Are you sure it's not graphing, and not that the graph is very close to the x-axis? Or else outside the borders of the graph? The inverse sine of 1/2 is a constant, about 0.5 in radians or 30 in degrees. In the first case at the standard window settings the graph is just above the end of the tick marks on the x-axis. In the latter case the graph is well above the top of the screen.Are you sure it's not graphing, and not that the graph is very close to the x-axis? Or else outside the borders of the graph? The inverse sine of 1/2 is a constant, about 0.5 in radians or 30 in degrees. In the first case at the standard window settings the graph is just above the end of the tick marks on the x-axis. In the latter case the graph is well above the top of the screen.
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Re: Theres no lines on the graph, i put in inverse of...
Change your mode (using the Mode button) from degrees to radians. You may also want to consider changing your window settings as you become more familiar with trigonometric functions.
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Not quite true. (Explanation at the end)
The sine and cosine of an angle have values in the closed interval [-1,1]. That means whatever the value of the angle, its sine or its cosine cannot be less than -1 or greater than 1.
When you consider the inverse problem where you want to find the angle whose sine or cosine is known, you must take the values of the sine or cosine in the interval [-1,1]. That is the domain of the functions arc sine and arc cosine. So if you take a value outside the domain [-1,1], and want to calculate the angle to which it belongs, the calculator signals that there a domain error, because there is no angle that answers the question. Limit the arguments of the sin^-1 and cos^-1 to values in [-1,1] For the inverses of arc tangent and arc cotangent functions there is no domain limitation because the range of tangent and cotangent is the whole real line ]-infinity, +infinity[.
I suspect that you are confusing things a bit.
The inverse sine, called the arcsine is a function defined in the closed interval [-1,1]. And so is the inverse cosine. Any value outside this interval will give you a non-real result (meaning a complex one).
There are no limitations on the domain of definition of the inverse hyperbolic sine or sinh^-1
If your input value is allowed to be complex, the arcsine function gives a complex value. See the screen capture
Does it refuse to do so or does it give an error message? Three common errors:
Not having the correct angle unit. Wrong result, No error message
Confusing reciprocal of sine (1/sin(x) with arc sine (x) ,sin^-1(x). Confusing the reciprocal of cosine, 1/cos(x) with arc cosine (cos^-1(x)). Wrong result, No error message
Taking the argument of the inverse sine and/or inverse cosine functions outside the interval [-1,1]. This gives a domain error.
The sine and cosine function have a range between [-1, 1]. The domain of their inverse functions is [-1,1]. So 20/1 which is 20 is out of the domain of definition of the functions. No limitations for tangent and cotangent.
If the angle unit is set to degrees and you use the standard window dimensions (Xmin=-10, Xmax=10) then all you will see is a straight line: Between 0 and 10 degrees, the sine is approximately equal to 0 and the cosine is about 1.
To see the features of the trigonometric functions you must do one of the following
Set angle to degree but set window dimensions to span the domain -180 to 180 or larger.
Set windows dimensions to the standard -10, 10 range but set angle degree to radians
The inverse sine is defined only for numbers in the range [-1, +1]. The sine of a number is at most 1.0. There is no number whose sine is 1.5, which is what you're asking for. The same applies to the cosine and inverse cosine.
The inverse sine and cosine equations mean "angle who's sine/cosine is:" The bounds of the arguments for these functions must be between 1 and -1, the maximum and minimum values of the sine and cosine functions, because no angle can have a sine or cosine that is greater than 1 or smaller than -1. Also it should be noted that the inverse sine/cosine function is not the same as 1/sine or 1/cosine, although the symbol makes it look like that.
Since you are familiar with sines, cosines, you know that their ranges (interval of values) varies from -1 to 1. The inverse functions of sine and cosine tkae their values in that very domain, [-1,1]. However you fed the arc sine function (sin^-1) a vlaue of (25/20.48) and that value is obviously larger outside the [-1,1] domain, hence the DOMAIN error message.
No such domain limitations exist for arc tangent (tan^-1) because the range of the tangent function spans the open interval ]negative infinity to positive infinity[.
The cotangent function is not implemented on the TI84PLUS native OS and consequently you will not find its inverse function on the calculator. However, there exits an equivalence (definition) between arc cotangent(x) and arc tangent (1/x). Thus cot^(-1)(X)= tan^(-1)(1/X),
Be careful. Since it is a multivariate function, some authors define the principal branch in the domain [0, PI], while others define the domain on [-Pi/2, Pi/2]. You will have to consult your textbook, or the documents you are using to find out the domain of the principal branch.
Are you sure it's not graphing, and not that the graph is very close to the x-axis? Or else outside the borders of the graph? The inverse sine of 1/2 is a constant, about 0.5 in radians or 30 in degrees. In the first case at the standard window settings the graph is just above the end of the tick marks on the x-axis. In the latter case the graph is well above the top of the screen.
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