what is the equation of the tangent line y = (sin(x))^2, when x = 5π/6
You are welcome but your differential equation below a few questions has me tearing my hair out, no hope.
I tried z = 3x+1 and it helped a little
y=(sinx)²
dy/dx=2cosxsinx
When x=5π/6=150°
Cos(5π/6)=cos(120+30)=-√3/2
Sin(120+30)=1/2
The slope is
M=-2(√3/4)=-√3/2
Consider the pair to be
(0,5π/6)
y-y1=m(x-x1)
Y=-√3/2(x-5π/6)
Y=-x√3/2-(5π√3)/12
Y=-(6x√3+5π√3)/12
Correction
(5π/6,0)=(x1,y1)
sin (5 pi/6) = + 0.5 so we know goes through ( 0.5)^2 = 0.25
or ( 5 pi/6 , 0.25)
dy/dx = slope = 2 sin x cos x
at that point slope = 2 (0.5)(-sqrt 3 / 2) = -0.5 sqrt 3
so
y = -0.5 sqrt 3 * x + b
0.25 = -0.5 sqrt 3 * (5 pi/6) + b
y = -0.5 sqrt 3 * x + (1/4) + (5pi/12) sqrt 3
check my arithmetic
Isaac
I think you mean
(x1 , y1) = ( 5 pi/6 , 1/4)
Thank you sir Damon
Y=(0.5)²
Ha sir I have been working on that problem for a very long time... please try harder for me I have gone beyond my limit
Noted will try to implement this and see where it's takes me... thank you so much
To find the equation of the tangent line to the curve y = (sin(x))^2 at a given point, we need to find both the slope and the y-intercept of the tangent line.
First, let's find the slope of the tangent line by taking the derivative of the function y = (sin(x))^2 with respect to x.
Using the chain rule, we have:
dy/dx = 2(sin(x))(cos(x))
Next, substitute the given value x = 5π/6 into the derivative to find the slope at that point.
dy/dx = 2(sin(5π/6))(cos(5π/6))
Now, simplify the expression:
Using the values from the unit circle, we know that sin(5π/6) = 1/2 and cos(5π/6) = -√3/2.
dy/dx = 2(1/2)(-√3/2)
= -√3/2
We have found the slope (-√3/2) of the tangent line at the given point.
To determine the y-intercept of the tangent line, we need to plug in the values of x and y at the given point into the point-slope form of the equation of a line:
y - y1 = m(x - x1),
where m is the slope and (x1, y1) is the given point.
In this case, the given point is (5π/6, (sin(5π/6))^2).
Plugging these values into the equation, we get:
y - (sin(5π/6))^2 = (-√3/2)(x - 5π/6).
Now, simplify the expression:
y - (1/4) = (-√3/2)(x - 5π/6).
This is the equation of the tangent line to the curve y = (sin(x))^2 at the point x = 5π/6.