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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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Mar 20th, 2002, 05:36 PM
#2
Hyperactive Member
Re: can u solve this in less than 8 steps? double angle trig Identities
(Sec[x])^2 + (Sec[x])^4
----------------------------
2 + (Sec[x])^2 - (Sec[x])^4
//--- Step 1 ---
= (Sec[x])^2 * [(Sec[x])^2 + 1]
-----------------------------------
[(Sec[x])^2 + 1] - [(Sec[x])^4 - 1]
//--- Step 2 ---
Dividing Both Numerator and Denominator by [(Sec[x])^2 + 1]
= [1 + (tan[x])^2] * 1
-------------------------
1 - [(Sec[x])^2 - 1]
//--- Step 3 ---
= [1 + (tan[x])^2]
---------------------
[1 - (tan[x])^2]
//--- Step 4 ---
Multiplying Both Numerator and Denominator by (Cos[x])^2
= (Cos[x])^2 + (Sin[x])^2
-----------------------------
(Cos[x])^2 - (Sin[x])^2
//--- Step 5 ---
= 1/Cos[2x]
//--- Step 6 ---
= Sec[2x]
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Mar 21st, 2002, 01:03 AM
#3
PowerPoster
Step 1
Scroll down
Step 2
Copy
Step 3
Paste
Sec[2x]
done !
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Mar 21st, 2002, 05:29 AM
#4
Step 1
Like BB Said!
Done!!!
Beat THAT!
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Mar 21st, 2002, 07:24 PM
#5
Thread Starter
Registered User
hm...
Think tank, thanks for the responce, but i noticed u solved the proof as if it where
(Sec[x])^2 + (Sec[x])^4
----------------------------
2 + (Sec[x])^2 - (Sec[x])^4
but it really is
sec^2x + sec^4x
-----------------------
2 + sec^2x - sec^4x
or is that also correct becuasde u wrote it with the ( ) 's?
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Mar 25th, 2002, 05:07 PM
#6
Addicted Member
Well, it depends on what do you consider as a step. I didn't follow your proof but I think a proof with the steps detailed is this:
//Step 1// multiply by [sec(x)]^-4
1+[sec(x)]^-2
----------------------------------------
2[sec(x)]^-4 + [sec(x)]^-2 - 1
//Step 2// convert sec to cos
1 + [cos(x)]^2
--------------------------------------------
2 [[cos(x)]^2]^2 + [cos(x)]^2 - 1
//Step 3// factor denominator 2[w]^2 + w - 1 == (1+w)(2w-1)
1
---------------------
2[cos(x)]^2 - 1
//Step 4// known identity : 2[cos(x)]^2 - 1 = cos(2x)
1
--------
cos(2x)
Done !
Actually, copying & pasting is faster, but.....
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