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Hello
How do you do
Find the exact value of the following
cos (15 "Pie" / 2 )
Sorry dont know how to write pie on a keyboard.
What is the question asking me to do, just put this expression into a calculator, is there a method to doing this kind of question?
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Keep on subtracting 2π until you get a number that is greater than or equal to 0 and less than 2π.
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You should know how to derive "common" values for sine and cosine. For this problem, you use a right triangle with legs of length 1. Find the length of the hypotenuse, and since the triangle is isosceles, the other two angles are pi/2. Now you can find sine and cosine of pi/2. But you must figure out the sign of the angles. Use the mnemonic:
ASTC - All students take calculus
In quadrant 1, All values of trig functions are positive. In quadrant 2, only sine is positive. In quadrant 3, only tangent is positive, and in quadrant 4, only cosine is positive.
No calculator needed.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Find the length of the hypotenuse, and since the triangle is isosceles, the other two angles are pi/2.
No, they are not.
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as fairfax said, change 15pi/2 to 3pi/2 -->then take cos(3pi/2)= 0
teacher resources--my online edu dir
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Ricky wrote:Find the length of the hypotenuse, and since the triangle is isosceles, the other two angles are pi/2.
No, they are not.
Whoops, my mistake. Thanks Jane. For the angles pi/2 (and multiples of it), draw the unit circle remembering that the x position of a point on the circle is cos(theta) and the y position is sin(theta). The point (0, 1) (the very top of the circle) corresponds to theta=pi/2, and so cos(pi/2) = 0 and sin(pi/2) = 1.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Remember that memorising the unit circle isn't necessary, the fact that it follows a clear pattern should make it easy enough to obtain the trigonometric values needed for the question.
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