Trig identities.

  • Thread starter Cephalopod
  • Start date
  • #1
Hi, I'm confused about using trig identities.

Homework Statement



Match the trigonometric function with one of the following: (a) -1, (b) cos(x), (c) cotx (d) 1, (e) -tan(x), (f) sin(x)

(1-cos^2x)(cscx)

Homework Equations



None that I know of.

The Attempt at a Solution



I multiply it through, which gives me:

csc(x) - cos^2(x)(cscx)

I divide out csc(x) which gives me:

csc(x)(1 - cos^2(x)(1))

(got me nowhere really)

edit: I just realized that I can do

1-cos^2(x)=sin^2(x)

edit2: am I wrong in thinking that since cosecant is the reciprocal of sine, that in csc(sin^2x) one sine cancels out, leaving me with sin(x)?

I might of just solved my own problem :P can anybody confirm? Thanks
 
Last edited:

Answers and Replies

  • #2
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Yes, that looks right. Good job.
 
  • #3
HallsofIvy
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My first thought with something like that would be to write everything in terms of sine and cosine. Here csc(x)= 1/sin(x) so that problem is (1- cos2(x)/sin(x)= 1/sin(x)- cos2(x)/sin(x). But cos2(x)= 1- sin2(x) so that second fraction is (1- sin2(x))/sin(x)= 1/sin(x)- sin(x).
1/sin(x)- cos2(x)/sin(x)= 1/sin(x)- 1/sin(x)+ sin(x)= sin(x), just as you say.
 

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