Graphing Inverse Trigonometric Functions
Key Questions
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Throughout the following answer, I will assume that you are asking about trigonometry restricted to real numbers.
Using Domain of
arc sin x Find
arc sin (3) .3 is not in the domain ofarc sin , (3 is not in the range ofsin ) soarc sin (3) does not exist.Using Range of
arc sin x Find
arc sin (1/2) .Although there are infinitely many
t withsin t = 1/2 , the range ofarc sin restricts the value to thoset with(-pi)/2 <= t <= pi/2 , So the value we want ispi/6 .Using the quadrant
This is the same as using the range, but it involves thinking about the problem more geometrically.Find
arc sin (1/2) .arc sin (1/2) is a number (an angle) in quadrant I or IV. It is thet with smallest absolute value (the shortest path from the initial side).Again,
arc sin (1/2) = pi/6 .
Jim H
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Mar 19 2015
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For a trig function, the range is called "Period"
For example, the function
f(x) = cos x has a period of2pi ; the functionf(x) = tan x has a period ofpi . Solving or graphing a trig function must cover a whole period.The range depends on each specific trig function.
For example, the inverse functionf(x) = 1/(cos x) = sec x has as period2pi .Its range varies from (+infinity) to Minimum
1 then back to (+infinity), between (-pi/2 andpi/2 ).Its range also varies from (-infinity) to Max -1 then back to to (-infinity), between (
pi/2 and3pi/2 ).Nghi N. · 2 · Apr 7 2015
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Since the graphs of
f(x) andf'(x) are symmetric about the liney=x , start with the graph of a trigonometric function with an appropriate restricted domain, then reflect it about the liney=x .(Caution: Their domains must be restricted to an appropriate interval so that their inverses exist.)
Let us sketch the graph of
y=sin^{-1}x .The graph of
y=sinx on[-pi/2, pi/2] looks like:
By reflecting the graph above about the line
y=x ,
The curve in purple is the graph of
y=sin^{-1}x .The graphs of other inverse trigonometric functions can be obtained similarly.
I hope that this was helpful.
Wataru
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Oct 29 2014