# Math Help - Absolute extreme values of sin and cos

1. ## Absolute extreme values of sin and cos

Hi there just wondering if someone could answer a question I have.
I've been given the function:

f(x)= cos x + sin x with a domain of [0,π] π <--- is pi (3.14)

and it wants me to determine the coordinates of the point where the tangent is horizontal.

I know first off I must take the derivative of this and I get

f ' (x)= cos x - sin x (i've just rearanged so the negative isn't infront)

0=cos x - sin x (set to zero)

sin x = cos x move sin over to make it equal

(sin x = cos x) / cos x divide both sides by cos

tan x = 1 this gives us the idenity (sinx/cosx =tanx) and 1

**this is where I get stuck I don't understand where I go from here to determine the coordinates if someone could show me it be much appreciated. Thank you

2. Originally Posted by darkfenix21
Hi there just wondering if someone could answer a question I have.
I've been given the function:

f(x)= cos x + sin x with a domain of [0,π] π <--- is pi (3.14)

and it wants me to determine the coordinates of the point where the tangent is horizontal.

I know first off I must take the derivative of this and I get

f ' (x)= cos x - sin x (i've just rearanged so the negative isn't infront)

0=cos x - sin x (set to zero)

sin x = cos x move sin over to make it equal

(sin x = cos x) / cos x divide both sides by cos

tan x = 1 this gives us the idenity (sinx/cosx =tanx) and 1

**this is where I get stuck I don't understand where I go from here to determine the coordinates if someone could show me it be much appreciated. Thank you
from the unit circle ...

$
\sin{x} = \cos{x}$
at $x = \frac{\pi}{4} + k\pi \, ; \, k \in \mathbb{Z}
$

3. Ahh that makes much more sense to me. I guess I went to far ahead with simplifying it for my own good.