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
what is the equation of the tangent line y = (sin(x))^2, when x = 5π/6
9 answers
Correction
(5π/6,0)=(x1,y1)
(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
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)
I think you mean
(x1 , y1) = ( 5 pi/6 , 1/4)
Thank you sir Damon
Y=(0.5)²
Y=(0.5)²
You are welcome but your differential equation below a few questions has me tearing my hair out, no hope.
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
I tried z = 3x+1 and it helped a little
Noted will try to implement this and see where it's takes me... thank you so much