Final answer:
For (a), assuming a typo and that the expression should be π/6 + 5/6, f(π/6 + 5/6) equals sin(π), which is 0. For (b), f(5/6) + f(13/6) simplifies to 2sin(5/6) because 13/6 is equivalent to 2π + 1/6 and sine function has a period of 2π.
Step-by-step explanation:
To evaluate the expressions using the functions provided, we can use identities from trigonometry. We have f(x) = sin x, g(x) = cos x, h(x) = tan x, and k(x) = 2x. For the given expressions, we must remember that the arguments for trigonometric functions are in radians.
(a) For f(\(ð + 5/6)), it seems there might be a typo, as \(ð is not a standard notation or number. Assuming it should be π/6, then f(π/6 + 5/6) = sin(π) because π/6 + 5/6 = 1π (or 6π/6), and we know sin(π) = 0.
(b) The second expression can be evaluated directly: f(5/6) + f(13/6). Since 13/6 is the same as 2π + 1/6 (because 2π is one full rotation in radians), f(5/6) = sin(5/6), and f(13/6) = sin(5/6) because sin has a period of 2π. Therefore, f(5/6) + f(13/6) = sin(5/6) + sin(5/6) = 2sin(5/6).