Post by Ken OliverYou most likely used the tangent double angle formula: tan(2x) = 2tan(x)
/ ( 1-tan^2(x) )
This "identity" has different domains for LHS and RHS. The value x=90
makes sense in the LHS, but makes the RHS undefined.
So after you tan(2x) in your problem to the RHS, 90 will no longer show
up as a solution.
Special handling for the "lost root" when using this identity or other
solving methods { sines and cosines } must be used to avoid missing the
90 deg solution.
You have a good point. And perhaps the best approach to the problem was
already given here by phillipe. But let me mention another method of
solution, which is probably not really appropriate for billq.
If we allow the range of the tangent and cotangent functions to be the
one-point extension of the reals, then tan(90 degrees) is defined to be
unsigned infinity, the identity cot(x) = 1/tan(x) is valid for all real x,
etc. In particular, we have an alternative version of the double-angle
formula for tangent which is valid for _all_ real x:
tan(2x) = 2/(cot(x) - tan(x))
Using that, the equation to be solved becomes
2/(cot(x) - tan(x)) = cot(x)
Then, taking reciprocals of both sides and multiplying through by 2, we get
cot(x) - tan(x) = 2tan(x)
Perhaps we would now like to divide both sides by tan(x), but we must be
careful! [Our situation is similar to that in the method outlined by
philippe, where, if we had divided both sides by cos(x), we would have lost
roots.] So, before dividing, note that we have roots whenever tan(x) is
infinite. Thus, x = 90 degrees is a solution.
Having noted that (together with the fact that we do not have any solution
when tan(x) is zero), we may now safely divide through by tan(x), obtaining
(cot(x))^2 - 1 = 2
cot(x) = +/- Sqrt(3)
and so on, to get the other solutions.
David Cantrell
Post by Ken OliverPost by billqHello,
When I solve tan2x= cotx I get tanx = +/-1/sqrt(3) which gives angle 30
and 150degrees as solutions. The answer key also adds 90degree as a
possible solution. Where does the 90 degree come from?