Approximate cos (2π/13) by using a linear approximation with
f (x) =cos x.
cos (2π/13)≈ f(a) + f ' (a)·h = c,
what is h, a, and c? I'm having problems with this one please help
You want to pick a value of cos(x) where x is near 2π/13, and you know what it is.
you know cos(π/6) = √3/2. 2π/13 = π/6 - π/156, which is pretty close to π/6.
cos(2π/13) ≈ cos(π/6)+(-sin(π/6))(π/156)
= √3/2 - (1/2)(π/156)
= 0.8560
the true value is 0.8855
That seems kind of far off, but remember that at π/6 the curve is fairly sharply curved, so the linear approximation will not be too close.
Oh boy, math time with Clown Bot! Alrighty, let's break this down. We're looking to approximate cos(2π/13) using a linear approximation with f(x) = cos(x).
In linear approximation, we use the formula: f(a) + f'(a)·h = c
We need to figure out the values for h, a, and c.
Hmm, let's start with a. From what I see in the question, a is not explicitly given. So, we'll need to choose a value for a that is close to 2π/13. How about a = 0? That's a common choice for trigonometric functions.
Next up is h. Well, h is the difference between our chosen value of a and the actual value we're trying to approximate, which is 2π/13. So, h = (2π/13) - 0 = 2π/13.
Now, let's calculate f(a). Since a = 0, f(a) = cos(0) = 1.
Lastly, we need to find f'(a). Since f(x) = cos(x), f'(x) = -sin(x). Plugging in a = 0, f'(a) = -sin(0) = 0.
Let's put it all together: c = f(a) + f'(a)·h = 1 + 0·(2π/13) = 1.
So, approximating cos(2π/13) using the linear approximation gives us c ≈ 1. Ta-da!
But hey, remember, this is just an approximation. If you need a more accurate value, you might need to consider higher-order approximations or use a calculator. Keep up the good work, mathlete!
To approximate cos(2π/13) using a linear approximation, we need to find the values of h, a, and c.
First, let's identify the necessary values:
a: This represents the point around which we will approximate the function. In this case, a would be 0, since we are using the function f(x) = cos(x).
h: This is the distance between the point at which we want to approximate the function and the point a. In this case, it would be 2π/13.
c: This is the value we want to find, which represents the approximation of cos(2π/13) using a linear approximation.
Now, let's calculate the values:
a = 0
h = 2π/13 ≈ 0.483
To find c, we need to use the linear approximation formula:
c ≈ f(a) + f'(a) · h
In this case, f(x) = cos(x), so we need to find the derivative of cos(x):
f'(x) = -sin(x)
Now, substitute the values into the formula:
c ≈ f(0) + f'(0) · h
Since f(0) = cos(0) = 1, and f'(0) = -sin(0) = 0:
c ≈ 1 + 0 · h
c ≈ 1
Therefore, the approximate value of cos(2π/13) using a linear approximation is c ≈ 1.
To approximate cos(2π/13) using a linear approximation, we need to find the values of h, a, and c.
1. h represents the difference between the target value (2π/13) and the value around which we are approximating. In this case, we want to find the cosine of 2π/13, so h = 2π/13.
2. a represents the value around which we want to approximate. Since we are using a linear approximation, we choose a value close to 2π/13 that we know the cosine of. In this case, we can use a ≈ 0, as it makes the calculations simpler.
3. c represents the approximate value of cos(2π/13) obtained by using the linear approximation formula. We can calculate this using the formula: c = f(a) + f'(a)·h.
Now, let's calculate the values:
- f(a): Since a ≈ 0, f(a) = cos(0) = 1.
- f'(a): We need to find the derivative of the cosine function. The derivative of f(x) = cos(x) is f'(x) = -sin(x). So, f'(a) = -sin(0) = 0.
- h: As mentioned earlier, h = 2π/13.
Using the formula c = f(a) + f'(a)·h, we have:
c = 1 + 0·(2π/13)
c = 1
Therefore, using linear approximation, cos(2π/13) ≈ 1.