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(tan(theta)-cot(theta))/(tan(theta)+cot(theta) = 1-cos^2(the
- Thread starter camgrover
- Start date
Re: Seemingly Impossible trig identitiy
Hello, camgrover!
Did you copy it wrong? . . . It is not an identity.
\(\displaystyle \text{The left side is: }\:\frac{\dfrac{\sin\theta}{\cos\theta} - \dfrac{\cos\theta}{\sin\theta}}{\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta}}\)
\(\displaystyle \text{Multiply top and bottom by }\sin\theta\cos\theta\!:\;\;\frac{\sin^2\!\theta - \cos^2\!\theta}{\sin^2\!\theta + \cos^2\!\theta} \;=\; \frac{-(\cos^2\!\theta-\sin^2\!\theta)}{1}\)
\(\displaystyle \text{Therefore, we have: }\;-\cos2\theta\)
Hello, camgrover!
Did you copy it wrong? . . . It is not an identity.
\(\displaystyle \frac{\tan\theta - \cot\theta}{\tan\theta+\cot\theta} \:= \:\underbrace{1-\cos^2\!\theta}_{???}\)
\(\displaystyle \text{The left side is: }\:\frac{\dfrac{\sin\theta}{\cos\theta} - \dfrac{\cos\theta}{\sin\theta}}{\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta}}\)
\(\displaystyle \text{Multiply top and bottom by }\sin\theta\cos\theta\!:\;\;\frac{\sin^2\!\theta - \cos^2\!\theta}{\sin^2\!\theta + \cos^2\!\theta} \;=\; \frac{-(\cos^2\!\theta-\sin^2\!\theta)}{1}\)
\(\displaystyle \text{Therefore, we have: }\;-\cos2\theta\)
D
Deleted member 4993
Guest
are you sure it is not:camgrover said:Ok, I've been stuck for like 3 hours. This seems impossible:
(tan(theta)-cot(theta))/(tan(theta)+cot(theta) = 1-cos^2(theta)
Help...please
(tan(theta)-cot(theta))/(tan(theta)+cot(theta) = 1 - 2 * cos^2(theta)