I need some help with these two calculator trig problems

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    Calculator Trig
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Discussion Overview

The discussion revolves around solving two trigonometric problems: one involving the equation tan2A = cot40 and the other concerning cosec^2x = 1. The scope includes mathematical reasoning and problem-solving techniques related to trigonometric identities and functions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Some participants suggest using a calculator to find cot(40) in order to solve for 2A.
  • There is a discussion about the definition of cosec x as 1/sin x and its implications for solving cosec^2x = 1.
  • One participant proposes that if cosec^2x = 1, then sin^2x must also equal 1, leading to potential values for x.
  • Another participant questions how to find cot(40) and discusses the relationship between cotangent and tangent.
  • It is noted that sin x can take both positive and negative values, suggesting that x could lie in multiple quadrants.
  • A later reply confirms the ability to take the square root of sin^2x, expressing gratitude for the clarification.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and approaches to the problems, with some agreeing on the definitions and methods while others seek clarification. No consensus is reached on the solutions to the problems.

Contextual Notes

There are unresolved assumptions regarding the use of calculators and the interpretation of trigonometric identities. The discussion does not clarify all mathematical steps involved in reaching conclusions.

omicron
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I need some help with these two questions:
[tex]tan2A=cot40[/tex](40degrees)
and
[tex]cosec^2x=1[/tex]
 
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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?
 
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?
 
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]
 
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?
 
It is in the 1st and 2nd quadrants?
 
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:
 
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:
Oh so now i know. I didn't know u could [tex]\sqrt{sin^2x}[/tex]. Thanks to everyone that helped. :smile:
 

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