Sine of Theta plus Pi halves. If we use one of our trig identities and expand it out, we get sine Theta cosine Pi halves plus cosine Theta sine Pi halves. And we know that cosine Pi halves is 0 And we know that sine Pi halves is one. So the original function is equivalent to cosine Theta and the derivative of cosine Theta is negative sine Theta. Or we could have done this by the chain rule F prime of Theta derivative of sine is cosine Theta plus Pi halves. And then we have to take the derivative of the inside. The derivative of Theta plus Pi halves is 1. So using the expansion formula on cosine Theta plus Pi halves we have cosine Theta cosine Pi halves minus sine Theta sine Pi halves. Once again cosine Pi halves is 0, sine Pi halves is 1. So we get negative sine Theta either method of finding the derivative. So next thing we do is we're going to set the first derivative equal to 0. So 0 equals negative sine Theta, 0 equals sine Theta. That occurs at both zero and at π. So then we need to figure out whether they're maxes or mins. We also are going to include the end point, so we get zero Pi and five Pi force in between 0 and π. If we think about the negative of sine Theta, we know that zero to π is going to be negative and π to π force is going to be positive. So the original function is decreasing from zero to π and increasing from π to five Pi force. So we have local maxes at zero and at 5 Pi force and the local min π. To find out the absolute maxes and mins, we have to put those three locations back into the original function. So F of 0 is 1F, of π is -1, F of five Pi force is negative root 2 / 2, so we have an absolute Max of one. When Theta is zero, we have an absolute min of -1 when theta's π and we have a local Max of negative root 2 / 2 when theta's five Pi force.