23that conversion is ok , it is valid for all angles
i think where u r making mistake is : taking both the natural nos as n . Take one as m and other as n. there isnt any reason for them to be equal , they may be related by some other equation.
ques- solve for θ:
tan2θ.tanθ = 1
i found two ways of solving this but both of them are giving different answers
I - by using tan2θ = 2tanθ1-tan2θ
and solving for tanθ ,i got
tanθ = ±1√3
ie θ = nΠ± Π6
II - tan2θ = cotθ
tan2θ = tan(Π2 - θ)
2θ = nΠ+Π2 - θ
ie θ = Π3(n + 12)
both the answers are coming out to be different .....
plz someone tell where is my mistake????
I know 2nd method is wrong but plz explain how
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11 Answers
39The pi/2-theta conversion part....I think it works only for first quadrant angles. Here we are considering all four quadrants.
Do correct me if I'm wrong.
23
1yes qwerty that is right ..
let the answer be:
θ = Π3(m + 12)
but the problem is that both answers give different set of angles ,for example taking
m = 1 (in second solution) gives θ = Π2
whereas the first solution's answer can never yield θ = Π2 for any value of 'n'
341Perhaps it would have been simpler to just use
\tan \theta \tan 2\theta =1 \Rightarrow \tan 3\theta=0
1yes euclid
tan3θ = ∞
using this the solution would be same as second one
θ = Π3(x + 12) .......where x is any integer
but the answer given is, θ = nΠ± Π6 same as the first solution
1There is a small mistake that can happen .
tan θ x tan 2 θ = 1 may imply that ;
tan 3 θ = ∞ ,
if and only if , tan θ + tan 2 θ ≠0 ,
But in this case , can this happen ?
No , this can't happen as tan θ and tan 2 θ must have the same sign .
So the arguments hold .
1thanx ricky
i think i got my mistake
the set of angles, θ = nΠ± Π6 is a subset of
the set θ = Π3(m + 12)
the extra values of θ that second set contains are = Π2 , 5Π2 , 9Π2 .........
what i did in solution II can't be used because converting
tan2θtanθ=1 to tan2θ = cotθ would be like
tan2θcotθ = 1
and at Î 2 , 5Î 2 , 9Î 2 .........
tan2θcotθ = 00 ....... an undefined term
so we can't bring cot in Dr
but still im not able to find why tan3θ = ∞ can't be used