Find tangent line to curve given a point..... very close just need alittle help.

Question

Find tangent line to the curve, given a point:

 

Y= cosx - sinx      (Pi, -1)

 

I used the difference rule:

(d/dx cosx) - (d/dx sinx)

 

-sinx - cosx

 

Then idk where to go from here... Im pretty sure I plug "Pi" in for x but I don't know what to do trig wise.

the answer is Y= x-Pi-1

Thanks!

Answer 1

Find tangent line to the curve, given a point:

f(x) = cos(x) – sin(x), (pi, -1).

 

f'(x) = – sin(x) – cos(x), which is what you got, Nicholas.

 

The slope of the tangent line at the given point is:

 

m = f'(pi) = – sin(pi) – cos(pi)

 

Trig refresher:

 

Given any point, P(x,y), define A(0,0), B(x,0), θ is angle between AB and AP, r = √(x^2+y^2).

 

We define (r cos(θ),r sin(θ)) = (x,y).

 

If (x,y) is on a Unit Circle centered at the origin, r = 1, so (cos(θ),sin(θ)) = (x,y).

 

Once around the Unit Circle is 2 pi radians, so pi radians is halfway around the Unit Circle starting from (1,0), moving CCW, and ending at (–1,0) = (cos(pi),sin(pi)).

 

Now we can find our slope.

 

m = f'(pi) = – sin(pi) – cos(pi) = – (0) – (–1) = 1

 

Using (pi, -1) and the point-slope form we get the tangent line:

 

y – (–1) = 1 (x – pi)

 

y = x – pi – 1