I need some help with these two calculator trig problems

In summary, the first problem involves finding the value of 2A given that tan 2A = cot 40 = tan 50. The second problem involves using the definition of cosec x as 1/sin x to solve for the value of x. By simplifying the equation, it can be concluded that x lies in all four quadrants, with a value of 90 degrees.
  • #1
omicron
46
0
I need some help with these two questions:
[tex]tan2A=cot40[/tex](40degrees)
and
[tex]cosec^2x=1[/tex]
 
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  • #2
omicron said:
I need some help with these two questions:
[tex]tan2A=cot40[/tex](40degrees)
and
[tex]cosec^2x=1[/tex]

Exactly what help do you need?
 
  • #3
Can you use your calculator (which looks to me like the only way to do the first one)?
If you can, use to find cot (40) and then to find 2A.


As for the second problem, do you know that cosec x is defined as 1/sin x?

If cosec^2 x= 1, what is sin^2 x?
 
  • #4
Exactly what help do you need?
Doing the question. I just don't know.

How do u find cot(40)? Don't u have to change it to [tex]\frac{1}{tan\theta}[/tex] or something like that?

As for the second problem, do you know that cosec x is defined as 1/sin x?
So are u saying that [tex]cosec^2x=1[/tex] can be also written as [tex]\frac{1}{sin^2x} = 1[/tex]
 
  • #5
omicron said:
Doing the question. I just don't know.

How do u find cot(40)? Don't u have to change it to [tex]\frac{1}{tan\theta}[/tex] or something like that?


So are u saying that [tex]cosec^2x=1[/tex] can be also written as [tex]\frac{1}{sin^2x} = 1[/tex]

Yup. And then what can you conclude from that, concerning x?
 
  • #6
It is in the 1st and 2nd quadrants?
 
  • #7
omicron said:
I need some help with these two questions:
[tex]tan2A=cot40[/tex](40degrees)
and
[tex]cosec^2x=1[/tex]

tan 2A = cot40 =tan 50
=> 2A = [50 +n(180)]degrees
=>A=25+90n deg

cosec^2 x = 1 =>sinx=+-1=>x=180n+(-1)^n *(+-90) deg

:bugeye:
 
  • #8
For the 2nd question...

[tex] cosec^2x=1 [/tex] can be expressed as,

[tex] 1/sin^2x=1 [/tex] while multiplying [tex] sin^2x [/tex] both sides, we have,

[tex] 1=sin^2x [/tex] and by square rooting both sides, we now have,

[tex] sinx= \pm1 [/tex]

and so, since sin x is both positive and negative, it must lie in all quadrants with alpha 90 degrees.
 
Last edited:
  • #9
Oh so now i know. I didn't know u could [tex]\sqrt{sin^2x}[/tex]. Thanks to everyone that helped. :smile:
 

1. What are the two calculator trig problems?

The two calculator trig problems are specific mathematical equations or questions that involve trigonometric functions (such as sine, cosine, and tangent) that require the use of a calculator to solve.

2. Why do I need help with these two calculator trig problems?

Trigonometry can be a challenging subject for many students, and using a calculator to solve problems can add an extra layer of difficulty. It is common for students to need help understanding concepts and techniques for solving calculator trig problems.

3. How can I solve these two calculator trig problems?

The best way to solve calculator trig problems is to first make sure you understand the concepts and formulas involved. Then, use your calculator to input the numbers and functions correctly. It can also be helpful to double check your work and use different methods or approaches if you are having trouble.

4. Are there any tips or tricks for solving calculator trig problems?

Yes, there are various tips and tricks that can help with calculator trig problems. Some include using the correct mode (degrees or radians) on your calculator, being familiar with common trig identities, and practicing with similar problems to build your skills and confidence.

5. Where can I find additional resources for help with these two calculator trig problems?

There are many online resources available for help with calculator trig problems, such as tutorial videos, practice problems, and interactive tools. You can also seek assistance from a tutor, teacher, or classmate for further guidance and clarification.

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