If sin x = \(\frac{4}{5}\), cos y = \(-\frac{12}{13}\), where x and y both lie in second quadrant, find the value of sin(x + y)?

Given:

sin x = \(\frac{4}{5}\) and cos y = \(-\frac{12}{13}\)

Formula Used:

sin(x + y) = sin x cos y + cos x sin y

sin²x + cos²x = 1

Calculation:

We know that,

sin2x + cos2x = 1

⇒ \((\frac{4}{5})^2 + cos^2x = 1\)

⇒ \(\frac{16}{25} + cos^2x = 1\)

⇒ cos²x = \(1 - \frac{16}{25}\)

⇒ cos²x = \(\frac{9}{25}\)

⇒ cos x = \(± \sqrt{\frac{9}{25}}\)

⇒ cos x = \(± \frac{3}{5}\)

Since x in II^nd  quadrant. 

So, cos x is negative.

⇒ cos x = \(- \frac{3}{5}\)

⇒ sin²y + cos²y = 1

⇒ sin²y = 1 - cos²y

⇒ sin²y = \(1 - (\frac{-12}{13})^2\)

⇒ sin²y = \(1 - \frac{144}{169}\)

⇒ sin²y = \(\frac{25}{169}\)

⇒ sin y = \(± \sqrt{\frac{25}{169}}\)

⇒ sin y = \(± \frac{5}{13}\)

Since y lies in II^nd quadrant,

So, sin y is positive.

⇒ sin y = \(\frac{5}{13}\)

Now, Putting the values of sin x, cos x, sin y, cos y

⇒ sin (x + y) = sin x cos y + cos x sin y

⇒ sin (x + y) = \(\frac{4}{5} \times (\frac{-12}{13}) + (\frac{-3}{5}) \times \frac{5}{13}\)

⇒ sin (x + y) = \(\frac{ - 48}{65} - \frac{15 }{65}\)

⇒ sin (x + y) = \(\frac{-48 - 15}{65}\)

⇒ sin (x + y) = \(-\frac{63}{65}\)

∴ The value of sin(x + y) is \(-\frac{63}{65}\)