really hard verifying trig identity.

spillthestars

New member
Joined
Apr 29, 2006
Messages
4
I'm having really bad trouble with a few trig identities..if anyone could help that'd be great.

Problem 1 -

2sinx cosx
---------------- = sinx + cosx + 1
sinx + cosx -1


Problem 2 -

sinx - cosx - 1 - cosx + 1
----------------- = ------------
sinx + cosx - 1 sinx

(Note the negative sign in the second fraction can be used for the bottom or the top)


I really appreciate this if anyone can help me out
 

Gene

Senior Member
Joined
Oct 8, 2003
Messages
1,904
1) multiply thr RHS by
(sin(x)+cos(x)-1)/(sin(x)+cos(x)-1)

2) your note is not correct unless you mean
Code:
sin(x) - cos(x) - 1   -(cos(x) + 1) 
-------------------- = ------------ 
sin(x) + cos(x) - 1        sinx
 

spillthestars

New member
Joined
Apr 29, 2006
Messages
4
I don't understand what you are saying sorry.
I don't understand the layout of what you wanted me to try.

SO i did that and got this for the right side, now I'm not sure where to go...


2sinx + 2cos -1
----------------
sinx + cosx - 1
 

Gene

Senior Member
Joined
Oct 8, 2003
Messages
1,904
When you do the multiplication the denominators become the same and the numerator goes
((sin(x)+cos(x))+1)* ((sin(x)+cos(x))-1)=
((sin(x)+cos(x))²-1 =
((sin²(x)+2sin(x)cos(x)+cos²(x))-1=
2sin(x)cos(x)

(The last step uses sin²+cos²=1)
 
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