If Y equals sine X, then we know Y prime equal cosine X and we want to find the equation of the tangents at these three locations. So I need the slope at all three locations. So Y prime and negative π Y prime of 0 and Y prime of three Pi halves -1 one and 0 respectively. I also need the original Y value at these three locations, so 00 and -1 so at the point negative π, zero my X&YI know my slope is -1. Putting it in point slope form and then solving I get Y equal negative X - π for the next one. My point was 00 with a slope of one. Putting it in slope point slope form y -, 0 equal 1 X -0 I get Y equal X. The last one, the .3 Pi halves common -1. With a slope of 0 we get Y equal -1. The last part of this says we want to graph the curve. So we're going to graph a sine curve, and we want to graph their tangents at negative π at 0 and at 3 Pi halves. So at negative π we should expect a slope of -1. At zero, we should expect a slope of +1 and three Pi halves we should expect a slope of 0.