Pre-Calc Homework Help: Simplify (sec^2 x)(csc x)/(csc^2 x)(sec x)"

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Homework Help Overview

The discussion revolves around simplifying a trigonometric expression involving secant and cosecant functions. The original poster presents the expression (sec^2 x)(csc x)/(csc^2 x)(sec x) and seeks assistance in simplifying it.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore simplification techniques, including cancellations and relationships between trigonometric functions. Some express uncertainty about their direction and seek validation for their reasoning.

Discussion Status

Several participants have offered insights into potential simplifications and relationships between secant and cosecant. There is a mix of attempts to clarify the expression and explore further simplifications, with no explicit consensus reached on the final form.

Contextual Notes

Participants mention challenges with understanding the problem due to teaching quality and express confusion about the application of trigonometric identities in different contexts.

math_fortress
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Simplify the given expression:
2) (sec^2 x)(csc x)/(csc^2 x)(sec x)
 
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What have you tried?
 
well...i've done this:

(tan^2 x + 1)(csc x)/(cot^2 x + 1)(sec x)

but...i don't know if I'm going in the right direction, for my teacher is horrible, and i don't know where to go from here if i am going in the right direction

any help would be greatly appreciated
 
OK: Let's make the EASIEST cancellations first:
If you look at the expression like this:
[tex]\frac{sec^{2}(x)csc(x)}{sec(x)csc^{2}(x)}[/tex]

isn't there a couple of cancellations which immediately spring to your mind?
 
so is it (sec x)/(csc x) ?
 
math_fortress said:
so is it (sec x)/(csc x) ?
Precisely!
Now, knowing the relation between sec and cos and csc and sin, can you simplify even further?
 
well, since sin/cos = tan...then would sec/csc = 1/tan ?

That's all i can think of
 
No, you have:
[tex]\frac{\frac{1}{\cos(x)}}{\frac{1}{\sin(x)}}=\frac{1}{\cos(x)}\frac{1}{\frac{1}{\sin(x)}}=\frac{\sin(x)}{\cos(x)}=tan(x)[/tex]
 
alright...thanks, I'm starting to get it a little better...
 
  • #10
Wait...this baffles me...

26.) Find the exact value of sin 5pi/12

Is there any way to do this logically w/ a calculator or anything?
 
  • #11
Of course there is! (hopefully you meant without a calculator) That's why the angle is given to you in radians, as a rational multiple of [itex]\pi[/itex].

Draw the unit circle: what coordinate points do certain angles represent? [itex]\pi, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}[/itex] etc.
 
  • #12
but 5pi/12 isn't on my unit circle...the one's you listed are though...i just don't get how exactly you can find 5pi/12 with information of pi/2, etc...
 
  • #13
would 1/4 make sense for the answer since i got the sin of 5pi/6 to equal 1/2?
 
Last edited:
  • #14
Try to do this using the trig identity
[tex]\cos{2a}=1-2\sin^{2}{a}[/tex]
 
  • #15
What? So you're saying 1/4 isn't right then?
Do I even need to use trig identities for this type of question?

(I'm not arguing with that identity..i'm just very confused :bugeye:)
 
Last edited:
  • #16
1/4 is incorrect. Sometimes identities are necessary.
 
  • #17
Yeah, sorry if I gave you the wrong idea. 1/4 is incorrect. You are making the assumption that if I halve the angle, I halve the sine. You can see why that would only be true for a linear relationship right? (which sine is not). I think Sirus has the right technique, since the trig identity involves a term with twice the angle and another with just the angle itself. We know how to work with [itex]\frac{\pi}{6}[/itex], and multiples of it, so find the cosine of the angle [itex]\frac{\5pi}{6}[/itex] and work from there.
 
  • #18
Another way is to convert the radians to degrees and work from there:

sin5pi/12=sin5(180)/12=sin75=sin(30+45)

now you can just use the formula for the sum of angles:
sin (a+b) = cos(b) sin(a) + sin(b) cos(a)
 
  • #19
use the sum formulas of trig:
sin (a+b)=sin(a) cos(b)+cos(a) sin (b)

now, sin 5pi/12=sin (2pi/12 + 3pi/12), so the exact value of sin 5pi/12 is?
 
  • #20
alright...root 2/4 + root 6/4


Thanks for all help
 

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