anonymous
  • anonymous
what is the domain of validity for the trigonometric identity cscθ=1/sinθ HELP PLEASE
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@jim_thompson5910 @amistre64 I've been stuck on this problem for so long and I really need the answer to this and a few others!
anonymous
  • anonymous
I don't know @jim_thompson5910
anonymous
  • anonymous
sin? theta? i'm really not sure

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
0? I don't know. There aren't numbers in the equation..
anonymous
  • anonymous
I don't know how to do that.. Sorry, I've been doing math homework all day and my brain is kind of fried, lol.
anonymous
  • anonymous
I'm sorry, I don't know how to write the problem and do it in order to solve it. I suck at math.
anonymous
  • anonymous
No
anonymous
  • anonymous
my computer isn't loading the page
anonymous
  • anonymous
all real numbers except multiples of pi, because that's when the denominator would equal 0 (csc = 1/sin)
anonymous
  • anonymous
thank you :D @mangorox
anonymous
  • anonymous
No problem @icedancerfigureskater
anonymous
  • anonymous
Can I have help with just a few more? @mangorox
anonymous
  • anonymous
I'll try to help
anonymous
  • anonymous
Ok. What is the radian measure of an angle whose tangent is -0.25? My answers are: -1.33, 0.24, 2.9, and 4.95
anonymous
  • anonymous
@mangorox
anonymous
  • anonymous
Ok let me see...
anonymous
  • anonymous
i'm sorry I don't know the answer to this question :(
anonymous
  • anonymous
It's okay, do you think you can help me with a question using unit circles and inverse functions? @mangorox
anonymous
  • anonymous
I can try, it depends on the question.
anonymous
  • anonymous
Okay. Use the unit circle to find the inverse function value in degrees. cos^-1 (sqrt3/2)
anonymous
  • anonymous
Does this help?: the unit circle can help you find the correct quadrants, but you have to memorize the actual degrees. You should know the sin, cos and tan of 0º, 30º, 45º, 60º, and 90º there is a pattern for sin: 0 1/2 sqrt(2)/2 sqrt(3)/2 1 if we use sqrt(0) is 0 and sqrt(1) is 1 and the sqrt(4) is 2, the pattern is sqrt(0)/2 sqrt(1)/2 sqrt(2)/2 sqrt(3)/2 sqrt(4)/2 cosine goes in the other direction tangent is sin/cos
anonymous
  • anonymous
Kind of, but I don't know how to get my answer
anonymous
  • anonymous
ok what is the sin(60º) ? the answer is : memorize it. See above. it is a number between -1 and 1. In this case 0.866025403... and on and on rather than type all the numbers people give the equivalent sqrt(3)/2 type sin(60 degrees)= in the google search window. It will give you that number type sqrt(3)/2= in the google search window. It will give the same number
anonymous
  • anonymous
This might help also: http://www.khanacademy.org/math/trigonometry/v/basic-trigonometry
anonymous
  • anonymous
the answer they want (I am assuming) is sin−1(3√2)=60º and 120º
anonymous
  • anonymous
the answer is 60 degrees, thank you.
anonymous
  • anonymous
No problem!
anonymous
  • anonymous
Do you know how to solve this one? @mangorox Simplify the trigonometric expression. sin^2 θ /1-cosθ
anonymous
  • anonymous
Sorry, I don't know how to solve that one
anonymous
  • anonymous
Ok, that's fine.
anonymous
  • anonymous
17. Verify the pythagorean identity. 1+cot^2 θ =csc^2 θ @DLe @mangorox @jim_thompson5910 @amistre64 @dan815
anonymous
  • anonymous
cot^2 θ = (cos^2 θ) / (sin^2 θ) so, 1 + (cos^2 θ) / (sin^2 θ) = [sin^2 θ + cos^2 θ] / (sin^2 θ) = 1 / (sin^2 θ) = csc^2 θ qed
anonymous
  • anonymous
Thank you! @mangorox I only have one more that I need help with if you can. 18. Verify the identity. tan θ + cot θ = 1/ sinθ cosθ
anonymous
  • anonymous
tan A + cot A =?= 1/(sinA cosA) tan A = sinA / cosA cot A = cosA / sinA So, sinA/cosA + cosA/sinA = 1/(sinAcosA) multiply both sides by sinAcosA sinAcosA (sinA/cosA) + sinAcosA(cosA/sinA) = 1 sin^A + cos^2A = 1
anonymous
  • anonymous
Thank you so very much!!! @mangorox
anonymous
  • anonymous
No problem! :) @icedancerfigureskater

Looking for something else?

Not the answer you are looking for? Search for more explanations.