Pre Calculus
posted by Peter .
Establish the identity
(cosx/2sinx/2)^2=1sin x

Pre Calculus 
Reiny
Hint:
sin 2A = 2sinAcosA
treat x/2 as A
Respond to this Question
Similar Questions

Trigonometry.
( tanx/1cotx )+ (cotx/1tanx)= (1+secxcscx) Good one! Generally these are done by changing everything to sines and cosines, unless you see some obvious identities. Also generally, it is best to start with the more complicated side … 
Trig........
I need to prove that the following is true. Thanks (cosx / 1sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds of problems when I was still teaching. it requires a little "trick" L.S. =cosx/(1sinx) multiply top and … 
PreCalc
Establish the identity. sinx + cosx/sinx  cosx = 1+2sinxcosx/2sin^2x1 
Math  Pre Clac
Prove that each of these equations is an identity. A) (1 + sinx + cos x)/(1 + sinx + cosx)=(1 + cosx)/sinx B) (1 + sinx + cosx)/(1  sinx + cosx)= (1 + sin x)/cosx Please and thankyou! 
Pre Calculus
Establish the following identity sin 2x/sinx  cos2x/cosx = secx 
Trigonometry Check
Simplify #3: [cosxsin(90x)sinx]/[cosxcos(180x)tanx] = [cosx(sin90cosxcos90sinx)sinx]/[cosx(cos180cosx+sinx180sinx)tanx] = [cosx((1)cosx(0)sinx)sinx]/[cosx((1)cosx+(0)sinx)tanx] = [cosxcosxsinx]/[cosx+cosxtanx] = [cosx(1sinx]/[cosx(1+tanx] … 
PreCalculus
How would you prove/Verify this Identity. x=theta (cosx+sinxsin^3x)/(sinx)=cotx+cos^2x 
PreCalc : Verify the Identity
Verify the Identity: sin(x+π)/cos(x+3π/2) =tan^2xsec^2x I've done: sinxcosπ+cosxsinπ / cosxcos(3π/2)  sinxsin(3π/2) sinx(1) + cosx(0) / cosx(0) sinx(1) sinx/sinx What do I do from here? 
Precalculus/Trig
I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1  cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1cosx Simplified: cosx + sin^3x/sin^3x = cscx/1cosx I don't know … 
Calculus 2 Trigonometric Substitution
I'm working this problem: ∫ [1tan^2 (x)] / [sec^2 (x)] dx ∫(1/secx)[(sin^2x/cos^2x)/(1/cosx) ∫cosxsinx(sinx/cosx) ∫cosx∫sin^2(x)/cosx sinx∫(1cos^2(x))/cosx sinx∫(1/cosx)cosx sinx∫secx∫cosx …