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.
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
1 answer