The question asks to find the sine of 5pi/7. My teacher says to add/subtract an n value of pi to get a value on the unit circle; and then I find the sine. I've been at this problem for awhile and there has to be a better way than guessing how many pi's it will take to get a value on the circle.
What is the answer?
I tried to add 2 pi.
5pi/7 + 14pi/7 = 19pi/7 (Not on the unit circle!)
Then 4 pi.
5pi/7 + 28pi/7 = 33pi/7
The idea is that its supposed to reduce to a value on the unit circle. . . But I tried some other numbers, and I can't get it!
sin(5pi/7)
=sin(7pi-2pi/7)
=sin(7pi/7-2pi/7)
=sin(pi-2pi/7)
=sin(pi)cos(2pi/7)-cos(pi)sin(2pi/7)]
=0-(-1)sin(2pi/7)
=sin(2pi/7)
as for the unit circle that ur teacher mentioned it is just a convention to remember what sign is taken by sin, cos and tan in which quadrant. the value neednot satisfy any circle's equation.
s | A
2 | 1
---------------
T | C
3 | 4
for ur question we found out that sin(5pi/7)=sin(pi-2pi/7) which means that it lies on the quadrant 2 where sin is always positive . and thats wat we found out by expanding it using sine(a+b)=sinacosb+cosasinb.was this helpful????please comment.