ratios of double angles (1 Viewer)

[Jkitty]

Q. simplify each of the following in terms of the indicated angle:

j. 1-cos(theta)/1+cos(theta);theta/2

I have tried squaring it and other methods but im not getting the answer tan^2 (theta)/2


Thank you

 

[Librah]

Q. simplify each of the following in terms of the indicated angle:

j. 1-cos(theta)/1+cos(theta);theta/2

I have tried squaring it and other methods but im not getting the answer tan^2 (theta)/2


Thank you

are you saying 1-cos0 OVER (fraction) on 1+cos0.

 

[Jkitty]

Umm using 0 as theta here. Isn't it just 1-cos(0/2)sin(0/2) and 1+cos(0/2)sin(0/2). Wait are you saying 1-cos0 OVER (fraction) on 1+cos0.

fraction

 

[Librah]

Ah oops i mistook sin double angle for cos ^_^

 

[Librah]

Yer ok i got it. 1-cos0 / 1+cos0 = 1- (cos^2(0/2) - sin^2(0/2) ) / 1+(cos^2(0/2) - sin^2(0/2) ) = 1- (1-2sin^2(0/2) ) / 1+ (2cos^2(0/2) - 1)

 

[hypermax]

the question isn't hard. Just do the double angles for cos(theta/2) which is 1-2sin^2(theta/2)

you should get 1-(1-2sin^2(theta/2)/1+1-2sin^2(theta/2)=2sin^2(theta/2)/2-2sin^2(theta/2). Clean it up factor the 2 out you get

sin^2(theta/2)/cos^2(theta/2)

 

[Jkitty]

thanks

 

[Drongoski]

I always remember:



which is often treated as a standard identity.

 

[dan964]

You could use 't' results instead, which simplifies down to 2t^2/2