If sin a = 4/5, pi/2<a<pi; cos b = 2/3, 0<b<pi/2 then what is cos (a+b)?

Answer 1

sin(a) = 4/5, π/2 < a < π (located on the 2nd qudrant)

y = 4, r = 5

Using r² = x² + y²

5² = x² + 4²

5²- 4² = x²

x² = 25 - 16

x² = 9

x = - 3 (x on second quadrant is negative)

∴ cos (a) = x / r = -3/5

cos (b) = 2/3, 0 < a < π/2 (located on the 1st quadrant)

x=2, r=3

Using r² = x² + y²:

3² = 2² + y²

3² - 2² = y²

y² = 9 - 4

y² = 5

y = 5^1/2 (y is positive on the 1st quadrant)

∴ sin (b) = 5^1/2 / 3

The sum formula for cosine:

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

Substitute:

cos(a+b) = (-3/5)(2/3) - (4/5)(5^1/2/3)

cos(a+b) = - 6/15 - (4•5^1/2/15)

Factor out -2/15:

cos(a+b) = (- 2/15)(3 + 2•5^1/2)

cos(a+b) ≈ -0.99628