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Mar 19th, 2002, 10:45 PM
#1
Thread Starter
Registered User
can u solve this in less than 8 steps? double angle trig Identities
hello everyone!
i solved this trig identity but it took 8 steps! and i was thinking htere had to be an easier way and would like to find out...
here is what i did...
sec2x = sec^2x + sec^4x
----------------------
2 + sec^2x - sec%^4x
= sec^2x(1 + sec^2x)
-----------------------------
(2-sec^2x)(1+sec^2x)
= sec^2x
-----------
2-sec^2x
= 1/cos^2x
------------
2-1/cos^2x
= 1/cos^2x
---------------------------
2cos^2x - 1
-----------------
cos^2x
(1/cos2x)(cos^2x/2cos^2x -1)
= 1
-----------
2 cos^2x-1 //double angle, cos2x = 2cos^2x-1
= 1
--------
cos2x
= sec2x
if any of u can do it a differnet and faster way post it, thanks~!
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