If Theta =(pi )/(6) find the exact value of each expresslo (a) sin2Theta = (b) sin(-Theta )= square (c) sin^2Theta =

(a) To find the exact value of $sin2\Theta$, we can use the double-angle formula for sine, which states that $sin2\Theta = 2sin\Theta cos\Theta$. <br /><br />Given that $\Theta = \frac{\pi}{6}$, we can substitute this value into the formula:<br /><br />$sin2\Theta = 2sin\left(\frac{\pi}{6}\right)cos\left(\frac{\pi}{6}\right)$<br /><br />We know that $sin\left(\frac{\pi}{6}\right)frac{1}{2}$ and $cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.<br /><br />Substituting these values, we get:<br /><br />$sin2\Theta = 2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$<br /><br />Simplifying, we have:<br /><br />$sin2\Theta = \frac{\sqrt{3}}{2}$<br /><br />Therefore, the exact value of $sin2\Theta$ is $\frac{\sqrt{3}}{2}$.<br /><br />(b) To find the exact value of $sin(-\Theta)$, we can use the property of states $sin(-\Theta) = -sin(\Theta)$.<br /><br />Given that $\Theta = \frac{\pi}{6}$, we can substitute this value into the formula:<br /><br />$sin(-\Theta) = -sin\left(\frac{\pi}{6}\right)$<br /><br />We know that $sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.<br /><br />Substituting this value, we get:<br /><br />$sin(-\Theta) = -\frac{1}{2}$<br /><br />Therefore, the exact value of $sin(-\Theta)$ is $-\frac{1}{2}$.<br /><br />(c) To find the exact value of $sin^{2}\Theta$, we can square the value of $sin\Theta$.<br /><br />Given that $\Theta = \frac{\pi}{6}$, we can substitute this value into the formula:<br /><br />$sin^{2}\Theta = \left(sin\left(\frac{\pi}{6}\right)\right)^2$<br /><br />We know that $sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.<br /><br />Substituting this value, we get:<br /><br />$sin^{2}\Theta = \left(\frac{1}{2}\right)^2$<br /><br />Simplifying, we have:<br /><br />$sin^{2}\Theta = \frac{1}{4}$<br /><br />Therefore, the exact value of $sin^{2}\Theta$ is $\frac{1}{4}$.