View Full Version : Trig Identities
sahill
Nov 11, 2004, 06:58 PM
This may be a typo, but if not, is this solvable?
1-cot^2theta=?
CroCivic91
Dec 3, 2004, 03:01 AM
1-cot^2theta = ?
It is kind of important what you want this expression to be equal to...
Lets assume you want it to be equal to 0...
1-cot^2theta = 0 is equal to (1-cot(theta))*(1+cot(theta)) = 0
Now, 2 numbers multiplied can produce 0 only if one (or both) of them are equal to 0.
So, first lets assume that 1-cot(theta)=0
That means that cot(theta) = 1 and that means that theta = pi/4 rad + k*pi, for any natural k
Now, lets assume that 1+cot(theta)=0
That means that cot(theta)=-1 and that means that theta = -pi/4 + k*pi, for any natural k
Kresho
Glipto
Mar 9, 2005, 06:30 PM
Hmmm... is not an ecuation.
I think...
Cos^2 theta + Sin^2 theta = 1 (Pitagoras)
1 - Cos^2 theta = Sin^2 theta
CroCivic91
Mar 10, 2005, 01:15 AM
I would agree with you, but his original post said:
1-cot^2theta=?
cot != cos
reinsuranc
Mar 10, 2005, 11:10 AM
sin^2theta + cos^2theta = 1
divide both sides by sin^2theta:
(sin^2theta)/(sin^2theta) + (cos^2theta)/(sin^2theta) = 1/sin^2theta
1 + cot^2theta = csc^2theta
Glipto
Mar 10, 2005, 09:56 PM
OOPS!! I NEED GLASSES!!
I agree with Reinsuranc, if the sign is positive the identity is beautiful.
MathMaven53
Mar 23, 2005, 01:16 PM
1- cot^2 t = 1 - cos^2 t/sin^2 t
= (sin^2 t - cos^2 t)/sin^2 t
= - cos 2t/sin^2 t
Note that cos^2 t - sin^2 t = cos 2t
2sin^2 t = 1 - cos 2t
1- cot^2 t = - cos 2t/sin^2 t
= (-2 cos 2t)/(2sin^2 t
= (-2 cos 2t)/(1 - cos 2t)
sammay_mon
Jan 19, 2006, 01:57 PM
Hay I'm new to this
Do I ask a "?" and someone'll help me?