SOLUTION: Write the given expression in terms of x and y only. sin(tan^−1(x) − tan^−1(y)) Find the exact value of the expression. tan(sin^−1(2/3)−cos^&

Algebra ->  Trigonometry-basics -> SOLUTION: Write the given expression in terms of x and y only. sin(tan^−1(x) − tan^−1(y)) Find the exact value of the expression. tan(sin^−1(2/3)−cos^&      Log On


   



Question 1031179: Write the given expression in terms of x and y only.
sin(tan^−1(x) − tan^−1(y))


Find the exact value of the expression.
tan(sin^−1(2/3)−cos^−1(1/3))





Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find the exact value of the expression.
tan(sin^−1(2/3)−cos^−1(1/3))
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

tan%28sin%5E%28-1%29%282%2F3%29+-+cos%5E%28-1%29%281%2F3%29%29.


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


So I will do it now.  We need to calculate  tan%28alpha+-+beta%29,  where  alpha = arcsin%282%2F3%29  and  beta = arccos%281%2F3%29.

So we have the angle  alpha  in Q1  with sin%28alpha%29 = 2%2F3,  and the angle  beta  in Q1  with cos%28beta%29 = 1%2F3.

Since tan%28alpha+-+beta%29 = sin%28alpha-beta%29%2Fcos%28alpha-beta%29, the plan is to calculate sin%28alpha-beta%29 and cos%28alpha-beta%29.  //Making a good plan is the second half of the success. 

To calculate  sin%28alpha-beta%29  and  cos%28alpha-beta%29,  we will use well known formulas of Trigonometry


sin%28alpha-beta%29 = sin%28alpha%29%2Acos%28beta%29+-+cos%28alpha%29%2Asin%28beta%29  and  cos%28alpha-beta%29 = cos%28alpha%29%2Acos%28beta%29+%2B+sin%28alpha%29%2Asin%28beta%29.    (1)


   (Regarding these formulas, see the lesson  Addition and subtraction formulas  in this site). 


Looking into the formulas (1), you see that we need to know/to have the values  cos%28alpha%29  and  sin%28beta%29  
in addition to the given values of  sin%28alpha%29  and  cos%28beta%29.  It is easy.  

First, cos%28alpha%29 = sqrt%281+-+sin%5E2%28alpha%29%29 = sqrt%281+-+%282%2F3%29%5E2%29 = sqrt%281-4%2F9%29 = sqrt%285%29%2F3. Second, sin%28beta%29 = sqrt%281+-+cos%5E2%28beta%29%29 = sqrt%281+-+%281%2F3%29%5E2%29 = sqrt%281-1%2F9%29 = sqrt%288%29%2F3 = %282%2Asqrt%282%29%29%2F3.

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%28alpha-beta%29 and cos%28alpha-beta%29 according to (1).  //Implementing the plan accurately is the third half of the success.


sin%28alpha-beta%29 = %282%2F3%29%2A%281%2F3%29-%28sqrt%285%29%2F3%29%2A%28%282%2Asqrt%282%29%29%2F3%29 = 2%2F9+-+%282%2Asqrt%2810%29%29%2F9 = %282-2%2Asqrt%2810%29%29%2F9.   (by the way, it shows that alpha-beta lies in Q4).

cos%28alpha-beta%29 = %28sqrt%285%29%2F3%29%2A%281%2F3%29+%2B+%282%2F3%29%2A%28%282%2Asqrt%282%29%29%2F3%29 = %28sqrt%285%29%2B4%2Asqrt%282%29%29%2F9. 

And finally  tan%28alpha-beta%29 = %282-2%2Asqrt%2810%29%29%2F%28sqrt%285%29%2B4%2Asqrt%282%29%29.    (2)

If you rationalize the denominator in (2),  you will get the answer  tan%28alpha-beta%29 = -%282%2A%28sqrt%285%29-sqrt%282%29%29%29%2F3.

Answer.  -%282%2A%28sqrt%285%29-sqrt%282%29%29%29%2F3.