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Given that angle A lies in Quadrant III and sin(A)= −17/19, evaluate cos(A).

User Alejandra
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1 Answer

1 vote

As we know;


  • sin^2(x)+cos^2(x)=1

We will use this equality. We take the square of the sine of the given angle and subtract it from
1.


  1. sin^2(A)=(-(17)/(19) )^2=(289)/(361)

  2. sin^2(A)+cos^2(A)=1

  3. sin^2(A)=1-cos^2(A)

  4. (289)/(361)=1-cos^2(A)

  5. cos^2(A)=1-(289)/(361) =(72)/(361)

  6. √(cos^2(A)) =cos(A)

  7. \sqrt{(72)/(361) }=(6√(2) )/(19)

In the third region the sign of cosines is negative. Therefore, our correct answer should be as follows;


  • cos(A)=-(6√(2) )/(19)

User Sunhee
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