Question 1190009
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Please help me with this question, If SinA=2/3,and A is obtuse, find the exact value of,
1)cosA, 2)sin2A, 3) tan2A
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            The solution by @MathLover1 is  INCORRECT.


            I came to bring a correct solution.



<pre>
(1)  cos(A) = {{{-sqrt(1-sin^2(A))}}} = {{{-sqrt(1 - (2/3)^2)}}} = {{{-sqrt(1-4/9)}}} = {{{-sqrt(5/9)}}} = {{{-sqrt(5)/3}}}.


     Since A is obtuse, angle A is in QII, where cosine is negative, so we use sign  " - "  at the square root.


     It is first difference from the solution by @MathLover1.




(2)  sin(2A) = 2*sin(A)*cos(A) = {{{2*(2/3)*(-sqrt(5)/3)}}} = {{{-(4*sqrt(5))/9}}}.


     Notice sign  " - " in my answer.  It is the second difference from the solution by @MathLover1.




(3)  tan(2A) = {{{sin(2A)/cos(2A)}}} = {{{sin(2A)/(cos^2(A) - sin^2(A))}}} =  {{{ ((-4*sqrt(5)/9))/((-sqrt(5)/3)^2 - (2/3)^2)}}} = {{{((-4*sqrt(5)/9))/((5/9 - 4/9))}}} = {{{((-4*sqrt(5)/9))/((1/9))}}} = {{{-4*sqrt(5)}}}.


     Notice sign  " - " in my answer.  It is the third difference from the solution by @MathLover1.
</pre>

Solved &nbsp;&nbsp;&nbsp;&nbsp;(correctly).