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
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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) (Show Source): You can put this solution on YOUR website!
We are letting v be defined as and u be defined as .
We know that
In this example, we are letting
Then
and
And so u^2 + v^2 = .5^2 + .866025^2 = .25 + .75 = 1
Claiming that as they've done means that , which can never happen. Thus their logic is flawed, and that is the error.
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