View Full Version : Need to verifying Trig identities
lwilli2
Mar 17, 2008, 03:50 PM
I'm stuck!
Can someone please help?
I need to verify?
tanx + cotx = 2tanx - cscx secx
and
(csc(squared)x ) (cos(squared)x) - cos(squared)x = -1
Any help is greatly appreciated Thanks
ISneezeFunny
Mar 17, 2008, 03:57 PM
Table of Trigonometric Identities (http://www.sosmath.com/trig/Trig5/trig5/trig5.html)
Hope that helps at all.
galactus
Mar 17, 2008, 04:03 PM
Recheck. Neither of those identities is valid.
#1:
tan(x)+cot(x)
=\frac{sin(x)}{cos(x)}+\frac{cos(x)}{sin(x)}
=\frac{sin^{2}(x)+cos^{2}(x)}{sin(x)cos(x)}
=\frac{1}{sin(x)cos(x)}
=sec(x)csc(x)
#2 is not true. Recheck and make sure you typed it correctly.
lwilli2
Mar 17, 2008, 04:09 PM
I'm stuck!
Can someone please help?
I need to verify?
tanx + cotx = 2tanx - cscx secx
and
(csc(squared)x ) (cos(squared)x) - cos(squared)x = -1
Any help is greatly appreciated Thanks
Thanks galactus
I did double check #2. I do not think it is an identity.
They way you worked out #1. My teacher said that we should pick one side and try to manipulate it so it it equals the other side.
so your
secX cscX does not = 2tanX - cscX secX
or am I missing something?
galactus
Mar 17, 2008, 04:17 PM
If the identity were valid, you could plug in something and get the same answer from both
sides. It would appear the 2tan(x) is extraneous. It works out to be just
csc(x)sec(x).
galactus
Mar 18, 2008, 06:30 AM
Here is a valid identity.
cot^{2}(x)-cot^{2}(x)sec^{2}(x)=-1
=\frac{cos^{2}(x)}{sin^{2}(x)}-\frac{\sout{cos^{2}(x)}}{sin^{2}(x)}\cdot\frac{1}{ \sout{cos^{2}(x)}}
=\frac{cos^{2}(x)}{sin^{2}(x)}-\frac{1}{sin^{2}(x)}
=\frac{\overbrace{cos^{2}(x)-1}^{\text{-sin^2(x)}}}{sin^{2}(x)}
=\frac{-sin^{2}(x)}{sin^{2}(x)}=-1