trig

Hi there! I NEED SERIOUS HELP, PLEASE!!! i have such a hard time with verifying identities! The question is:

[(sin(theta/2)) / csc(theta/2)] + [(cos (theta/2) / sec(theta/2)] = 1

I have a few ideas on how to solve this, but am mainly not sure how to get rid of (theta/2). I am taking trig online and use professor notes and the textbook, in addition to google searches.

I would normally try to add both fractions if they both had the same denominator, but am unsure how to find a common denominator in this question. Do I even need to be finding a common denominator??



If I am shown how to get rid of (theta/2) to make "2sin theta", or somehow "sin^2 theta", I would then do [(1/csc theta) / (1/sin theta)] + [(1/sec theta) / (1/cos theta)] .... which even then, I am not sure if that will get me anywhere.

OR

Do I cross multiply numerators together and denominators together?? Making it
[(sin(theta/2)cos(theta/2)] / [(csc(theta/2)sec(theta/2)] ?

***I would mainly appreciate a formula to mimic, and any form of help is GREATLY appreciated it. Thank you for your time and help in advance!

Stephani

asked by stephani
  1. While this might appear to be an exercise in half-angle or double-angle formulas, it is really just a check to see whether you remember your basic trig definitions:

    cscθ = 1/sinθ and secθ = 1/cosθ
    You happen to be using θ/2, but the principle holds.
    so, since cscθ=1/sinθ, 1/cscθ = sinθ

    [(sin(θ/2)) / csc(θ/2)] + [(cos (θ/2) / sec(θ/2)] = 1
    [sin(θ/2) * sin(θ/2)] + cos(θ/2) * cos(θ/2)] = 1
    sin^2(θ/2) + cos^2(θ/2) = 1
    as we all know, this is true.

    posted by Steve

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