I need to do this without a calculator. I've tried addition/subtraction on the unit circle to get 17pi/5, but none of the combinations give me clean numbers to work with.
Also, cos^-1 if arc cosine, I believe (if you're unfamiliar with those signs).
2007-08-19 13:43:22 · 7 answers · asked by bubblebobboy 1 in Science & Mathematics ➔ Mathematics
We would like to use the fact that arccos(cosx) = x for 0<=x<=pi. Because 17pi/5 > pi, we cannot use this principle directly.
Observe that cosx is a periodic function with fundamental period 2pi. Hence cos(x + n2pi) = cosx whenever n is an integer (i.e. n= 0,1,-1, 2,-2,...)
Note that 17pi/5 -10pi/5 = 7pi/5 Hence cos(17pi/5) = cos(7pi/5). Observe that cosx is symmetric about the y-axis. In other words cos(-x) = cos(x). Hence cos(7pi/5) = cos(-7pi/5) = cos (- 7pi/5 + 2pi) = cos(10pi/5 - 7pi/5) = cos(3pi/5).
Now 0< 3pi/5< pi So arccos(cos[3pi/5]) = 3pi/5 by the above principle
**arccosx is the inverse of cosx for 0<=x<=pi**
2007-08-19 14:03:24 · answer #1 · answered by guyava99 2 · 0⤊ 0⤋
Value Of Cos 1
2016-12-15 04:25:32 · answer #2 · answered by Anonymous · 0⤊ 0⤋
cos 17pi/5 = cos 7pi/5 = cos 252 degrees
arccos( cos(252)) = 108 degrees
2007-08-19 14:13:54 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
cos(x) takes all values between 1 and -1 for x between 0 and Pi. The (numeric) calculator gives you the smallest x, for wich cos(x)=cos(17*Pi/5) which is x=3*Pi/5.
2007-08-19 14:00:46 · answer #4 · answered by Peter 1 · 0⤊ 0⤋
cos^-1(cos[17pi/5]) = 17pi/5
In the principal domain [0, pi), the answer is 3pi/5.
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Ideas: cos(17pi/5) = cos(-17pi/5) = cos(20pi/5 - 17pi/5) = cos(3pi/5)
2007-08-19 13:47:48 · answer #5 · answered by sahsjing 7 · 0⤊ 1⤋
We have 17π/5
cosine is periodic with period 10π/5
So cos(17π/5) = cos(7π/5)
= cos(-3π/5)
= cos(+3π/5) since cos(+x) = cos(-x)
Now arccos of this is simply
3π/5
2007-08-19 13:49:02 · answer #6 · answered by Dr D 7 · 0⤊ 1⤋
?
2016-05-17 11:13:59 · answer #7 · answered by kira 3 · 0⤊ 0⤋