# Absolute extreme values of sin and cos

• Apr 28th 2010, 06:31 AM
darkfenix21
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
• Apr 28th 2010, 10:28 AM
skeeter
Quote:

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 ...

$\displaystyle \sin{x} = \cos{x}$ at $\displaystyle x = \frac{\pi}{4} + k\pi \, ; \, k \in \mathbb{Z}$
• Apr 28th 2010, 10:46 AM
darkfenix21
Ahh that makes much more sense to me. I guess I went to far ahead with simplifying it for my own good.