Find, without using a calculator, the value of sin of 𝜃 over two given tan of 𝜃 equals negative fifteen-eighths where three 𝜋 over two is less than 𝜃 is less than two 𝜋.
If we start out here by considering this angle 𝜃, we’re told that it lies between three 𝜋 over two here on our unit circle and an angle of two 𝜋 which defines the boundary between the first and the fourth quadrants. 𝜃 then is somewhere in this fourth quadrant. Along with this, we know that the tangent of this angle equals negative fifteen-eighths. Recalling that in general the tangent of an angle equals the ratio of 𝑦 to 𝑥, we can see that when it comes to a ratio of negative 15 over eight in the fourth quadrant, we must be working with an 𝑥-value of positive eight and a 𝑦-value of negative 15. That is, a line from the origin to this point here defines our angle 𝜃 where we could sketch it in as beginning at the positive 𝑥-axis.
Knowing all this, we want to solve for the sin not of 𝜃 but of 𝜃 over two. To help us do this, we can recall the half-angle identity for the sine function. The sin of half of an angle 𝜃 is equal to plus or minus the square root of one minus the cos of 𝜃 all over two. In our case, because 𝜃 lies between three 𝜋 over two and two 𝜋. We can say that 𝜃 over two is somewhere between three 𝜋 over four and 𝜋. Since the sine of any of these angles is positive, we can write that the sin of 𝜃 over two is simply equal to the square root of one minus the cos of 𝜃 all over two. We want to figure out then what is the cos of 𝜃.
Considering this right triangle in pink and this angle interior to that triangle, we can say that because the angle outside of it is 𝜃, then we must be able to write this interior angle as two 𝜋 minus 𝜃. Knowing that 𝜃 is an angle in the fourth quadrant tells us that two 𝜋 minus 𝜃 is an angle in the first quadrant. We can see this on our sketch that two 𝜋 minus 𝜃 is less than 𝜋 over two radians. Since the cosine function is positive in both the first and the fourth quadrants, for our particular value of 𝜃, we can say that the cos of two 𝜋 minus 𝜃 equals the cos of 𝜃. This is important because it means if we solve for the cosine of this angle, we’ll have solved for the cos of 𝜃, which we can use in our equation for the sin of 𝜃 over two. So let’s consider just what is the cosine of this angle.
In general, given a right triangle where one of the other interior angles we call 𝜃, we can express the cosine of that angle in terms of a ratio of the length of the sides of the triangle, specifically the adjacent side length to the hypotenuse. Returning to our sketch, relative to the angle two 𝜋 minus 𝜃, the side of the triangle adjacent to that is this side here while the hypotenuse is this length here. We know the length of this adjacent side. It’s eight units. But we don’t yet know the hypotenuse length. However, we can solve for it by recognizing that this opposite side of our triangle has a length of magnitude 15. The hypotenuse then by the Pythagorean theorem has a length of the square root of 15 squared plus eight squared. That comes out to exactly 17.
So, we’ve solved for the cos of two 𝜋 minus 𝜃. And as we argued, that equals the cos of 𝜃 itself. We’re now ready to substitute this value into our equation for the sin of 𝜃 over two. The square root of one minus eight seventeenths all over two equals the square root of nine seventeenths over two or the square root of nine over 34. If we recognize the fact that nine is equal to three squared and then take the step of rationalizing our expression, multiplying numerator and denominator by the square root of 34, we get a result of three times the square root of 34 over 34. This is the sin of the angle 𝜃 over two.