Graffing sine and cosine functions?

**homeylova223** · January 8th 2011, 11:08 AM

Can anyone explain to me how to do this as I am confused?

Find the values of angle which each equation is true
1. cos= -1
The back of the book says the answer is pi+ 2pi(n)
Know I know the cos of pi which is 180 degrees is -1
But I am confused on how to get 2pi(n)

**homeylova223** · January 8th 2011, 11:10 AM

Sorry I spelled graphing wrong

**snowtea** · January 8th 2011, 11:11 AM

180 + 360 degrees is the same thing as 180 degrees (rotate a whole circle once)
so is 180 + 360n.

**pickslides** · January 8th 2011, 11:13 AM

The books answer is correct $\pi\pm 2n\pi$

$\cos x =-1$ has repeated solutions given the nature of the function.

The $\pm 2n\pi$ part comes from the period of the function and where the solution repeats, do you know how to find the period?

**homeylova223** · January 8th 2011, 11:26 AM

No I am not sure how to find the period. Can you also show me how you would do sin angle=1?

**pickslides** · January 8th 2011, 11:32 AM

Originally Posted by homeylova223

No I am not sure how to find the period. Can you also show me how you would do sin angle=1?

The period of $\sin bx$ and $\cos bx$ is $\frac{2\pi}{b}$

The solution for $\sin x = 1 \implies x = \frac{\pi}{2}$ Now what is the period?

**homeylova223** · January 8th 2011, 11:41 AM

The period is 2pi. I see know because it repeat itself every 2pi I was looking at a chart from my book and your example and I just noticed that. So in order to test my understanding would sin angle=1 be pi/2+pi(n)?

**rtblue** · January 8th 2011, 05:02 PM

Take a look at the graph of cosx. That should help you.

y=cosx

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