Any help? I missed these on a quiz today :(

[atac57]

1) If sin theta = -3/5 and pi < theta < 5pi/2,
compute cos theta and tan theta.

2) Prove sinx tanx = (1 - cos^2x)/cosx


For 1), it looks like theta will be in Quadrant 3, I just don't know how to get cos theta. Once you get that, we know what tan theta will be.

For 2) for the LHS, I got: (sinx cosx)/cosx * sinx/cosx and then I got stuck. I multiplied sinx/1 by cosx to get a common denominator and then tanx is just sinx/cosx.

I'm stumped, thank you!

 

[HallsofIvy]

Are you sure you copied (1) correctly? No, pi< theta< 5pi/2 is NOT "in Quadrant 3". That would be pi< theta< 3pi/2. pi< theta< 5pi/2 includes quadrants 3, 4, and 1. If you meant pi< theta< 3pi/2, then cos(theta) is positive and tan(theta) is negative. Since sin^2(\theta)+ cos^2(\theta)= 1, with sin(theta)= -3/5, then 9/25+ cos^2(theta)= 1. So what is cos^2(theta)? And, knowing that it is negative, what is cos(theta)? For (2), tan(x)= sin(x)/cos(x) so sin(x)tan(x)= sin^2(x)/cos(x). Now, what can you do with sin^2(x)?

 

[atac57]

My mistake, that was supposed to be pi < theta < 3pi/2.

1) So I did (-3/5) + cos^2(theta) = 1. Solved for cos^2(theta) = 1 - 9/25 => cos^2(theta) = 16/25. Square root that to cos(theta) = 4/5 and tan (theta) = -3/4. Is that correct?

2) I'm not sure how you went from sinx tanx to sin^2x/cosx.

 

[stapel]

2) I'm not sure how you went from sinx tanx to sin^2x/cosx.

Restate the tangent in terms of sines and cosines. See what you get.