Question

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

Answer 1

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)`