Question

If sinA=3/5 and cosB= 12/13
Find the value of sin(A+B) and cos(A-B),in first quadrant A

Answer 1

Answer:

Step-by-step explanation:

We may NOT assume A and B are complementary

angles. They in fact belong to different

right triangles.

Since sin A = 3/5, the opposite side of the

right triangle is 3 and the hypotenuse is 5.

The missing adjacent side must then be 4.

So cos(A) = 4/5

Likewise, since cosine of B is 12/13, this

right triangle has hypotenuse 13 and adjacent

side 12. The missing opposite side is

5. So the sin(B) = 5/13

Deriving the angle subtraction formula for sine:

sin(A-B) = sin (A + -B)

= sin(A)cos(-B) + sin(-B)cos(A)

= sin(A)cos(B) - sin(B)cos(A) negative sign while sine is an odd

function; so sine spits out the negative

= (3/5)(12/13) - (5/13)(4/5)

= 36/65 - 20/65

= 16/65

= 0.2461538

Alternatively, you get the same answer using inverse trig functions:

A = inverse-sine (3/5) = 36.87

B = inverse-cosine(12/13) = 22.619865

A-B = 14.250135

sin(A-B) = 0.246155576724839