## sierraurb one year ago PLEASE HELP!! A rain gutter is to be constructed of aluminum sheets 12 inches wide. After marking off a length of 4 inches from each edge, this length is bent up at an angle Θ. The area A of the opening may be expressed as the function: A(Θ) = 16 sin Θ • (cos Θ + 1). If Θ = 90°, what is the area of the opening? When i plug 90 in i keep getting 15, but my choices are: 21.8 in2 16.0 in2 7.9 in2 1.0 in2

1. sierraurb

@dangerousjesse any luck?

2. sierraurb

@phi @vineeth10

3. sierraurb

@ganeshie8 @MATHMALE

4. phi

first, you can do this without a calculator. But if you do use a calculator, be sure it is in degree mode. you should(!) know sin(90º)= 1 and cos(90º)= 0

5. sierraurb

@phi i just figured it out actually!! C right? and could you please help me with one more!! im stuck on it :/

6. sierraurb

Simplify $\sqrt{\left( 1-\sin \Theta \right)\left( 1+\sin \Theta \right)}$

7. sierraurb

@phi i have no idea what to do

8. phi

First, How did you get C 7.9? It seems if sin(90) is 1 and cos(90) is 0 you would do 16 sin Θ • (cos Θ + 1) 16 • 1 •(0+1) and now we add 0+1 (not *too* hard) and multiply that by 16*1 you don't get 7.9

9. sierraurb

For some reason i divided 16 in 2 idk why haha thank you!!

10. phi

11. sierraurb

b right?

12. sierraurb

any idea on how to solve the second one? i really have no idea what to do @phi

13. phi

do you know how to multiply the two binomials inside the square root (some people use FOIL)

14. phi
15. sierraurb

yeah when i foil i get sqrt(1-sin O)

16. sierraurb

is that right? @phi

17. phi

18. sierraurb

F: 1*1=1 O: 1*sinO=sinO I: -sinO*1=-sinO L: -sinO*sinO=-sin2 so its sqrt(1-sin^2) oops!

19. sierraurb

these are my options btw @phi ±sin Θ |cos Θ| ±tan Θ square root sine theta

20. phi

ok looks better there is a *very useful* identity you should remember for trig problems $\sin^2 x + \cos^2 x = 1$ which you can rewrite as $\sin^2 x = 1 - \cos^2 x$ or $\cos^2 x= 1 - \sin^2 x$

21. sierraurb

would it be B?

22. phi

yes. replace 1 -sin^2 with cos^2 inside the square root sqrt(cos^2) = cos is the only choice that looks close (they must only want the positive root) | cos Θ |

23. sierraurb

@phi thank you so so much!! i really truly appreciate it :)

24. phi

yw