1. Which expression is equivalent to cos 2theta for all values of theta ?
cos^2 theta – sin^2 theta ~
cos^2 theta – 1
1 – 2 sin^2 theta
2 sin theta cos theta
2. Use a half–angle identity to find the exact value of sin 105°.
-1/2(sqrt)(2 + Sqrt3)
1/2(sqrt)(2 + Sqrt3) ~
1/2(sqrt)(2 + Sqrt2)
11/2(sqrt)(1 + Sqrt3)
3.) Solve cos x tan x - sin^2x = 0 for all real values of x.
piK, pi/2 + 2piK
pi/2 + piK, 2pik
pi/2 + 2piK, 3pi/2 + 2piK
piK, pi/4 +2piK ~
4.) Solve 2 sin2x – sin x = 0 for principal values of x.
60° and 120°
0° and 150° ~
0° and 30°
60°
5.) Solve csc x + 2 = 0 for 0 < x < 2pi.
pi/6 and 5pi/6
pi/6 and 7pi/6
4pi/3 and 5pi/3
7pi/6 and 11pi/6 ~
6.) Solve 2 cos x – 1 = 0 for 0 <x< 2.
pi/6 and 5pi/6
pi/3 and 5pi/3 ~
pi/3 and 2pi/3
7pi/6 and 11pi/6
7.) Write the equation 3x + 4y – 7 = 0 in normal form.
3/5x + 4/5y + 7/5 = 0
-3/5x - 4/5y - 7/5 = 0
-3/5x - 4/5y + 7/5 = 0 ~
3/5x + 4/5y - 7/5 = 0
8.)Write the standard form of the equation of a line for which the length of the normal is 4 and the normal makes an angle of 45° with the positive x–axis.
sqrt2x + sqrt2y +8 = 0
2x + 2y – 8 = 0
sqrt2x + sqrt2y -8 = 0 ~
2x + 2y + 8 = 0
9.) Write the standard form of the equation of a line for which the length of the normal is 3 and the normal makes an angle of 135° with the positive x–axis.
sqrt2^x – sqrt2y + 6 = 0 ~
sqrt2^x + sqrt2y - 6 = 0
sqrt2^x – sqrt2y - 6 = 0
sqrt2^x + sqrt2y + 6 = 0
10.) Find the distance between P (–2, 1) and the line with equation x – 2y + 4 = 0.
4sqrt5/5
0 ~
-4sqrt5/5
5/4
the ones with~ are my answers
#3 should be piK, pi/2 + 2piK since sin=0 or 1
#4. should be 0° and 30° since they asked for principal values
1,2,5,6 look good
7-10 I'll have to check later, or another expert can come by.
1. The expression equivalent to cos 2theta for all values of theta is 2 sin theta cos theta.
2. Using a half-angle identity, the exact value of sin 105° is 1/2(sqrt)(2 + Sqrt3).
3. The values of x that solve cos x tan x - sin^2x = 0 for all real values of x are pi/2 + 2piK and 3pi/2 + 2piK.
4. The principal values of x that solve 2 sin2x – sin x = 0 are 0° and 150°.
5. The values of x that solve csc x + 2 = 0 for 0 < x < 2pi are 7pi/6 and 11pi/6.
6. The values of x that solve 2 cos x – 1 = 0 for 0 < x < 2 are pi/3 and 5pi/3.
7. The equation 3x + 4y – 7 = 0 in normal form is -3/5x - 4/5y + 7/5 = 0.
8. The standard form of the equation of a line for which the length of the normal is 4 and the normal makes an angle of 45° with the positive x-axis is sqrt2x + sqrt2y - 8 = 0.
9. The standard form of the equation of a line for which the length of the normal is 3 and the normal makes an angle of 135° with the positive x-axis is sqrt2x – sqrt2y + 6 = 0.
