SOLUTION: Explain the following apparent error in the sin function. The equation sin (π/6) = (3.141592654/6) = sin (.523598776) = .5 is true. But (.523598776, .5) is not a point

Algebra ->  Trigonometry-basics -> SOLUTION: Explain the following apparent error in the sin function. The equation sin (π/6) = (3.141592654/6) = sin (.523598776) = .5 is true. But (.523598776, .5) is not a point      Log On


   

Question 442652: Explain the following apparent error in the sin function.
The equation sin (π/6) = (3.141592654/6) = sin (.523598776) = .5 is true.
But (.523598776, .5) is not a point on the unit circle since
(.523598776)^2 + (.5)^2 = .274155678 + .25 = .524155678 =/= 1,
meaning that
u^2 + v^2 =/= 1 and that the point (.523598776, .5) is not on the unit circle.
The above statement is also true. In fact the only first-quadrant point on the unit circle for which v = .5 is (.866025404, .5) (approximately).
How could sin (.523598776) = .5 when the point (.523598776, .5) is not even approximately on the unit circle. Explain. (Hint: ask yourself - of what is the unit circle a graph?)
Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
We are letting v be defined as sin%28theta%29 and u be defined as cos%28theta%29.
We know that sin%28theta%29%5E2+%2B+cos%28theta%29%5E2+=+1
In this example, we are letting theta+=+pi%2F6+=+.523598776

Then v+=+sin%28pi%2F6%29+=+1%2F2+=+.5
and u+=+cos%28pi%2F6%29+=+sqrt%283%29%2F2%7D=+.866025
And so u^2 + v^2 = .5^2 + .866025^2 = .25 + .75 = 1
Claiming that sin%28theta%29%5E2+%2B+theta%5E2+=+1 as they've done means that theta%5E2+=+cos%28theta%29%5E2, which can never happen. Thus their logic is flawed, and that is the error.