Inverse trig

[John Smith]

I was assisting a student with finding all the answers to such questions as
sin x = 1/2 in the usual four quadrants from 0 to 360.

The student had been told not to use a calculator, which I thought was an
excellent idea, but I couldn't resist using a calculator myself just to see
which quadrant it used for the answer.

This led to the question of why does the calculator not use the usual four
quadrants for its inverse trig functions?

Clearly these functions require a range of -1 to +1 for inverse sin and cos
and -lots to +lots for inverse tan.

It seems to me that any three adjacent quadrants will satisfy this
requirement.

For inverse tan, the calculator is clearly using the range -90 to +90 which
includes what might be called the zeroth quadrant.

For inverse sin and cos it's using the range -90 to +180

Have I missed the good reason for this?

Old guy

[W. Dale Hall]

For the inverse function to be an actual (i.e., single-valued) function,
the original function needs to be one-to-one. Due to the periodicity of
the trigonometric functions, this requires placing some limits on the
domain over which the function is to be inverted.

The standard range for inverse sine is [-pi/2, pi/2]. (or [-90, 90] in
degrees). The standard range for inverse cosine is [0, pi], which in
degrees gives [0 180].

Note that for these ranges, the complementary function is non-negative.
That is, to get the cosine value from sine, just take
sqrt(1-sin^2(x)), and similarly to pass from sine to cosine. This makes
the compositions

cos(sin^(-1)(x)) = sqrt(1-x^2)
sin(cos^(-1)(x)) = sqrt(1-x^2)

I suppose these simplifications come in handy in calculus, so you don't
have to worry about carrying minus sign factors in computations.

The standard range for inverse tangent is (-pi/2, pi/2). It includes 0,
which is a useful feature, covers one full period of the tangent
function, and avoids the pole at odd multiples of pi/2.

Why? I would imagine that it is mostly convention, together with the
fact that these values cover the typical applications to trigonometry.

[1treePetrifiedForestLane]

>         cos(sin^(-1)(x)) = sqrt(1-x^2)
=?... anyway, it has nothing to do
with the regular tetragon ...
>         sin(cos^(-1)(x)) = sqrt(1-x^2)

[John Smith]

"W. Dale Hall" <wdh...@alum.mit.edu> wrote in message
news:QsudnYnWzuNmLwvM...@giganews.com...

Ok so cos is >= 0 from -90 to +90
And sin is >= 0 from 0 to 180

>
> I suppose these simplifications come in handy in calculus, so you don't
> have to worry about carrying minus sign factors in computations.
>
> The standard range for inverse tangent is (-pi/2, pi/2). It includes 0,
> which is a useful feature, covers one full period of the tangent
> function, and avoids the pole at odd multiples of pi/2.
>
> Why? I would imagine that it is mostly convention, together with the
> fact that these values cover the typical applications to trigonometry.

Ok thanks for your reply.

Old Guy.

[quasi]

John Smith wrote:

>I was assisting a student with finding all the answers to
>such questions as sin x = 1/2 in the usual four quadrants
>from 0 to 360.
>
>The student had been told not to use a calculator, which
>I thought was an excellent idea, but I couldn't resist
>using a calculator myself just to see which quadrant it
>used for the answer.
>
>This led to the question of why does the calculator no

>use the usual four quadrants for its inverse trig functions?

Because the 6 trig functions are not one-to-one, so they don't
have inverses.

To repair that, for each of the trig functions, there is a
somewhat standard choice of restricted domain such that, on the
restricted domain, the given trig function satisfies the
following conditions:

(1) It has full range, that is, has the same range as the
unrestricted version of the function.

(2) It's one-to-one (hence has an inverse).

Provided the above conditions are met, the choice of restricted
domain is somewhat arbitrary, but there are some motivating
preferences:

(3) The domain for each of the restricted trig functions should
include the open interval 0 to Pi/2, so that, as a minimum, all
acute angles are in the domain.

(4) It would be nice if the restricted trig function was
continuous. Equivalently, it would be nice if the domain was an
interval.

Of course, we would also like universal agreement on the choice
of restricted domain.

From the graph of y = sin(x), a natural choice of restricted
domain is evident, namely -Pi/2 <= x <= Pi/2.

Similarly, for y = cos(x), the natural choice of restricted
domain is 0 <= x <= Pi.

For y = tan(x), the natural choice of restricted domain is
-Pi/2 < x < Pi/2.

For y = cot(x), the natural choice of restricted domain is
0 < x < Pi.

