Solve the equation in the interval [0•,360•]
Csc theta = 1+cot theta
multipy both sides by sintheta
1=sinTheta+cosTheta
This is only possible at Theta=0, 90, 360
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.
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)
To solve the given equation, we will first simplify the equation using the trigonometric identities. Then, we will solve for theta within the given interval.
Let's start by simplifying the equation using the reciprocal and quotient identities:
Csc(theta) = 1 + Cot(theta)
Since Cosec(theta) is the reciprocal of Sin(theta) and Cot(theta) is the reciprocal of Tan(theta), we can rewrite the equation as:
1/Sin(theta) = 1 + Cos(theta)/Sin(theta)
Now, let's simplify further by multiplying through by Sin(theta) to get rid of the denominators:
1 = Sin(theta) + Cos(theta)
Next, we can square both sides of the equation to eliminate the square root:
1 = Sin^2(theta) + 2Sin(theta)Cos(theta) + Cos^2(theta)
Using the Pythagorean identity (Sin^2(theta) + Cos^2(theta) = 1), we can simplify further:
1 = 1 + 2Sin(theta)Cos(theta)
Now, subtracting 1 from both sides, we get:
0 = 2Sin(theta)Cos(theta)
Divide both sides by 2 to isolate Sin(theta)Cos(theta):
0 = Sin(theta)Cos(theta)
Now we have two possibilities for solving this equation:
1. If Sin(theta) = 0, then theta can be 0, 180, or any multiple of 180 degrees within the given interval.
2. If Cos(theta) = 0, then theta can be 90 or 270 degrees within the given interval.
Therefore, the solution for theta within the interval [0•,360•] is:
θ = 0°, 90°, 180°, 270°