Find sin(s+t) and (s-t) if cos(s)= 1/5 and sin(t) = 3/5 and s and t are in quadrant 1.

=Sin(s)cos(t) + Cos(s)Sin(t)
=Sin(1/5)Cos(3/5) + Cos(-1/5)Sin(3/5)
= 0.389418

Sin(s-t)
=sin(s)cos(t) - cos(s)sin(t)
=sin(-3/5)cos(1/5) - cos(1/5)sin(3/5)
=Sin-3/5 cos-3/5
= -0.46602

HELP PLEASE!!

no, no. Draw your triangles:

if cos(s) = 1/5, then sin(s) = √24/5 = 2√6/5
if sin(t) = 3/5, then cos(t) = 4/5

sin(s+t) = (2√6/5)(4/5) + (1/5)(3/5) = (8√6+3)/25

and similarly for sin(s-t)

To find the values of sin(s+t) and sin(s-t), we can use the following trigonometric formulas:

sin(s + t) = sin(s)cos(t) + cos(s)sin(t)
sin(s - t) = sin(s)cos(t) - cos(s)sin(t)

Given that cos(s) = 1/5 and sin(t) = 3/5, and both s and t are in the first quadrant, we can now substitute these values into the formulas.

Let's calculate sin(s + t):

sin(s + t) = sin(s)cos(t) + cos(s)sin(t)
= (1/5)(3/5) + (cos(s))(3/5) [substituting the given values]
= 3/25 + (1/5)(3/5) [evaluating the multiplication]
= 3/25 + 3/25
= 6/25

Therefore, sin(s + t) = 6/25.

Now, let's calculate sin(s - t):

sin(s - t) = sin(s)cos(t) - cos(s)sin(t)
= (1/5)(3/5) - (cos(s))(3/5) [substituting the given values]
= 3/25 - (1/5)(3/5) [evaluating the multiplication]
= 3/25 - 3/25
= 0

Therefore, sin(s - t) = 0.

To recap:
- sin(s + t) = 6/25
- sin(s - t) = 0

To solve this problem, we can use the double-angle identities for sine and cosine:

1. sin(s + t) = sin(s)cos(t) + cos(s)sin(t)
2. sin(s - t) = sin(s)cos(t) - cos(s)sin(t)

Given that cos(s) = 1/5 and sin(t) = 3/5, we can substitute these values into the formulas:

1. sin(s + t) = sin(s)cos(t) + cos(s)sin(t)
= (1/5)(3/5) + (1/5)(3/5)
= (3/25) + (3/25)
= 6/25
≈ 0.24

2. sin(s - t) = sin(s)cos(t) - cos(s)sin(t)
= (1/5)(3/5) - (1/5)(3/5)
= (3/25) - (3/25)
= 0

So, sin(s + t) is approximately 0.24 and sin(s - t) is 0.

In Quadrant 1, both s and t are positive, which means sin(s + t) is positive and sin(s - t) is 0.

I hope this helps, let me know if you have any further questions!

I support steve,he is correct.