Can someone check these for me? Please?
Use half-angle identity to find the exact value of cos165 degrees.
(-1/2) sqrt(2+sqrt(3))
Write the equation 2x+3y-5=0 in normal form.
(-2sqrt(13)/13)x- (3sqrt(13)/13)y+ (5 sqrt (13)/13) = 0
Find the distance between the lines with equations 3x-y=9 and y=3x-4.
sqrt (10)/2
To check the exact value of cos(165 degrees) using the half-angle identity, we need to use the half-angle formula for cosine:
cos(theta/2) = ± sqrt((1 + cos(theta))/2)
Since the given angle is 165 degrees, we can use the half-angle identity to find the exact value of cos(165 degrees).
First, we find the value of cos(330 degrees), which is twice the angle (165 degrees).
cos(330 degrees) = cos(2 * 165 degrees)
Using the double angle identity for cosine:
cos(2 * theta) = 2 * cos^2(theta) - 1
cos(330 degrees) = 2 * cos^2(165 degrees) - 1
Next, we substitute cos(165 degrees) with x.
cos(330 degrees) = 2 * x^2 - 1
Since we already know the value of cos(330 degrees) to be -0.5, we can substitute:
-0.5 = 2 * x^2 - 1
Rearranging the equation, we get:
2 * x^2 = 1 - 0.5
2 * x^2 = 0.5
Dividing both sides by 2:
x^2 = 0.25
Taking the square root of both sides:
x = ±sqrt(0.25)
So, we have two possible values for x = ±0.5.
Since cos(theta) is positive in the second and third quadrants, and 165 degrees falls in the second quadrant, we choose the negative value:
x = -0.5
Finally, substituting the value of x back into the half-angle identity equation:
cos(165 degrees) = -0.5 * sqrt(2 + sqrt(3))
Therefore, the exact value of cos(165 degrees) is:
cos(165 degrees) = -0.5 * sqrt(2 + sqrt(3))
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To write the equation 2x + 3y - 5 = 0 in normal form, we need to rearrange the equation to satisfy the condition: Ax + By + C = 0, where A, B, and C are constants.
Given the equation: 2x + 3y - 5 = 0
We can rearrange it as follows:
2x + 3y = 5
Now, to normalize the equation, we divide both sides by the square root of the sum of the squares of the coefficients of x and y.
Dividing by √(2^2 + 3^2) = √(13):
(2/√13)x + (3/√13)y = 5/√13
Multiplying both sides by √13 to remove the denominators, we get:
-2√13x - 3√13y + 5√13 = 0
Thus, the equation 2x + 3y - 5 = 0 in normal form is:
(-2√13/13)x - (3√13/13)y + (5√13/13) = 0
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To find the distance between two lines given their equations, you need to find the perpendicular distance from any point on one line to the other line.
The given equations of the lines are:
1) 3x - y = 9
2) y = 3x - 4
First, rewrite equation 2 in the standard form:
y - 3x = -4
Now, find the perpendicular distance between the two lines by using the formula:
distance = |Ax + By + C| / sqrt(A^2 + B^2)
For the equation 1) 3x - y = 9:
A = 3
B = -1
C = -9
distance = |(3)(0) + (-1)(-4) - (-9)| / sqrt((3)^2 + (-1)^2)
Simplifying this expression, we get:
distance = |4 + 9| / sqrt(9 + 1)
distance = 13 / sqrt(10)
To simplify further, we rationalize the denominator by multiplying the numerator and denominator by sqrt(10):
distance = (13 * sqrt(10)) / sqrt(10) * sqrt(10)
Canceling out the sqrt(10) terms in the denominator, we get:
distance = 13 * sqrt(10) / 10
So, the distance between the lines with equations 3x - y = 9 and y = 3x - 4 is:
distance = 13 * sqrt(10) / 10