whatis the possibilities for the square root of 2 times tan theta = 2 sin theta
√2tanØ = 2sinØ
√2 sinØ/cosØ - 2sinØ = 0
sinØ(√2/cosØ - 2) = 0
sinØ = 0 or √2/cosØ - 2 = 0
case 1:
sinØ = 0 , Ø = 0, π, 2π
case 2:
√2/cosØ -2=0
√2/cosØ = 2
2cosØ = √2
cosØ = √2/2
Ø = 45° or 315° or π/4, 7π/4
so for 0 ≤ Ø ≤ 2π
Ø = 0 , π/4 , π , 7π/4, 2π
An obvious solution is θ=0, since tanθ = sinθ = 0
Eliminating that root, we have
√2/cosθ = 2
cosθ = 1/√2
θ = π/4 or -π/4
Add 2kπ for other roots.
To find the possibilities for the angle theta that satisfy the equation √2 * tan(theta) = 2 * sin(theta), we can start by manipulating the equation algebraically.
First, let's rewrite the equation using the trigonometric identities:
√2 * tan(theta) = 2 * sin(theta)
√2 * sin(theta) / cos(theta) = 2 * sin(theta)
Now, we can divide both sides of the equation by sin(theta):
√2 / cos(theta) = 2
Next, let's square both sides of the equation to eliminate the square root:
(√2 / cos(theta))^2 = 2^2
(√2)^2 / cos^2(theta) = 4
2 / cos^2(theta) = 4
Now, let's rearrange the equation to solve for cos^2(theta):
cos^2(theta) = 2 / 4
cos^2(theta) = 1 / 2
Finally, taking the square root of both sides, we can find the possibilities for the cosine of theta:
cos(theta) = ± √(1 / 2)
Since the cosine function takes on positive and negative values for different angles, we have two possibilities for the cosine of theta:
1. cos(theta) = √(1 / 2)
2. cos(theta) = -√(1 / 2)
To find the values of theta that correspond to these cosine values, we can use the inverse cosine function (arccos). Evaluating the inverse cosine of both possibilities, we get:
1. theta = arccos(√(1 / 2))
2. theta = arccos(-√(1 / 2))
These values represent different angles that satisfy the original equation √2 * tan(theta) = 2 * sin(theta).