Few questions on proving identities

  • Thread starter Thread starter VanKwisH
  • Start date Start date
  • Tags Tags
    identities
Click For Summary

Homework Help Overview

The discussion revolves around proving a trigonometric identity involving secant and cosecant functions. The original poster presents a fraction that combines these functions and seeks guidance on how to prove the identity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants suggest breaking down the left and right sides into sine and cosine components. There are discussions about cross-multiplying and simplifying the expressions, with some participants questioning the correctness of their steps and interpretations of cross-multiplication.

Discussion Status

Participants are actively engaging with the problem, providing guidance on how to manipulate the expressions. There is a recognition of confusion regarding the process of cross-multiplication, and some participants are clarifying their understanding of the method. Multiple interpretations of the steps are being explored.

Contextual Notes

There is an indication that participants are working under the constraints of homework rules, which may limit the extent of guidance provided. Some assumptions about the definitions and operations involved in trigonometric identities are being questioned.

VanKwisH
Messages
107
Reaction score
0

Homework Statement



seca + csca
--------------- = seca*csca
sina + cosa

Homework Equations



no relevant equations

The Attempt at a Solution



How would i prove this ?? i know i break down the left side...
which is

1/cosa + 1/sina
---------------------
sina + cosa
 
Physics news on Phys.org
Now break down the right side to sines and cosines. From there, cross multiply the denominator of your right side to the left, from there it will simplify to 1.
 
on the right side it's 1/cosa * 1/sina = 1 /cosasina right??
 
Yes, that's correct. Now cross multiply.
 
hmmm i did cross multiply and then i got .....
((cosa*sina /cosa ) + ( cosa*sina /sina)) / sina+cosa for the left side
and then for the right side i got sina+cosa / cosa*sina for the right side ... is that right??
 
Yes, that's correct. Now simplify and L=R.
 
alright for the left side i was able to get one but for the right side it doesn't seem right ...

sina + cosa / cosa*sina ... that doesn't equal 1 does it.?
 
Sorry, all the writing confused me. The right side should only be sina+cosa, right? B/c you cross multiplied and so sinacosa disappears from the right.
 
OOOOOOOOOOOO alright ... but that doesn't equal 1 though ... only sin^2a + cos^2 a = 1
 
  • #10
O wait but since i cross multiplied the denominator on the left side is gone as well right??
 
  • #11
VanKwisH said:
hmmm i did cross multiply and then i got .....
((cosa*sina /cosa ) + ( cosa*sina /sina)) / sina+cosa for the left side
and then for the right side i got sina+cosa / cosa*sina for the right side ... is that right??
I'm not sure what you mean by "cross multiply". To me, it means multiply both sides by the common denominator: here, that is (sin a cos a)(sin a+ cos a). You seem to have multiplied one side by (sin a+ cos a) and the other by (sin a cos a). Of course, that will not give you the same thing- you are multiplying the two sides by different things.
 
  • #12
HallsofIvy said:
I'm not sure what you mean by "cross multiply". To me, it means multiply both sides by the common denominator: here, that is (sin a cos a)(sin a+ cos a). You seem to have multiplied one side by (sin a+ cos a) and the other by (sin a cos a). Of course, that will not give you the same thing- you are multiplying the two sides by different things.


yahhh ... i messed up but i fixed it now and i got it ...
but i think cross multiplying is different for example

sinx cosx
------ ---------
cosx sinx

from what i know ... u would multiply the sinx on top by the one on the bottom ...
and the cosx by the cosx ... therefore u would have sin^2x and cos^2x ...
that's how i interpret it...unlesss i am completely wrong ...
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
13K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
25
Views
4K