Is there a value of x such that sin x = 3/2 ?
2 Answers
No
Explanation:
Within the realm of real numbers (your standard 1,2,3.1,pi,0 etc),
This is because the sin function will only output numbers ranging from
Visually, we can show this by graphing
graph{sin(x) [-10, 10, -5, 5]}
Yes, but only for complex values of
Explanation:
As a real valued function of real numbers,
So there is no real number
However, consider the following:
e^(itheta) = cos theta + i sin theta
So:
e^(itheta) - e^(-itheta) = (cos theta + i sin theta) - (cos(-theta) + i sin(-theta))
color(white)(e^(itheta) - e^(-itheta)) = (cos theta + i sin theta) - (cos(theta) - i sin(theta))
color(white)(e^(itheta) - e^(-itheta)) = 2i sin theta
So we find:
sin x = (e^(ix)-e^(-ix))/(2i)
We can use this definition of
Then we want to solve:
3/2 = sin x = (e^(ix)-e^(-ix))/(2i)
Multiply both ends by
3 = e^(ix)/i - 1/(ie^(ix)) = e^(ix)/i + i/e^(ix) = t+1/t
where
Multiply both ends by
12t = 4t^2+4
Subtract
0 = 4t^2-12t+4
color(white)(0) = (2t)^2-2(2t)(3)+3^2-5
color(white)(0) = (2t-3)^2-(sqrt(5))^2
color(white)(0) = ((2t-3)-sqrt(5))((2t-3)+sqrt(5))
color(white)(0) = (2t-3-sqrt(5))(2t-3+sqrt(5))
Hence:
e^(ix)/i = t = 1/2(3+-sqrt(5))
So:
e^(ix) = 1/2(3+-sqrt(5))i
So:
ix = ln (1/2(3+-sqrt(5))i) + 2npii" "n in ZZ
So:
x = -i ln (1/2(3+-sqrt(5))i) + 2npi" "n in ZZ