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

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

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Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
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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