Can sin x be > 1? My son's pre-calculus teacher sent home a worksheet for sums and differences of angles. One of the angles showed sin x = 5/3, and that angle x was assigned to quadrant II.
The only way I calculate that happening is if cos x = -4i/3 (imaginary).
Other students were saying that cos x = -4/3.
I don't follow. By the Pythagorean theorem, x^2 + y^2 = r^2. The only way that x^2 + 25 = 9 is if x^2 = -16. That forces x = -4i.
Did the teacher mess up, or am I confused? I took math up to this level in high school (even making a perfect score on the ACT Math back in the 70s), but have had no math for more than forty years.
+23254 The value of sin(x) is pinned between -1 and + 1. It can be -1 at 270° ± k·360° (where k is any integer).
It can be +1 at 90° ± k·360° (where k is any integer).
For all other values of x, sin(x) must be between these two values.
If you think back to a right triangle, sin is defined to be the length of the opposite side divided by length of the hypotenuse. In a right triangle, the opposite side cannot be longer than the hypotenuse, thus the value of sin can never by greater than 1.
+23254 The value of sin(x) is pinned between -1 and + 1. It can be -1 at 270° ± k·360° (where k is any integer).
It can be +1 at 90° ± k·360° (where k is any integer).
For all other values of x, sin(x) must be between these two values.
If you think back to a right triangle, sin is defined to be the length of the opposite side divided by length of the hypotenuse. In a right triangle, the opposite side cannot be longer than the hypotenuse, thus the value of sin can never by greater than 1.
