Solving The Trigo Dilemma for cos(-A)

[Kyoma]

Given that sinA= $$\frac{-1}{\sqrt{5}}$$ where A is more than 180 degrees and less than 270 degrees. Find the value of cos(-A).

Without using Calculator,

Since cos(-A) = cosA, and that A is in the 3rd quadrant, then after solving for the hypotenuse, adjacent and opposite, I got:

$$\frac{-2}{\sqrt{5}}$$

With Calculator,

A= Inverse Sin($$\frac{-1}{\sqrt{5}}$$) = -26.57 (4 T.C.)
Subst -26.57 into cos(-A), I got:

$$\frac{2}{\sqrt{5}}$$

One is positive, another is negative. Which is which?

 

[Hurkyl]

With Calculator,

A= Inverse Sin($$\frac{-1}{\sqrt{5}}$$) = -26.57 (4 T.C.)

That angle very clearly does not satisfy the system of equations and inequalities you were trying to solve...

 

[Kyoma]

Then is it possible to get Angle A?

 

[verty]

Yes, it is possible. Think about the domain and range of the arcsin function, then use trigonometric identities to get the correct answer.

 

[mathman]

A= -26.57 (4 T.C.)

I'll assume A is in degrees. That angle is in the fourth quadrant, not the third.