1<e^(4x-1)<9
Taking the natural logarithm of both sides, we have ln(1) < 4x-1 < ln(9).
Since ln(1) = 0, this simplifies to:
0 < 4x-1 < ln(9).
Adding 1 to all three parts of the inequality, we get:
1 < 4x < 1 + ln(9).
Dividing all three parts by 4, we have:
1/4 < x < (1 + ln(9))/4.
Therefore, the solution for the inequality 1 < e^(4x-1) < 9 is 1/4 < x < (1 + ln(9))/4.
Find the exact value of this expression in radians.
cos(2 arcsin(15/17))
To find the exact value of cos(2 arcsin(15/17)), let's start by finding the value of arcsin(15/17).
We know that sin(arcsin(x)) = x, so sin(arcsin(15/17)) = 15/17.
Now, let's express cos(2 arcsin(15/17)) in terms of sin(arcsin(15/17)).
We know that cos(2θ) = 1 - 2sin^2(θ). Therefore,
cos(2 arcsin(15/17)) = 1 - 2(sin(arcsin(15/17)))^2
= 1 - 2(15/17)^2
= 1 - 2(225/289)
= 1 - 450/289
= 289/289 - 450/289
= (-161)/289
Therefore, the exact value of cos(2 arcsin(15/17)) is -161/289 in radians.
the angle y whose sine is -(sqrt2)/2 and which is in the interval [-pi/2,pi/2] is y = x radians
To find the angle y whose sine is -(sqrt2)/2 and which is in the interval [-pi/2,pi/2], we need to find the angle in the appropriate range whose sine is -(sqrt2)/2.
Since the sine of -(sqrt2)/2 is negative, we know that the angle y must be in the second or third quadrant.
The angle in the unit circle with a sine of -(sqrt2)/2 can be found by taking the inverse sine of -(sqrt2)/2:
y = arcsin(-(sqrt2)/2)
By using a calculator, we find that the possible angles for y are -3pi/4 and -pi/4.
However, since y must be in the range [-pi/2, pi/2], the angle -3pi/4 is not in the correct range.
Therefore, the angle y whose sine is -(sqrt2)/2 and which is in the interval [-pi/2,pi/2] is y = -pi/4 radians.
if y = arccos(x), then cos(y) = x, 0 <= y <= pi. therefore, to find y = arccos(-(sqrt2)/2), we must find the angly y whose cosine is (-(sqrt2)/2). there are many possible angles with this cosine, but the range of y = arccos(x) is restricted to the interval x, and so y must be in this interval
You are correct. I apologize for the confusion in my previous response.
To find the angle y whose cosine is -(sqrt2)/2, we need to find the angle in the appropriate range whose cosine is -(sqrt2)/2.
Since the cosine of -(sqrt2)/2 is negative, we know that the angle y must be in the second or third quadrant.
The angle in the unit circle with a cosine of -(sqrt2)/2 can be found by taking the inverse cosine of -(sqrt2)/2:
y = arccos(-(sqrt2)/2)
Using a calculator, we find that the possible angles for y are 3pi/4 and 5pi/4.
However, since y must be in the range [0, pi], the angle 5pi/4 is not in the correct range.
Therefore, the angle y whose cosine is -(sqrt2)/2 and which is in the interval [0, pi] is y = 3pi/4 radians.
if y = arccos(x), then cos(y) = x, 0 <= y <= pi. therefore, to find y = arccos(-(sqrt2)/2), we must find the angly y whose cosine is (-(sqrt2)/2). there are many possible angles with this cosine, but the range of y = arccos(x) is restricted to the interval x
You are correct again. I apologize for the confusion caused by my previous response.
To find the angle y whose cosine is -(sqrt2)/2, we need to find the angle in the appropriate range whose cosine is -(sqrt2)/2.
Since the cosine of -(sqrt2)/2 is negative, we know that the angle y must be in the second or third quadrant.
The angle in the unit circle with a cosine of -(sqrt2)/2 can be found by taking the inverse cosine of -(sqrt2)/2:
y = arccos(-(sqrt2)/2)
Using a calculator, we find that the possible angles for y are 3pi/4 and 5pi/4.
However, since the restricted range for y = arccos(x) is [0, pi], the angle 5pi/4 is not in the correct range.
Therefore, the angle y whose cosine is -(sqrt2)/2 and which is in the interval [0, pi] is y = 3pi/4 radians.