What is the length of the vector [5 cos è, 5 sinè]? If the vector [5,0] is rotated 36 degrees in the counter clockwise direction, what are the components of the resulting vector?
What is the length of the vector [5 cos è, 5 sinè]?
5 I suspect although I do not know your notation
5 cos 36 i or 5 cos 36 in x direction
5 sin 36 j or 5 sin 36 in y direction
To find the length of a vector [5 cos è, 5 sin è], we can use the formula for the magnitude of a vector in two dimensions:
Magnitude = √(x^2 + y^2)
In this case, x = 5 cos è and y = 5 sin è. Substituting these values into the formula, we get:
Magnitude = √((5 cos è)^2 + (5 sin è)^2)
= √(25 cos^2 è + 25 sin^2 è)
= √(25 (cos^2 è + sin^2 è))
Since cos^2 è + sin^2 è = 1 (due to the trigonometric identity), the magnitude simplifies to:
Magnitude = √(25)
= 5
So, the length of the vector [5 cos è, 5 sin è] is 5.
To find the components of the vector [5, 0] after rotating 36 degrees counterclockwise, we can use the rotation formulas:
x' = x cos θ - y sin θ
y' = x sin θ + y cos θ
In this case, x = 5 and y = 0, and θ = 36 degrees. Substituting these values into the formulas, we get:
x' = 5 cos 36 - 0 sin 36
y' = 5 sin 36 + 0 cos 36
Using a calculator, we can calculate the values of cos 36 and sin 36 as approximately 0.809 and 0.588, respectively. Therefore,
x' = 5 * 0.809 - 0 * 0.588
= 4.045
y' = 5 * 0.588 + 0 * 0.809
= 2.940
So, the components of the resulting vector after rotating [5, 0] by 36 degrees counterclockwise are approximately [4.045, 2.940].
To find the length of a vector, we need to calculate the magnitude of the vector. In this case, the vector [5 cos è, 5 sin è] has components expressed in terms of cosine and sine functions.
The magnitude of a vector can be calculated using the Pythagorean theorem, which states that the magnitude of a vector is the square root of the sum of the squares of its components.
So, for the given vector [5 cos è, 5 sin è], the length can be found using the following formula:
Magnitude = √((5 cos è)^2 + (5 sin è)^2)
Simplifying the equation:
Magnitude = √(25 cos^2 è + 25 sin^2 è)
Magnitude = 5√(cos^2 è + sin^2 è)
Now, according to the trigonometric identity, cos^2 è + sin^2 è = 1. Therefore, the magnitude further simplifies to:
Magnitude = 5√(1)
Magnitude = 5
So, the length of the given vector [5 cos è, 5 sin è] is 5.
Moving on to the second part of the question, we are asked to find the components of the resulting vector when the vector [5,0] is rotated 36 degrees counterclockwise.
To rotate a vector counterclockwise by an angle, we can use rotation formulas based on trigonometry. These formulas can be used to find the new components of the vector after rotation using the original components.
Let's denote the original vector as [a, b]. To rotate this vector counterclockwise by an angle θ, the new components can be calculated as follows:
New x-component = a * cos θ - b * sin θ
New y-component = a * sin θ + b * cos θ
In this case, the original vector is [5, 0], and we want to rotate it counterclockwise by 36 degrees.
Plugging in the values:
New x-component = 5 * cos 36° - 0 * sin 36°
New y-component = 5 * sin 36° + 0 * cos 36°
Now, use trigonometric values to evaluate cos 36° and sin 36°:
cos 36° ≈ 0.809
sin 36° ≈ 0.587
Calculating the values:
New x-component = 5 * 0.809 - 0 * 0.587
≈ 4.045
New y-component = 5 * 0.587 + 0 * 0.809
≈ 2.935
Therefore, the components of the resulting vector after rotating [5,0] counterclockwise by 36 degrees are approximately [4.045, 2.935].