Given that `tan x = (12)/(5), cos y = (-3)/(5),` and the angles x and y are in the same quadrants, calcualte without the use of tables the values of `(i) sin (x + y), (ii) cos ""(y)/(2).`

Answer

To solve the problem, we need to calculate \( \sin(x+y) \) and \( \cos\left(\frac{y}{2}\right) \) given that \( \tan x = \frac{12}{5} \) and \( \cos y = -\frac{3}{5} \).

### Step 1: Determine the values of \( \sin x \) and \( \cos x \)

Given \( \tan x = \frac{12}{5} \), we can visualize this as a right triangle where the opposite side is 12 and the adjacent side is 5.

Using the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

Now we can find \( \sin x \) and \( \cos x \):
\[ \sin x = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{12}{13} \]
\[ \cos x = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{5}{13} \]

Since \( x \) is in the third quadrant, both sine and cosine will be negative:
\[ \sin x = -\frac{12}{13}, \quad \cos x = -\frac{5}{13} \]

### Step 2: Determine the value of \( \sin y \)

We know \( \cos y = -\frac{3}{5} \). To find \( \sin y \), we use the identity:
\[ \sin^2 y + \cos^2 y = 1 \]
Substituting the value of \( \cos y \):
\[ \sin^2 y + \left(-\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2 y + \frac{9}{25} = 1 \]
\[ \sin^2 y = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25} \]
\[ \sin y = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5} \]

Since \( y \) is also in the third quadrant, \( \sin y \) will be negative:
\[ \sin y = -\frac{4}{5} \]

### Step 3: Calculate \( \sin(x+y) \)

Using the sine addition formula:
\[ \sin(x+y) = \sin x \cos y + \cos x \sin y \]
Substituting the values we have:
\[ \sin(x+y) = \left(-\frac{12}{13}\right)\left(-\frac{3}{5}\right) + \left(-\frac{5}{13}\right)\left(-\frac{4}{5}\right) \]
Calculating each term:
\[ \sin(x+y) = \frac{36}{65} + \frac{20}{65} = \frac{56}{65} \]

### Step 4: Calculate \( \cos\left(\frac{y}{2}\right) \)

Using the half-angle formula:
\[ \cos\left(\frac{y}{2}\right) = \sqrt{\frac{1 + \cos y}{2}} \]
Since \( \cos y = -\frac{3}{5} \):
\[ \cos\left(\frac{y}{2}\right) = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{2}{5}}{2}} = \sqrt{\frac{2}{10}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]
Since \( y \) is in the third quadrant, \( \frac{y}{2} \) is in the second quadrant, where cosine is negative:
\[ \cos\left(\frac{y}{2}\right) = -\frac{1}{\sqrt{5}} \]

### Final Answers:
1. \( \sin(x+y) = \frac{56}{65} \)
2. \( \cos\left(\frac{y}{2}\right) = -\frac{1}{\sqrt{5}} \)