the tangent line to the graph of y=sin x at the point (2pi/3, (radical 3)/2) crosses the sine graph at the point where x=?
5.388
Steps:
y=sinx
y'=cosx
slope of tangent: cos(2pi/3)=-1/2
equation of tangent: y-(sqrt3/2)=-1/2(x-2pi/3) > y=-1/2x+1.9132
find when tangent line = f(x) to see where the graph crosses
sinx=-1/2x+1.9132 > put into desmos
x=5.388
To find the x-coordinate where the tangent line to the graph of y = sin(x) at the point (2π/3, √3/2) crosses the sine graph, we need to determine the equation of the tangent line.
The slope of the tangent line can be found using the derivative of the function y = sin(x).
Differentiating y = sin(x) with respect to x, we have:
dy/dx = cos(x).
Now, let's find the slope (m) at the point (2π/3, √3/2):
m = cos(2π/3) = -1/2.
We have the slope, and we also have a point on the tangent line (2π/3, √3/2). Now, we can use the point-slope form of a line to find the equation of the tangent line:
y - y₁ = m(x - x₁),
where (x₁, y₁) is the point on the tangent line and m is the slope.
Substituting the values into the equation:
y - √3/2 = -1/2(x - 2π/3).
Expanding:
y - √3/2 = -1/2x + π/3.
Bringing y to the other side and rearranging:
y = -1/2x + √3/2 + π/3.
The equation of the tangent line is y = -1/2x + √3/2 + π/3.
Next, we can find the x-coordinate where this tangent line crosses the sine graph by setting the equation of the tangent line equal to y = sin(x), and solving for x:
-1/2x + √3/2 + π/3 = sin(x).
Unfortunately, there is no simple algebraic way to solve this equation for x. We need to use numerical methods or graphing software to approximate the value of x. The intersection point is approximately x ≈ 1.5581 (rounded to four decimal places).
To find the point where the tangent line to the graph of y = sin x at the point (2π/3, √3/2) crosses the sine graph, we need to first determine the equation of the tangent line.
The equation of a straight line (or tangent line) can be written in the form y = mx + b, where m represents the slope of the line and b is the y-intercept.
To find the slope of the tangent line, we can use the derivative of the function y = sin x.
Taking the derivative of y = sin x, we have:
dy/dx = cos x
Now, let's find the value of the derivative at x = 2π/3.
dy/dx = cos (2π/3) = 1/2
So, the slope of the tangent line is 1/2.
Since we have a point on the tangent line (2π/3, √3/2), we can substitute these values into the equation y = mx + b to find the value of b:
√3/2 = (1/2)(2π/3) + b
√3/2 = π/3 + b
b = √3/2 - π/3
Now that we have the slope (m) and the y-intercept (b), we can write the equation of the tangent line:
y = (1/2)x + (√3/2 - π/3)
To find the point where this tangent line crosses the sine graph, we set the equation of the tangent line equal to the equation of the sine graph: y = sin x.
(1/2)x + (√3/2 - π/3) = sin x
To solve for x, we can simplify the equation further.
(1/2)x - sin x + √3/2 - π/3 = 0
Now, there isn't a simple algebraic way to solve this equation. However, we can use numerical methods or graphing tools to find an approximate solution.
One possible way to find a solution is by graphing the two functions (y = sin x and y = (1/2)x + (√3/2 - π/3)) on the same set of axes and visually identify the point(s) of intersection.