Given that sin(a) = 3/5 and cos(b) = − 1/3 , with a and b both in the interval π/2 , π , find the exact values of sin(a + b) and cos(a − b).

Answer 1

If the angle is between π/2 and π and sin(a) = 3/5 then the triangle looks like this:

That means, using the Pythagorean Theorem, the missing side is 4 (actually -4 because it's in the negative x direction) so the cos(a) = -4/5

If the angle is between π/2 and π and cos(b) = -1/3 then the triangle looks like this:

√

That means, using the Pythagorean Theorem, the missing side is √8 (or 2√2) so the sin(b) = 2√2/3.

Using the angle sum theorem, sin(a + b) = sin(a)cos(b) + cos(a)sin(b) and plugging in the values we get sin(a + b) = (3/5)(-1/3) + (-4/5)(2√2/3) simplifying we get sin(a + b) = -3/15 - 8√2/15 or (-3 - 8√2)/15

Then using the angle difference theorem, cos(a - b) = cos(a)cos(b) + sin(a)sin(b) and plugging in the values we get cos(a - b) = (-4/5)(-1/3) + (3/5)(2√2/3) simplifying we get cos(a - b) = 4/15 + 6√2/15 or (4 + 6√2)/15

Answer 2

sin(a) = 3/5 ---> cos(a) = 4/5 by pythagorean

cos(b) = -1/3 --->sin(b) = 2*sqrt(2) by pythagorean

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

= (3/5)( -1/3) + (2 * sqrt(2)) ( 4/5)

= (-1/5) + (8/5)(sqrt(2))

(8./5)(sqrt(2)) - 1/5

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

= (4/.5)(-1/3) + (3/5)(2 * sqrt(2))

= (3/5)(sqrt(2) - 4/15