Missing solution with tan
I've noticed that when I am solving trig equations which may be in terms of sin and cos, when I try to solve it in terms of tan only, I end up dropping a solution and I'm really confused about why? Thinking about it now I feel like I do know but I would like to clarify this...
For example: https://ibb.co/kV7gCCty
For example: https://ibb.co/kV7gCCty
Reply 1
21
Original post
by TwisterBlade596I've noticed that when I am solving trig equations which may be in terms of sin and cos, when I try to solve it in terms of tan only, I end up dropping a solution and I'm really confused about why? Thinking about it now I feel like I do know but I would like to clarify this...
For example: https://ibb.co/kV7gCCty
For example: https://ibb.co/kV7gCCty
Pretty much line 1 for theta=pi/2 you have
0 = 0
as sin(2theta) = cos(theta) = 0. So dividing by 0 aint a great idea. If you do this, you need to state cos(theta) != 0 and cos(2theta) != 0 and explicitly consider these seperately.
Id have done sin(2theta) = 2cos(theta)sin(theta) so youd get something like
cos(theta) ( ...) = 0
at the start so the theta=pi/2 is one solution then the () = 0 gives the others. So factorise, dont divide.
Reply 2
Original post
by mqb2766Pretty much line 1 for theta=pi/2 you have
0 = 0
as sin(2theta) = cos(theta) = 0. So dividing by 0 aint a great idea. If you do this, you need to state cos(theta) != 0 and cos(2theta) != 0 and explicitly consider these seperately.
Id have done sin(2theta) = 2cos(theta)sin(theta) so youd get something like
cos(theta) ( ...) = 0
at the start so the theta=pi/2 is one solution then the () = 0 gives the others. So factorise, dont divide.
0 = 0
as sin(2theta) = cos(theta) = 0. So dividing by 0 aint a great idea. If you do this, you need to state cos(theta) != 0 and cos(2theta) != 0 and explicitly consider these seperately.
Id have done sin(2theta) = 2cos(theta)sin(theta) so youd get something like
cos(theta) ( ...) = 0
at the start so the theta=pi/2 is one solution then the () = 0 gives the others. So factorise, dont divide.
Was what I was doing dividing by 0? I'm a bit confused.
Reply 3
When you solve in terms of tan, you're basically ignoring solutions where cos(theta) is 0. So you miss out on things like 0 = pi/2, which works for sin/cos but not for tan.
Reply 4
Original post
by Adam_jarWhen you solve in terms of tan, you're basically ignoring solutions where cos(theta) is 0. So you miss out on things like 0 = pi/2, which works for sin/cos but not for tan.
Because tan is undefined at that angle?
Reply 5
21
Original post
by TwisterBlade596Because tan is undefined at that angle?
Consider the two equations
sin(theta) = cos(theta)
sin(2theta) = cos(theta)
In both cases we can divide by cos(theta), but we should exclude cos(theta)=0 from the subsequent analysis. So
tan(theta) = 1, cos(theta) != 0
sin(theta) = 1/2, cos(theta) != 0
Without working out the left hand equations, consider the excluded condition in both cases so theta = pi/2 (for simplicity, just the principle angle). Subbing it into the two original equations we have
1 = 0
0 = 0
In the first equation, theta=pi/2 is not a solution, so we can simply ignore the excluded condition and proceed. tan(pi/2)=inf (roughly, better to say its undefined) and thats not equal to 1 and the value corresponding to cos(theta)=0 has no bearing on these solutions. For the second equation, theta=pi/2 is one solution as both sides are zero, but sin(theta)=1/2 does not contain this solution as its been divided out.
When you divide by cos(theta), for example, any subsequent analysis which reasons about the equation when there is a solution at theta=pi/2 is going to be erroniouus as youve divided out that solution and are just considering the other ones. That value is not in the domain. Either explicitly exclude the =0 value from the subsequent analysis and handle it seperately or factorise rather than dividing. So the second equation becomes
cos(theta)(2sin(theta) - 1) = 0
and it should be obvious that the factor/root
2sin(theta) - 1 = 0
is independent of the factor/root
cos(theta) = 0
It could be argued that tan=sin/cos is an identity transformation, but as the first equation shows, the values where tan(theta) is undefined may or may not correspond to a solution. You simply dont know and need to consider the case where cos(theta)=0 seperately.
Edit - as a simple example to work through yourself, why not solve
x^2 = x
using the ways mentioned above and think about what works/why.
(edited 8 months ago)
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