Please help explain how to graph Cotangent and Sin x + cos x

Printable View

- Oct 26th 2007, 07:56 AMQuanHelp please: How to graph Sin x + Cos x, and Cotangent
Please help explain how to graph Cotangent and Sin x + cos x

- Oct 26th 2007, 09:33 AMred_dog
For cotangent use the values for $\displaystyle \displaystyle\frac{\pi}{6}, \ \frac{\pi}{4}, \ \frac{\pi}{3}, \ \frac{\pi}{2}, \ \frac{2\pi}{3}, \ \frac{3\pi}{4}, \ \frac{5\pi}{6}$ and the vertical asympthotes.

For the second one we have

$\displaystyle \sin x+\cos x=\sin x+\sin\left(\frac{\pi}{2}-x\right)=2\sin\frac{\pi}{4}\cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}\cos\left(\frac{\pi}{4}-x\right)$ - Oct 26th 2007, 10:39 AMticbol
1) to graph cot(x).

You should know how to graph tan(x).

cot(x) = tan(90 -x) ----in degrees.

cot(x)

= tan(90 -x)

= tan[-(x -90)]

= -tan(x -90) ------the tangent of a negative angle is the negative value of the tangent of that angle. View the angle to be in the 4th quadrant, where tangent is negative.

So, you graph -tan(x -90) instead of cot(x).

That is a basic tangent curve that is shifted 90 degrees to the right, and it is "inverted" because of the negative sign.

---------------------------------------------

2) to graph sin(x) +cos(x)

cos(x) = sin(90 -x) ----in degrees.

cos(x)

= sin(90 -x)

= sin[-(x -90)]

= -sin(x -90) -----same reasoning as in tan[-(x -90)].

So,

sin(x) +cos(x)

= sin(x) -sin(x -90)

Using the trig identity

sinA -sinB = 2cos[(A+B)/2]sin[(A-B)/2],

= 2cos[(x +x -90)/2]sin[(x -x +90)/2]

= 2cos(x -45)sin(45)

= 2cos(x -45)*[1 / sqrt(2)]

= sqrt(2)cos(x -45) --------------***

So graph sqrt(2)cos(x -45) instead of sin(x) +cos(x).

That is a cosine curve that has an amplitude of sqrt(2), and a horizontal shift of 45 degrees to the right.