I was able to do this before but can't follow my steps.
Some help would be awesome!
Thank you so much!
2007-12-26 15:36:56 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Hi,
cos(2x) = 2 cos² x - 1 so
cos(2x ) secx = (2 cos² x - 1)sec x =
2 cos² x - 1
---------------- =
......cos x
2 cos² x.........1
-----------..-..-------- =
cos.x..........cos.x
2 cos x.........1
-----------..-..-------- =
.....1............cos.x
2 cos x - sec x
I hope that helps!! :-)
2007-12-26 15:50:57 · answer #1 · answered by Pi R Squared 7 · 0⤊ 1⤋
Cos2x= 2 cos2x – 1
Therefore cos2xsecx
= (2 cos2x – 1) secx
= (2 cos2x – 1) 1/cosx
= 2 cos2x x 1/cosx – 1/cosx
= 2cosx – 1/cosx
= 2cosx - secx
2007-12-26 18:20:30 · answer #2 · answered by KJ_Jockey 2 · 0⤊ 0⤋
cos2x = 2cos^2x - 1 , secx = 1 / cosx
so
[2cos^2(x) - 1] / cosx = 2cosx - secx
2cos^2(x) / cosx] - ( 1/cosx) = 2cosx - secx
2cosx - secx = 2cosx - secx
2007-12-26 15:50:04 · answer #3 · answered by LE THANH TAM 5 · 0⤊ 0⤋
Proving?
cos(2x) secx = 2cosx - secx
cos(2x) secx = 2cosx - 1/cosx
cos(2x) secx = 2cos^2x -1 / cosx (note: 2cos^2x-1 = cos2x)
cos(2x) secx = cos2x / cosx (note: 1/cosx = secx)
cos(2x) secx = cos(2x) secx
2007-12-26 15:41:42 · answer #4 · answered by ¿ /\/ 馬 ? 7 · 0⤊ 1⤋
You need to use:
Cos(2x)=Cos^2x-Sin^2x=2Cos^2x-1
since sin^2x+cos^2x=1
secx=1/cosx
So, the left hand side becomes:
[2cos^2x-1]cosx=
2cosx-1/cosx=2cosx - sec x
2007-12-26 15:53:21 · answer #5 · answered by kuiperbelt2003 7 · 0⤊ 0⤋
cos(2x)secx
= (2cos^2x -- 1) / cosx
= 2cosx -- secx
2007-12-26 15:40:30 · answer #6 · answered by sv 7 · 0⤊ 1⤋
cos2(x)secx â¡ 2cosx - secx
Taking LHS
cos(2x) = 2cos²x - 1
sec(x) = 1/cos(x)
2cos²(x)-1/cosx
2cos²(x)/cos(x) - 1/cos(x)
2cos(x) - sec(x) (Proved)
2007-12-26 15:51:55 · answer #7 · answered by Murtaza 6 · 0⤊ 0⤋
LHS
(2 cos ² x - 1) / (cos x)
2 cos x - 1 / cos x
2 cos x - sec x
RHS
2 cos x - sec x
LHS = RHS
2007-12-30 05:40:05 · answer #8 · answered by Como 7 · 1⤊ 0⤋