Question 1178933
.


            If at the test you will solve the problem as  @MathLover1 does it and teaches you,

            the unsatisfactory score is  GUARANTEED  to you,


            because the solution by  @MathLover1 is  INCORRECT.


            I came to bring you the correct solution.



<pre>
tan(a) = -2 means that the angle "a" is in the second quadrant, QII,  OR  in the fourth quadrant, QIV.


tan(a) = -2   means that the opposite leg of the right angled triangle has the length 2, while the adjacent leg is of the length 1.


It implies that the hypotenuse is  {{{sqrt(1*2 + 2^2)}}} = {{{sqrt(5)}}} units long, and therefore


    |sin(a)| = {{{2/sqrt(5)}}}.


Now we should determine  <U>THE  SIGN</U>  of sin(a).



        There are  TWO CASES,  and we should consider them  SEPARATELY.



<U>CASE 1.  Angle "a" is in QII</U>


    in this case, sin(a) is positive, hence  sin(a) = {{{2/sqrt(5)}}} = {{{(2*sqrt(5))/5}}}.



<U>CASE 2.  Angle "a" is in QIV</U>


    in this case, sin(a) is negative, hence  sin(a) = {{{- 2/sqrt(5)}}} = {{{-(2*sqrt(5))/5}}}.




<U>ANSWER</U>.  If tan(a) = -2, then there are two possibilities for  "a"  and for  sin(a).


         If angle "a" is in QII, then sin(a) = {{{2/sqrt(5)}}} = {{{(2*sqrt(5))/5}}}.


         If angle "a" is in QIV, then sin(a) = {{{-2/sqrt(5)}}} = {{{(-2*sqrt(5))/5}}}.
</pre>

Solved &nbsp;(correctly, &nbsp;as it &nbsp;SHOULD &nbsp;BE); &nbsp;&nbsp;answered and explained.  &nbsp;&nbsp;&nbsp;&nbsp;And completed.



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For your safety, &nbsp;ignore the post by &nbsp;@MathLover1.



Also, &nbsp;&nbsp;ignore the post by @ewatrrr, &nbsp;&nbsp;since it is &nbsp;&nbsp;IRRELEVANT &nbsp;&nbsp;to the right solution.



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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;You, &nbsp;the visitor, &nbsp;should know it --- it is the truth of the life.



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