eqn of tangent line w/trig fcns

[Guest]

Problem is: Find the eqn of the tangent line to y=f(x)=tan(x) at the point (pi/8, .4142).

Ok, this is really a shot in the dark:

first we need f' tan(x) = sec^2 x
and pi/8 = .3926,

my calc (ti89) doesn't have a sec button I don't think, but I think that 1/cos(x) = the same thing?

If it does sec ^2 x = 1.1714 - I think! which is of course m,

so plugging into y=mx+b, we get .4142 = (1.1714)(.3926)+b,
b = -.0457,

so y = 1.1714x - .0457????

 

[tkhunny]

Problem is: Find the eqn of the tangent line to y=f(x)=tan(x) at the point (pi/8, .4142).

Ok, this is really a shot in the dark:

first we need f' tan(x) = sec^2 x
and pi/8 = .3926,

my calc (ti89) doesn't have a sec button I don't think, but I think that 1/cos(x) = the same thing?

If it does sec ^2 x = 1.1714 - I think! which is of course m,

so plugging into y=mx+b, we get .4142 = (1.1714)(.3926)+b,
b = -.0457,

so y = 1.1714x - .0457????

I do not understand your desire to use approximations when exact values are readily available.

tan(pi/8) = sqrt(2) - 1

You should be able to calculate this easily, knowing tan(pi/4) = 1. Did the problem statement tell you to use 0.4142, or did it say only x = pi/8?

Good call on finding sec(x) = 1/cos(x). If you look hard enough, I am confident you will find the sec(x) function in your calculator, somewhere. There must be a list of functions you can access.

I get [sec(pi/8)]² = 2*(2-sqrt(2))

So, we have a point (pi/8,sqrt(2)-1) and we have a slope 2*(2-sqrt(2)).

We should be able to write down the equation without further difficulty.

y - (sqrt(2)-1) = 2*(2-sqrt(2))*(x-pi/8), which leads pretty much to your solution. Good work. I get y = 1.1716*x - 0.0459, rounding at the very end instead of using rounded values in intermediate steps.