1. Use the unit circle to find the inverse function value in degrees. tan^-1 √3
2. Use a calculator to find the value of the inverse function in radians
sin^-1 (0.65)
0.71
-3.85
0.86
-0.61
14 answers
0.86 (rounded to two decimal places)
1. Use the unit circle to find the inverse function value in degrees. tan^-1 √3
tan^-1 √3 is the angle in radians whose tangent is √3. In the unit circle, this corresponds to an angle of π/3 or 60 degrees.
Therefore, tan^-1 √3 = 60 degrees.
Therefore, tan^-1 √3 = 60 degrees.
1. Use the unit circle to find the inverse function value in degrees. tan^-1 √3
120
90
60
30
120
90
60
30
tan^-1 √3 is the angle in radians whose tangent is √3. In the unit circle, this corresponds to an angle of π/3 or 60 degrees.
Therefore, tan^-1 √3 = 60 degrees. Answer: 60
Therefore, tan^-1 √3 = 60 degrees. Answer: 60
9. 4cos 0 + 1 = 2cos
2pi/3 , 4pi/3
pi/6, 5pi/6
7pi/6 , 11pi/6
pi/3 , 6pi/3
10. cos 0- tan 0 cos0 -0
0, pi/4 , pi, 5pi/4
pi/4 5pi/4
pi/2, 3pi/4 , 3pi/2, 7pi/4
pi/2, 7pi/6, 3pi/2, 11pi/6
2pi/3 , 4pi/3
pi/6, 5pi/6
7pi/6 , 11pi/6
pi/3 , 6pi/3
10. cos 0- tan 0 cos0 -0
0, pi/4 , pi, 5pi/4
pi/4 5pi/4
pi/2, 3pi/4 , 3pi/2, 7pi/4
pi/2, 7pi/6, 3pi/2, 11pi/6
9. We can simplify the left-hand side of the equation using the identity cos(2π - θ) = cos(θ):
4cos(0) + 1 = 4(1) + 1 = 5
Now we can solve for cos(2θ):
2cos(2θ) = 5 - 2 = 3
cos(2θ) = 3/2
This equation has no real solutions, since the cosine function has a maximum value of 1 and a minimum value of -1. Therefore, there is no solution to the equation 4cos(0) + 1 = 2cos(2π/3) or 4cos(0) + 1 = 2cos(4π/3).
Answer: None of the given options.
10. We can simplify the left-hand side of the equation using the tangent identity tan(θ) = sin(θ)/cos(θ):
cos(0) - tan(0) = cos(0) - sin(0)/cos(0) = 1 - 0/1 = 1
cos(0) - tan(0) = 1
Now we can simplify the right-hand side of the equation:
cos(0) - 0 = 1 - 0 = 1
cos(0) - 0 = 1
Therefore, the equation is true for all values of θ.
Answer: All of the given options.
4cos(0) + 1 = 4(1) + 1 = 5
Now we can solve for cos(2θ):
2cos(2θ) = 5 - 2 = 3
cos(2θ) = 3/2
This equation has no real solutions, since the cosine function has a maximum value of 1 and a minimum value of -1. Therefore, there is no solution to the equation 4cos(0) + 1 = 2cos(2π/3) or 4cos(0) + 1 = 2cos(4π/3).
Answer: None of the given options.
10. We can simplify the left-hand side of the equation using the tangent identity tan(θ) = sin(θ)/cos(θ):
cos(0) - tan(0) = cos(0) - sin(0)/cos(0) = 1 - 0/1 = 1
cos(0) - tan(0) = 1
Now we can simplify the right-hand side of the equation:
cos(0) - 0 = 1 - 0 = 1
cos(0) - 0 = 1
Therefore, the equation is true for all values of θ.
Answer: All of the given options.
9. 4cos 0 + 1 = 2cos
A. 2pi/3 , 4pi/3
B. pi/6, 5pi/6
C. 7pi/6 , 11pi/6
D pi/3 , 6pi/3
A. 2pi/3 , 4pi/3
B. pi/6, 5pi/6
C. 7pi/6 , 11pi/6
D pi/3 , 6pi/3
We can simplify the left-hand side of the equation using the identity cos(2π - θ) = cos(θ):
4cos(0) + 1 = 4(1) + 1 = 5
Now we can solve for cos(2θ):
2cos(2θ) = 5 - 2 = 3
cos(2θ) = 3/2
This equation has no real solutions, since the cosine function has a maximum value of 1 and a minimum value of -1. Therefore, there is no solution to the equation 4cos(0) + 1 = 2cos(2π/3) or 4cos(0) + 1 = 2cos(4π/3).
Answer: None of the given options.
4cos(0) + 1 = 4(1) + 1 = 5
Now we can solve for cos(2θ):
2cos(2θ) = 5 - 2 = 3
cos(2θ) = 3/2
This equation has no real solutions, since the cosine function has a maximum value of 1 and a minimum value of -1. Therefore, there is no solution to the equation 4cos(0) + 1 = 2cos(2π/3) or 4cos(0) + 1 = 2cos(4π/3).
Answer: None of the given options.
i think you are wrong
I apologize for the mistake. To solve the equation:
4cos(0) + 1 = 2cos(2θ)
Since cos(0) = 1, we can simplify to:
4 + 1 = 2cos(2θ)
5 = 2cos(2θ)
cos(2θ) = 5/2
This equation has no real solutions, since the cosine function has a maximum value of 1 and a minimum value of -1. Therefore, there is no solution to the equation 4cos(0) + 1 = 2cos(2π/3) or 4cos(0) + 1 = 2cos(4π/3).
Answer: None of the given options.
4cos(0) + 1 = 2cos(2θ)
Since cos(0) = 1, we can simplify to:
4 + 1 = 2cos(2θ)
5 = 2cos(2θ)
cos(2θ) = 5/2
This equation has no real solutions, since the cosine function has a maximum value of 1 and a minimum value of -1. Therefore, there is no solution to the equation 4cos(0) + 1 = 2cos(2π/3) or 4cos(0) + 1 = 2cos(4π/3).
Answer: None of the given options.
10. cos 0- tan 0 cos0 -0
0, π/4 , π, 5π/4
π/4 5π/4
π/2, 3π/4 , 3π/2, 7π/4
π/2, 7π/6, 3π/2, 11π/6
0, π/4 , π, 5π/4
π/4 5π/4
π/2, 3π/4 , 3π/2, 7π/4
π/2, 7π/6, 3π/2, 11π/6
4cosθ + 1 = 2cosθ
2cosθ = -1
cosθ = -1/2
so choice A
2cosθ = -1
cosθ = -1/2
so choice A
Yes, you are correct. The solution to equation 4cosθ + 1 = 2cosθ is cosθ = -1/2.
Therefore, the answer for question 9 is A. 2π/3, 4π/3.
Therefore, the answer for question 9 is A. 2π/3, 4π/3.
1 answer