10. The distance between P (-2, 1) and the line with equation x - 2y + 4 = 0 is 4sqrt5/5.
To find the equivalent expression for cos 2theta, we can use the double angle identity for cosine, which states that cos 2theta = cos^2 theta - sin^2 theta. So the correct answer is cos^2 theta – sin^2 theta.
To find the exact value of sin 105° using a half-angle identity, we can use the formula sin(theta/2) = sqrt((1 - cos theta)/2). Since 105° is not a special angle, we can rewrite it as 210°/2. Applying the formula, we have sin(105°) = sqrt((1 - cos 210°)/2). Simplifying further, we find sin(105°) = sqrt((1 + sqrt(3))/2).
To solve the equation cos x tan x - sin^2x = 0, we can use the Pythagorean identity for tangent, which states that tan x = sin x / cos x. Substituting this into the equation, we have cos x (sin x / cos x) - sin^2x = 0. Simplifying further, we find sin x - sin^2x = 0. Factoring out sin x, we get sin x (1 - sin x) = 0. This equation is true when sin x = 0 or 1 - sin x = 0. Solving for x in both cases, we find that x can be any multiple of pi or pi/4 plus any multiple of 2 pi.
To solve the equation 2 sin^2x – sin x = 0 for principal values of x, we can factor out sin x from the equation to get sin x (2 sin x - 1) = 0. This equation is true when sin x = 0 or 2 sin x - 1 = 0. Solving for x in both cases, we find x = 0° and x = 30°.
To solve the equation csc x + 2 = 0 for 0 < x < 2pi, we can start by finding the reciprocal of csc x, which is sin x. Then, the equation becomes sin x + 2 = 0. Solving for sin x, we find sin x = -2. However, there are no values of sin x that lie between -1 and 1, so there are no solutions for this equation.
To solve the equation 2 cos x - 1 = 0 for 0 < x < 2pi, we can start by isolating cos x. Adding 1 to both sides of the equation, we get 2 cos x = 1. Dividing by 2, we find cos x = 1/2, which is a special angle. The solutions for cos x = 1/2 are x = pi/3 and x = 5pi/3.
To write the equation 3x + 4y – 7 = 0 in normal form, we need to rearrange the equation to have the coefficients of x and y divided by the square root of their squares. Dividing each term by the square root of the sum of their squares, we get (3/5)x + (4/5)y - 7/5 = 0.
To write the standard form of the equation of a line for which the length of the normal is 4 and the normal makes an angle of 45° with the positive x-axis, we can use the formula for the normal slope, which is the negative reciprocal of the line's slope. Let's assume the slope of the line is m. Since the normal makes an angle of 45° with the positive x-axis, the slope of the normal is tan 45° = 1. So, the slope of the line is -1. Now, we can use the point-normal form of the equation of a line to calculate it in standard form. Let's assume the point of the line is (x₀, y₀). The equation becomes: (y - y₀) = -1(x - x₀). Rearranging this equation to standard form, we get sqrt(2)x + sqrt(2)y - 8 = 0.
To write the standard form of the equation of a line for which the length of the normal is 3 and the normal makes an angle of 135° with the positive x-axis, we can follow a similar process as in the previous question. The normal slope is -tan 135° = 1. The line slope is the negative reciprocal, which is -1. Let's assume the point of the line is (x₀, y₀). The equation becomes: (y - y₀) = -1(x - x₀). Rearranging this equation to standard form, we get -sqrt(2)x - sqrt(2)y - 6 = 0.
To find the distance between point P(-2, 1) and the line with equation x - 2y + 4 = 0, we can use the formula for the distance between a point (x₁, y₁) and a line Ax + By + C = 0, which is given by the formula: d = |Ax₁ + By₁ + C| / sqrt(A² + B²). Plugging the values into the formula, we get d = |-2 - 2(1) + 4| / sqrt(1² + (-2)²) = |-2 - 2 + 4| / sqrt(1 + 4) = |0| / sqrt(5) = 0 / sqrt(5) = 0. Therefore, the distance between point P and the line is 0.