Answer check from quiz

[jenn9580]

I took a quiz today & i'm not confident in my answer & I ran out of time.
cos(theta)=-4/9 in III quadrant. Find sin(theta).

x=-4 x^2 + y^2 = r^2
y=-(square root 65) 16 + y^2 = 81
r=9 y^2 = 65
y = square root 65
My answer: sin(theta) = -sq root 65/9

Am i correct? thanks

 

[Mrspi]

I took a quiz today & i'm not confident in my answer & I ran out of time.
cos(theta)=-4/9 in III quadrant. Find sin(theta).

x=-4 x^2 + y^2 = r^2
y=-(square root 65) 16 + y^2 = 81
r=9 y^2 = 65
y = square root 65
My answer: sin(theta) = -sq root 65/9

Am i correct? thanks

Yes, you are correct. I had a bit of trouble in the beginning deciphering what you had done. So, I'll show you how I would have approached this problem.

I guess I would go with the Pythagorean Identity:

sin^2 x + cos^2 x = 1

You know that cos x = -4/9, and that x (I'm using that instead of theta....much easier to type!) is a third-quadrant angle. If x is a third-quadrant angle, then sin x < 0 also.

Substitute the value you are given for cos x. cos x = -4/9

sin^2 x + (-4/9)^2 = 1

sin^2 x + (16/81) = 1

sin^2 x = 1 - (16/81)

sin^2 x = 65/81

Take the square root of both sides, and remember that since x terminates in quadrant III, sin x < 0.

So,

sin x = (-sqrt(65)) / 9

 

[jenn9580]

thanks! my biggest problem is my confidence. I can rest easy for a few days!