0. Code

>	restart; with(plots):

Warning, the name changecoords has been redefined

>

ArcsinTracePlot := proc(t)     local a,c1,c2,c3,A,B,C,E,F,G,GG,H,J;     theta||1 := t;     theta||2 := arcsin( sin(t));      a := sin(theta||1):      c1 := COLOR(RGB, 1, .75, .2 ):      c2 := COLOR(RGB, .7, .4, .1):      c3 := COLOR(RGB, 1, .2, .1):      A  := plot(sin(x), x = -Pi..2*Pi,        color = c3, tickmarks = [1,1]):      B  := plot(sin(x), x = -Pi/2..Pi/2,    color = c2, thickness = 4):      C  := polygonplot( [[-Pi/2,-1],[Pi/2,-1],[Pi/2,1],[-Pi/2,1],[-Pi/2,-1]],                color = c1, style = patchnogrid ):      E  := plot( [[-Pi/2,-1],[Pi/2,-1],[Pi/2,1],[-Pi/2,1],[-Pi/2,-1]],                color = orange, linestyle = 3):      F  := plottools[arrow]( [theta||1, a], [theta||2,a],                .05, .10, .15,  color = red):      G  := plottools[arrow]( [theta||1, 0], [theta||1,a],                .15,.25,.10,  color = red):     GG  := plottools[arrow]( [theta||2,a], [theta||2, 0],                .15,.25,.10,  color = red):        H := textplot([theta||1, -.15,  evalf(theta||1,3)],align={ABOVE, CENTER}):     J := textplot([  -.2, a,  evalf(a, 3)],align={ABOVE, LEFT}):     display([F,G,GG, A,B,C,E,H,J ]); end proc :

 1.  Simple Inverse Trig Expressions

A simple inverse trig expression involves find an angle which gives a specific value.

      Example 1.1  : Compute  

In this case, we are trying to find an angle theta, where . Perhaps you can guess

>	sin(theta) = sqrt(3)/2; solve(%, theta);

>	arcsin( sqrt(3)/2);

In other cases, it's not simple to guess the answer.

      Example 1.2  : Compute  

>	arcsin( 2/3); evalf(%);

       Example 1.3  : Simplify  arcsin( 2 )

Notice that the only valid input to arcsine and arccosine are numbers in the range [-1,1]. Maple will solve equations for values outside that range using more advanced mathematics (a branch of mathematics called Complex Analysis , which you will study if you major in mathematics) and comes up with a complex number.

>	arcsin(  2); evalf(%);

 

      Example 1.4  : Simplify  

>	arccos( 2/3); evalf(%);

      Example 1.5  : Simplify  arctan( 100 )

>	arctan( 100 ); evalf(%);

 2.  Compose Function with Its Inverse Function

This is the simpler of two cases. Inverse trig functions are always one-to-one and yield unique angles. When a trig function is applied to that angle, there is a unique result. In other words, a trig function composed with its inverse cancel one another's transformations, and render the original value.

That is, sin(arcsin(x)) = x, in the same way that  and

      Example  : Simplify sin( arcsin( .819 ))

>	sin( arcsin( .819 ));

And while we're at it...

>	cos( arccos( .819 ));

>	tan( arctan( .819 ));

 3.  Compose Function with Another Inverse Function

What about a trig function applied to another inverse trig function?

       Example 3.1  : sin( arccos( .819) )

>	sin( arccos( .819 ) );

 
Maple gives us the answer promptly. However, to better understand the process, we can follow these steps.

>	c := cos( arccos( .819 ));

>	Well_Known_Formula := sin(x)^2 + cos(x)^2 = 1;

>	subs( cos(x) = c, Well_Known_Formula);

>	soln := solve( %, sin(x) );

We have a value for sin(x) ... in fact, two values... which is one answer too many. How do we tell which of the two possible answers is the real answer?

A positive cosine = .819 is in the first quadrant, rather than the second quadrant. The sine in the first quadrant is positive.

>	arccos( .819 ) < Pi/2; evalf(%);

>	abs( soln[1]);

This type of problem can be solved even when dealing with unknowns.

       Example 3.2  :  tan( arcsin( x/3 ) )

>	tan( arcsin( x/3 ) );

To solve this problem on paper, its best to draw a triangle which has an opposite side of length x and a hypotenuse of 3. Such an angle will have sine  :

>	sin(theta) = x/3;

Then we find the third side of the triangle using the theorem of Pythagoras. Then the tangent is easy to find as the opposite side over the adjacent side.

>	third_side := sqrt( 3^2 - x^2);

>	tan(theta) = x/third_side;

 Here are other examples ...

>	tan( arcsin( x/3 ) );

>	cos( arctan( x/3 ) );

>	sin( arccos( x/3 ) );

 4.  Compose Inverse Function with Function

This is the more mysterious case. It seems to work well in some cases - where the arcsine undoes the work of the sin. But in other cases it does not seem to work as well.

>	arcsin( sin( Pi/6) );

>	arcsin( sin( 7*Pi/6) );

>	arcsin( sin( 19*Pi/6 ) );

The reason for this is that the sine can accept any real number as its input. If we were to use the sine with restricted domain, the inverse would cancel the function. The restricted domain for sine is . Any value in that domain will work well.

>	arcsin( sin( Pi/3) ); arcsin( sin( -Pi/5 ) ); arcsin( sin( 0 ) );

But any angle outside this domain,  will not come back to where it started.

>	arcsin( sin( 2*Pi/3) ); arcsin( sin( -5*Pi/6 ) ); arcsin( sin( 2*Pi ) );

Here is a diagram which demonstrates why arcsin( sin( 7Pi./6)) = - Pi/6.

>	ArcsinTracePlot(7*Pi/6);

>	ArcsinTracePlot(5*Pi/6);

>	ArcsinTracePlot(-5*Pi/6);

 5.  More Complicated Inverse Function Expressions

We can also do some more complicated problems.

            Example 5.1  : Simplify cos( arcsin( 3/4) + arcsin( 2/3) )

>	cos( arcsin( 3/4) + arcsin( 2/3) );

>	theta := arcsin(3/4); theta := evalf(%);

>	phi := arcsin(2/3); phi := evalf(%);

>	cos( theta + phi);

While that gives us the answer, there is another way of doing this problem which might work out better for paper-and-pencil solving - plus it gives us an exact radical form of the answer.

>	CosineAdditionIdentity :=          cos( alpha + beta) = cos(alpha)*cos(beta) - sin(alpha)*sin(beta);

>	subs( {  alpha = arcsin( 3/4), beta = arcsin( 2/3) }, CosineAdditionIdentity );

Now we have reduced the one hard problem to four simple problems that we have already seen how to solve.

>	simplify(%);

>	evalf(rhs(%));

>