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Given sinθ = -2/3, π, < θ < 3π/2, find the values of the remaining trigonometric functions of θ. Help please! We didn't go over this in class & this is all foreign for me.
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|dw:1350472777669:dw|Now solve by ur selves
How did you get -sqrt (5)? I don't understand the concept, are we trying to find reference angles?
\[\sin(\theta)=\frac{ opposite }{ hypotenus }=\frac{ -2 }{ 3 }\]
Use Pythagoras theorem
OH! I kinda get it! So it's like SOH CAH TOA, right? & we're looking for CAH & TOA?
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The theorem is x^2 + y^2 = 1 or the A^2 + B^2 = C^2, I'm mixing up some stuff.
\[(-2)^{2}+m^{2}=3^{2}...thus...m=\sqrt{5}\] But m is -ve since theta is third quadrant angle.
|dw:1350473502023:dw|Now can u do that?
I get it better!
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