If `sinx=3/5,cosy=-(12)/(13),`where x and y both lie in second quadrant, find the value of `sin (x + y)dot`

Options

A. 56/65
B. -56/65
C. 33/65
D. -33/65

Answer

To find the value of \( \sin(x + y) \) given \( \sin x = \frac{3}{5} \) and \( \cos y = -\frac{12}{13} \), where both \( x \) and \( y \) lie in the second quadrant, we can follow these steps:

### Step 1: Find \( \cos x \)

Since \( x \) is in the second quadrant, \( \cos x \) will be negative. We can use the Pythagorean identity:

\[ \cos^2 x + \sin^2 x = 1 \]

Substituting \( \sin x = \frac{3}{5} \):

\[ \cos^2 x + \left(\frac{3}{5}\right)^2 = 1 \]

\[ \cos^2 x + \frac{9}{25} = 1 \]

\[ \cos^2 x = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \]

Taking the negative square root (since \( x \) is in the second quadrant):

\[ \cos x = -\sqrt{\frac{16}{25}} = -\frac{4}{5} \]

### Step 2: Find \( \sin y \)

Since \( y \) is also in the second quadrant, \( \sin y \) will be positive. We can use the Pythagorean identity:

\[ \sin^2 y + \cos^2 y = 1 \]

Substituting \( \cos y = -\frac{12}{13} \):

\[ \sin^2 y + \left(-\frac{12}{13}\right)^2 = 1 \]

\[ \sin^2 y + \frac{144}{169} = 1 \]

\[ \sin^2 y = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169} \]

Taking the positive square root (since \( y \) is in the second quadrant):

\[ \sin y = \sqrt{\frac{25}{169}} = \frac{5}{13} \]

### Step 3: Use the sine addition formula

Now we can use the sine addition formula:

\[ \sin(x + y) = \sin x \cos y + \cos x \sin y \]

Substituting the known values:

\[ \sin(x + y) = \left(\frac{3}{5}\right) \left(-\frac{12}{13}\right) + \left(-\frac{4}{5}\right) \left(\frac{5}{13}\right) \]

Calculating each term:

\[ \sin(x + y) = -\frac{36}{65} - \frac{20}{65} \]

Combining the fractions:

\[ \sin(x + y) = -\frac{36 + 20}{65} = -\frac{56}{65} \]

### Final Answer

Thus, the value of \( \sin(x + y) \) is:

\[ \sin(x + y) = -\frac{56}{65} \]

---