the question is
find the equation of the normal line and tangent line to the graph of y = cos^2 x at x = Pi/3
how do i do this question without knowing the equation (something like 2x^2 + 2y^2 = 9xy that i can diffrentiate or derivate) ?????
the question is
find the equation of the normal line and tangent line to the graph of y = cos^2 x at x = Pi/3
how do i do this question without knowing the equation (something like 2x^2 + 2y^2 = 9xy that i can diffrentiate or derivate) ?????
You can differentiate $\displaystyle \cos^2(x)$ directly using the chain rule:
$\displaystyle \frac{d}{dx} \cos^2(x) = -2 \cos(x) \sin(x)$
If you prefer to use an identity: $\displaystyle \cos^2(x) = \frac{\cos(2x)}{2}+\frac{1}{2}$
$\displaystyle \frac{d}{dx} \left[\frac{1}{2}\cos(2x)\right] = -\sin(2x)$
Note that $\displaystyle \frac{d}{dx}\, 0.5 = 0$
These answers are equal as $\displaystyle \sin(2x) = 2\sin(x) \cos(x)$
To find your gradient find $\displaystyle f' \left(\frac{\pi}{3}\right)$
here's what i tried
if i did something wrong, could you please point me out?
because i still am not sure if i did it right or wrong... the answer doesn't seem to right...
f(x)=y=cos^2 x
f(pi/3)= (cos pi/3)^2 = (1/2)^2 = 1/4
point: (pi/3 , 1/4) (x0,y0)
f'(x)= -2(cosx)(sinx)
f'(pi/3) = -2(1/2)(sqrt3/2) = -sqrt3 / 2 <-- slope of tangent
y=m(x-x0)+y0
y= -sqrt3/2 (x-pi/3)+1/4
= -(sqrt3/2 x)+[(2sqrt3 pi + 3)/12] <-- eq. of tangent
slope of normal line = -1/-sqrt3 / 2 = 2/sqrt3 = 2sqrt3/3
y=m(x-x0)+y0
y= 2sqrt3/3 (x-pi/3) + 1/4
= 2sqrt3/3x - [(8sqrt3 + 9)/36] <--- eq. of normal line