For csc(x) and sec(x), assuming conditions (1),(2),(3) are
satisfied, condition (4) can't be satisfied, so as the choice
is even more arbitrary. There are some trade-offs but no
universally agreement as to the best choice.

The choice below is the one used by Maple, TI-85-Plus and
Wikipedia:

For y = csc(x), choose the restricted domain
-Pi/2 <= x <= Pi/2 and x =/= 0.

For y = sec(x), choose the restricted domain
0 <= x <= Pi and x =/= Pi/2.

The above choices are natural in the sense that each restricted
domain is as connected as possible (each domain has a missing
point but no gap).

However, as I mentioned above, for csc(x) and sec(x), the
choice of restricted domain in not a universal standard.

For example, some Calculus and Precalculus textbooks make the
following choice instead:

For y = csc(x), choose the restricted domain (using interval
notation) (0,Pi/2] U (Pi,(3/2)*Pi].

For y = sec(x), choose the restricted domain (using interval
notation) [0,Pi/2) U [Pi,(3/2)*Pi).

The above choices apparently provide some advantages for
simplifications used in Calculus.

Bottom line:

For sin^(-1),cos^(-1),tan^(-1),cot^(-1), the choice of range
is standard, so no worry about those, but if you plan to use
csc^(-1) or sec^(-1), you need to know what range is being
assumed.

>Clearly these functions require a range of -1 to +1 for
>inverse sin and cos

Yes.

Restricted sin has range [-1,1] (the same as unrestricted sin),
and sin^(-1) has domain [-1,1].

Restricted cos has range [-1,1] (the same as unrestricted cos),
and cos^(-1) has domain [-1,1].

>and -lots to +lots for inverse tan.

Restricted tan has range (-oo,oo) (the same as unrestricted tan),
and tan^(-1) has domain (-oo,oo).

>It seems to me that any three adjacent quadrants will satisfy
>this requirement.

For the restricted trig functions, any 3 adjacent quadrants will
have full range but you can't use all 3 -- one must also ensure
that the restricted trig functions are one-to-one.

>For inverse tan, the calculator is clearly using the range
>-90 to +90 which includes what might be called the zeroth
>quadrant.

The range of tan^(-1) is the interval (-Pi/2,Pi/2), so
-90 to +90, exclusive.

>For inverse sin and cos it's using the range -90 to +180

No.

The range of sin^(-1) is the interval [-Pi/2,Pi/2], so
-90 to 90 inclusive.

The range of cos^(-1) is the interval [0,Pi], so
0 to 180 inclusive.

quasi

[quasi]

Change the word "Equivalently" above to "Alternatively".

>Of course, we would also like universal agreement on the choice
>of restricted domain.
>
>From the graph of y = sin(x), a natural choice of restricted
>domain is evident, namely -Pi/2 <= x <= Pi/2.
>
>Similarly, for y = cos(x), the natural choice of restricted
>domain is 0 <= x <= Pi.
>
>For y = tan(x), the natural choice of restricted domain is
>-Pi/2 < x < Pi/2.
>
>For y = cot(x), the natural choice of restricted domain is
>0 < x < Pi.
>
>For csc(x) and sec(x), assuming conditions (1),(2),(3) are

>satisfied, condition (4) can't be satisfied, so the choice

>is even more arbitrary. There are some trade-offs but no

>universal agreement as to the best choice.

[John Smith]

"quasi" <qu...@null.set> wrote in message
news:kl5lp8lkfrmrmi016...@4ax.com...

> John Smith wrote:
>
>>I was assisting a student with finding all the answers to
>>such questions as sin x = 1/2 in the usual four quadrants
>>from 0 to 360.

>>.....

>>
>>For inverse tan, the calculator is clearly using the range
>>-90 to +90 which includes what might be called the zeroth
>>quadrant.
>
> The range of tan^(-1) is the interval (-Pi/2,Pi/2), so
> -90 to +90, exclusive.
>

I was expecting Google to know that, or at least to ignore the question and
return just a search, but it didn't.

www.google.com/#q=tan(pi%2F2)

All the calculators I can find give a large multiple of 180/pi (i.e. 1
radian) for tan(89.99999999)

I made a spreadsheet for a student showing the graph of sin cos and tan
from -180 to +360.
I had to truncate tan at +/- 2.5 so as not to lose sin and cos in the
scaling.
The graph looks nice though and makes it clear what's happening in all
quadrants of interest.

Old Guy