Answer:
Explanation:
To prove that sin(51) + sin(81) - cos(21) = 0, we can use the trigonometric identity:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
We can rewrite 51 and 81 as the sum of two angles:
51 = 45 + 6
81 = 90 - 9
Using the above identity, we can write:
sin(51) = sin(45 + 6) = sin(45)cos(6) + cos(45)sin(6) = (2/2)sin(6) + (2/2)cos(6) = sin(6) + cos(6)
and
sin(81) = sin(90 - 9) = sin(90)cos(9) - cos(90)sin(9) = 0 + (-1)sin(9) = -sin(9)
Finally,
sin(51) + sin(81) - cos(21) = sin(6) + cos(6) - sin(9) - cos(21) = (sin(6) + cos(6)) - (sin(9) + cos(21))
We know that sin(90 - x) = cos(x) and cos(90 - x) = sin(x), so we can rewrite the right-hand side as:
(sin(6) + cos(6)) - (cos(90 - 9) + sin(90 - 21)) = (sin(6) + cos(6)) - (cos(9) + sin(69))
We also know that sin(180 - x) = -sin(x) and cos(180 - x) = -cos(x), so we can rewrite the right-hand side as:
(sin(6) + cos(6)) - (cos(9) + -sin(111)) = (sin(6) + cos(6)) - (-sin(111) + cos(9))
Since sin(6) + cos(6) = sin(6 + 90) = sin(96) and -sin(111) + cos(9) = sin(69 - 180) = -sin(111), we can simplify the expression further to:
sin(6) + cos(6) - (-sin(111)) + cos(9) = sin(96) + cos(9)
Since sin(x + y) = sin(x)cos(y) + cos(x)sin(y), we can write:
sin(96) + cos(9) = sin(60)cos(36) + cos(60)sin(36) = (2/2)sin(36) + (√3/2)cos(36) = sin(36) + (√3/2)cos(36)
Finally, since sin(2x) = 2sin(x)cos(x), we can write:
sin(36) + (√3/2)cos(36) = 2sin(18)cos(18) + (√3/2)cos(36) = 2(√2/2)(√2/2) + (√3/2)(√2/2) = (√2 + √6)/2 + (√6/2) = (√2 + √6)
So, sin(51) + sin(81) - cos(21) = (√2 + √6) ≠ 0.
Therefore, we have shown that sin(51) + sin(81) - cos(21) ≠ 0, and the statement "sin(51) + sin(81) - cos(21) = 0" is false.