Let's say I have cos(5pi/3). How do I know this equals 0.5 without using a calculator?
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Let's say I have cos(5pi/3). How do I know this equals 0.5 without using a calculator?
thanks but hopefully someone has a better explanation (:
and you wrote pi/6 twice...
I know how to do it I guess by converting it from radian to degrees: cos(5pi/3) = cos300=cos60=0.5
but isn't there a way to measure it by radian?
hehehehe... while pasting forgot to change that pi/3 :D:D:D
but if you draw your self that angle ... you'll realize that angle ofis the same as the angle of
and you know that angle that is
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because cos is positive in first and 4th quadrant :D:D:D:D get it ?
when drawn circle, drawand add it 5 times and you'll see that is the angle
:D
P.S. you realize thatso
is just for the
smaller than
so it have to be
The key concept her is to realize that 1 radian is the angle that is covered if you wrap a string of length 1 x radius around the perimeter of a circle. By convention we usually just consider a circle of radius 1, which would have a total perimeter length 2 pi. Hence a string of length 2 pi would wrap precisely once around this unit circle, and so the angle 2 pi radians is eqivalent to 360 degrees. Similarly, pi/2 radians is 180 degreees, pi/3 radians = 120 degrees, etc. Now, when first learning trig there is a natural tendency for some students to resist thinking in radians, and try to convert everything to degrees in their head. But trust me - you'd be much beter off getting to a point where you can "visualize" what pi/4 radians means without first thinking about degrees. Just consider that each quadrant of the circle is pi/2 radians, and so 2/3 pi is in the second quadrant, 5/3 pi is in the fourth, etc. So get used to visualizing how it is that tan(pi/4) = 1, and you'll be well on your way.
Without using a calculator...
That's
Hence, if you draw a unit-circle centred at the origin, you can draw a right-angled triangle
with hypotenuse going off at 60 degrees to the x-axis,
since Cos(angle) gives the x co-ordinate, and Cos(-A)=CosA.
This allows us to draw a regular hexagon, made of equilateral triangles as shown in the attachment.
From that, we can see that