Given that sin a = 3/5, cos b = 24/25. a is in (0,90) and b is in (180, 270 ). Calculate sin ( a + b ) and sin ( a - b ).

We notice that the angle a belongs to the first quadrant,
where the values of the functions sine and cosine are both positive. We also notice that
the angle b is in the third quadrant, where the values of the functions sine and cosine
are both negative.

We'll write the formula for sin
(a+b).

sin (a+b) = sin a*cos b + sin b*cos
a

We know the values for sin a and cos b but we'll have to
calculate the values for sin b and cos a.

We'll apply the
fundamental formula of trigonometry:

(sin a)^2 + (cos a)^2
= 1

(cos a)^2 = 1 - (sin
a)^2

Since a is in the first quadrant, when we'll calculate
the sqrt of 1 - (sin a)^2, we'll keep only the positive value for cos
a.

cos a = sqrt (1 - 9/25)

cos
a = sqrt 16/25

cos a =
4/5

(sin b)^2 + (cos b)^2 =
1

(sin b)^2 = 1 - (cos
b)^2

Since b is in the third quadrant, when we'll calculate
the sqrt of 1 - (cos b)^2, we'll keep only the negative value for sin
b.

sin b = -sqrt(1 -
24^2/25^2)

sin b = -
sqrt[(25-24)(25+24)/25^2]

sin b = -
sqrt(1*49/25^2)

sin b =
-7/5

Now, we can calculate sin
(a+b)

sin (a+b) = sin a*cos b + sin b*cos
a

sin (a+b) = (3/5)*(24/25) +
(-7/5)*(4/5)

sin (a+b) = 72/125 -
28/25

sin (a+b) =
(72-140)/25

sin (a+b) =
-68/25

We'll write the formula for sin
(a-b).

sin (a-b) = sin a*cos b - sin b*cos
a

sin (a-b) = 72/125 +
28/25

sin (a-b) =
(72+140)/25

sin (a-b) =
212/25