Question 1031179
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Find the exact value of the expression.
tan(sin^−1(2/3)−cos^−1(1/3))
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<pre>
{{{tan(sin^(-1)(2/3) - cos^(-1)(1/3))}}}.


When solving problems like this, <U>half of the success is to reformulate it reasonably</U>.


So I will do it now.  We need to calculate  {{{tan(alpha - beta)}}},  where  {{{alpha}}} = {{{arcsin(2/3)}}}  and  {{{beta}}} = {{{arccos(1/3)}}}.

So we have the angle  {{{alpha}}}  in Q1  with {{{sin(alpha)}}} = {{{2/3}}},  and the angle  {{{beta}}}  in Q1  with {{{cos(beta)}}} = {{{1/3}}}.

Since {{{tan(alpha - beta)}}} = {{{sin(alpha-beta)/cos(alpha-beta)}}}, the plan is to calculate {{{sin(alpha-beta)}}} and {{{cos(alpha-beta)}}}.  //<U>Making a good plan is the second half of the success</U>. 

To calculate  {{{sin(alpha-beta)}}}  and  {{{cos(alpha-beta)}}},  we will use well known formulas of Trigonometry


{{{sin(alpha-beta)}}} = {{{sin(alpha)*cos(beta) - cos(alpha)*sin(beta)}}}  and  {{{cos(alpha-beta)}}} = {{{cos(alpha)*cos(beta) + sin(alpha)*sin(beta)}}}.    (1)


   (Regarding these formulas, see the lesson  <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Addition-and-subtraction-formulas.lesson>Addition and subtraction formulas</A>  in this site). 


Looking into the formulas (1), you see that we need to know/to have the values  {{{cos(alpha)}}}  and  {{{sin(beta)}}}  
in addition to the given values of  {{{sin(alpha)}}}  and  {{{cos(beta)}}}.  It is easy.  

First, {{{cos(alpha)}}} = {{{sqrt(1 - sin^2(alpha))}}} = {{{sqrt(1 - (2/3)^2)}}} = {{{sqrt(1-4/9)}}} = {{{sqrt(5)/3}}}. Second, {{{sin(beta)}}} = {{{sqrt(1 - cos^2(beta))}}} = {{{sqrt(1 - (1/3)^2)}}} = {{{sqrt(1-1/9)}}} = {{{sqrt(8)/3}}} = {{{(2*sqrt(2))/3}}}.

Notice that the signs at square roots are chosen "+" since the angles {{{alpha}}} and {{{beta}}} both lie in Q1.

Now you have everything to calculate {{{sin(alpha-beta)}}} and {{{cos(alpha-beta)}}} according to (1).  //<U>Implementing the plan accurately is the third half of the success</U>.


{{{sin(alpha-beta)}}} = {{{(2/3)*(1/3)-(sqrt(5)/3)*((2*sqrt(2))/3)}}} = {{{2/9 - (2*sqrt(10))/9}}} = {{{(2-2*sqrt(10))/9}}}.   (by the way, it shows that {{{alpha-beta}}} lies in Q4).

{{{cos(alpha-beta)}}} = {{{(sqrt(5)/3)*(1/3) + (2/3)*((2*sqrt(2))/3)}}} = {{{(sqrt(5)+4*sqrt(2))/9}}}. 

And finally  {{{tan(alpha-beta)}}} = {{{(2-2*sqrt(10))/(sqrt(5)+4*sqrt(2))}}}.    (2)

If you rationalize the denominator in (2),  you will get the answer  {{{tan(alpha-beta)}}} = {{{-(2*(sqrt(5)-sqrt(2)))/3}}}.

<U>Answer</U>.  {{{-(2*(sqrt(5)-sqrt(2)))/3}}}.
</pre>