Asked by Stacey grahaM
Solve the equation in the interval [0•,360•]
Csc theta = 1+cot theta
Csc theta = 1+cot theta
Answers
Answered by
bobpursley
multipy both sides by sintheta
1=sinTheta+cosTheta
This is only possible at Theta=0, 90, 360
1=sinTheta+cosTheta
This is only possible at Theta=0, 90, 360
Answered by
MathMate
Reduce to sin and cosines:
1/sinθ=(sinθ+cosθ)/sinθ
Thus, if θ ≠ 0,π 2π ...
we get
sinθ+cosθ=1
Taking advantage of symmetry about π/4, where sinπ/4=cosπ/4, substitute θ=φ-π/4:
sin(φ-π/4)+cos(φ-π/4)=1
Expanding by sum/difference formulae,
sinφcosπ/4-cosφsinπ/4 + cosφcosπ/4+sinφsinπ/4=1
Since sinπ/4=cosπ/4, we cancel terms in cosφ to get
2sinπ/4 sinφ=1
φ=arcsin(sqrt(2)/2)=±π/4
θ=0 or π/2
The first value has been rejected since the beginning, so θ=π/2.
1/sinθ=(sinθ+cosθ)/sinθ
Thus, if θ ≠ 0,π 2π ...
we get
sinθ+cosθ=1
Taking advantage of symmetry about π/4, where sinπ/4=cosπ/4, substitute θ=φ-π/4:
sin(φ-π/4)+cos(φ-π/4)=1
Expanding by sum/difference formulae,
sinφcosπ/4-cosφsinπ/4 + cosφcosπ/4+sinφsinπ/4=1
Since sinπ/4=cosπ/4, we cancel terms in cosφ to get
2sinπ/4 sinφ=1
φ=arcsin(sqrt(2)/2)=±π/4
θ=0 or π/2
The first value has been rejected since the beginning, so θ=π/2.
Answered by
MathMate
from
φ=arcsin(sqrt(2)/2)=π/4 or 3π/4 ± 2kπ
θ=φ-π/4=0 or π/2 ±2kπ
Since θ=0 has been rejected since the beginning, we are left with
θ=π/2 (for solution between 0 and 360)
φ=arcsin(sqrt(2)/2)=π/4 or 3π/4 ± 2kπ
θ=φ-π/4=0 or π/2 ±2kπ
Since θ=0 has been rejected since the beginning, we are left with
θ=π/2 (for solution between 0 and 360)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.