{\circ})$<br />### We can rewrite $sin(320^{\circ})$ as $sin(360^{\circ} - 40^{\circ})$. Since $sin(360^{\circ} - x) = -sin(x)$, we have $sin(320^{\circ}) = -sin(40^{\circ})$.<br /><br />## Step 2: Relate $sin(40^{\circ})$ to $cos(40^{\circ})$<br />### Using the trigonometric identity $sin^2(x) + cos^2(x) = 1$, we can express $sin(40^{\circ})$ in terms of $cos(40^{\circ})$. We have $sin^2(40^{\circ}) = 1 - cos^2(40^{\circ})$. Since $40^{\circ}$ is in the first quadrant, $sin(40^{\circ})$ is positive. Therefore, $sin(40^{\circ}) = \sqrt{1 - cos^2(40^{\circ})}$.<br /><br />## Step 3: Substitute the given value<br />### We are given that $cos(40^{\circ}) = a$. Substituting this into the equation from Step 2, we get $sin(40^{\circ}) = \sqrt{1 - a^2}$.<br /><br />## Step 4: Calculate $sin(320^{\circ})$<br />### From Step 1, we know that $sin(320^{\circ}) = -sin(40^{\circ})$. Substituting the value of $sin(40^{\circ})$ from Step 3, we get $sin(320^{\circ}) = -\sqrt{1 - a^2}$.