Below is a graph of the function y = csc ( b x ) y=\csc\left(bx\right) y = csc ( b x ) . Determine the value of b .

Welcome back everyone. So let's give this problem a try. Here we're told below is a graph of the function y is equal to the cosecant of b x, and we're asked to determine the value of b. Now to solve this problem, what I'm first going to do is draw a graph of something I'm a bit more familiar with. Now what I recognize here is that the cosecant is a reciprocal identity, and it's a reciprocal identity to the sine function. So I'm going to draw a graph of y is equal to the sine of bx. Now Now the way that I can figure out what this graph looks like is by recognizing that the sine function is going to be a wave that starts at the origin. And it's going to have its peaks where we have these curves that point upward, these kind of smiley faces. And it's going to have the valleys where we have these curves that point downward, these kind of frowny faces. So when we have our sine function go up here, it's gonna reach a peak, then it's gonna come down here and valley, then it's going to go through here, and then it's gonna reach another peak there, and then reach another valley right about there. So this is what the sine function looks like. Now what I can see for the sine function is we can start here at the origin, and we go up, and then down on this wave, and then back up, and complete a full period at three pi over two on the x axis. So what that tells me is that our period for this function is three pi over two. Now the reason that this is helpful to find the period is because the period for a sine function is always going to be two pi over b. And since we just figured out that the period is three pi over two, I can set three pi over two equal to two pi over b. And all I need to do from here is solve for b. So what I'm first going to do is multiply both sides of the equation by b. This is gonna get the b's to cancel on the right side giving me three pi over two times b is equal to two pi. Now from here what I'm going to do is multiply both sides of the equation by one over pi. And this is going to just allow me to cancel this pi on both sides. Because notice that the pi's are going to cancel right there, and the pi's are going to cancel right here. So what we're going to have is three over two times b is equal to two. The way that I'm going to solve this problem is I'm going to take both sides of this equation and multiply it by the reciprocal of three over two, which is two over three. Now when I do this, this is going to allow me to cancel the threes right there, and the twos right here, leaving me with just b. And then on the right side of this equation we're going to have two times two, which is four divided by three. So the solution for b is four thirds, and that's the answer to this problem. So I hope you found this video helpful. Thanks for watching.

Below is a graph of the function y = csc ⁡ ( b x ) y=\csc\left(bx\right) y = csc ( b x ) . Determine the value of b .

b = 3 π 4 b=\frac{3\pi}{4} b = 4 3 π

12. Trigonometric Functions

Graphs of Other Trigonometric Functions

Business Calculus

12. Trigonometric Functions

Graphs of Other Trigonometric Functions

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Multiple Choice

Below is a graph of the function $y=\csc\left(bx\right)$ . Determine the value of b.

$b=\frac12$

$b=3$

$b=\frac43$

$b=\frac{3\pi}{4}$

Verified step by step guidance

Step 1: Recall the general form of the csc function, y = csc(bx), where b affects the period of the function. The period of the csc function is given by T = (2π)/b.

Step 2: Analyze the graph provided. The graph shows one complete cycle of the csc function between x = π/2 and x = 3π/2. This indicates that the period of the function is T = π.

Step 3: Use the formula for the period, T = (2π)/b, and substitute T = π into the equation. This gives π = (2π)/b.

Step 4: Solve for b by multiplying both sides of the equation by b and then dividing by π. This results in b = 2.

Step 5: Verify the solution by checking that the calculated value of b = 2 matches the observed period of the graph. The period of the function with b = 2 is indeed π, confirming the correctness of the solution.