Trig probs

[shaunjpeterson]

Im havin some serious problems with these questions:
1)√3 cot 3x + 1 = 0 where 0≤x≤2pie
2)lim sin^2 x/2
x→0 Sin x
3)Derivatives of y=2csc^3 (√x) and y= x/2 - sin2x/4
4)Derivatives using implicit differentation x=sin y + cosx and xy-y^3 = sinx
5)Find the solutions in the interval [0, 2pie] such that -1 + tan 2theta= 0
6) lim 1-tanx / sinx - cosx
x→ pie/4

 

[stapel]

Note: The Greek letter is "pi"; "pie" is something one eats.

1) Isolate the cosine. Use the basic reference-angle values you've memorized to figure out the angle values that 3x could be (but find all values between 0 and, say, 6pi). Then divide through by 3 to find x.

2) Sorry, but I can't read this. Please use the formatting suggestions found in the "Forum Help" pull-down menu at the very top of the page.

3) Where are you stuck in the differentiation process?

4) See (3).

5) See (1).

6) See (2).

When you reply, please include a clear statement of the steps you have tried thus far. Thank you.

Eliz.

 

[shaunjpeterson]

1) I get it to cotangent3x= √3/-3 and i fdont no what to do next.
2) Its a limit question what is the limit of (sin^2 x/2) / sinx as x nears 0. Im not sure what my first step should be?
3) I'm not sure what to do first turn csc^3 into something else or what?Do i cross multiply first for the second question?
4)Ive gotten the first one to 1= cos y * dy/dx + -sinx * dy/dx Is this right and then what to i do next.
5)Ove got it to tan 2theta = 1 now do i turn tan into something else or what.
6) what is the limit of 1-tanx/sinx - cosx as x nears pi/4. Do ichange tanx into sinx/cosx?

 

[stapel]

1) If cot(3x) = -1/sqrt(3), then tan(3x) = -sqrt(3). Then what does 3x equal?

2) What is the argument in the numerator? As written, it is "sin²(x)" divided by 2. Is this what you mean?

3) Do you not have differentiation rules for cosecants and secants? (Not all books present the same variety of rules, is why I ask.)

4) How did you get that the derivative, with respect to x, of "cos(x)" is "-(dy/dx)sin(x)"? I get the negative sine part, but where is the "dy/dx" coming from? (Without a clear listing of your steps, it's hard to follow your work.)

5) For what values of 2@ is tan(2@) equal to 1? Get a list of those values, and then divide through by 2 to find the solution values for @.

6) You have given the limit as being of 1 - tan(x)/sin(x) - cos(x). Since tan(x) = sin(x)/cos(x), then tan(x)/sin(x) simplifies as 1/cos(x), giving you 1 - 1/cos(x) - cos(x). Is this what you meant?

Eliz.