how do you put these in a fraction or square root??? can someone tell me how? i)sin120 ii)cos(-120) iii)tan135 iv)sin300 v)cos270 thx :)

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how do you put these in a fraction or square root??? can someone tell me how? i)sin120 ii)cos(-120) iii)tan135 iv)sin300 v)cos270 thx :)

Mathematics
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Use the unit circle
are the angles expressed in radians or degrees?
ill assume degrees; Q2 sin(120) = sin(180 - 60) = sin(60) = sqrt(3)/2 Q3 cos(-120) = cos(180 + 60) = -cos(60) = -(1/2) Q2 tan(135) = tan(180-45) = -tan(45) = -(1) Q4 sin(300) = sin(360-60) = -sin(60) = -sqrt(3)/2 yaxis cos(270)= cos(3*90) = cos(90) = 0

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all degrees
I thought so :) at my age you can never be to sure tho...
i dont get why you sqrt it tho :S and whats the cos(3*90) and why did you put something about y axis? soz :(
Q= quadrant that it ends up in which is important in order to determine whether or not the answer is a positive or negative. The cosine funtion actually tells you the value of the x coordinate. At 270 degrees you are pointing straight down on the y axis and x=0 so cos(90)-straight up, or cos(270)-straight dwon, equals 0.
lol... "sqrt" is how I do square root :)
i know its the square root but why did you have to do that? :S clueless :(
It is the definition of sin and cos. If you draw a right triangle with a 60 degree angle and a 30 degree angle you end up with a hypotenuse of (2) and a leg that is equal to (1). If we use the pythagorean theorum; x^2 + y^2 = r^2 we can find the length of the other leg. x^2 + (1)^2 = (2)^2 x^2 +1 = 4 x^2 = 4-1 x^2 = 3 ; sqrt both sides x= sqrt(3) That is it in a nutshell. It is just what happens when trying to figure out the measurements of a triangle. Sin = opposite leg/hypotenuse; sin(60) = sqrt(3)/2 Same thing for cosine; which is adjacent leg/hypotenuse for any given right triangle.

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