sin 3/5 ans cos12/13 find sin(a+b) and cos(a-b)
Answers
Answer:
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