I have the values of Sin, Cos and Tan values of general angles (0,30,45,60,90,120,135,150,180,210.... up till 360)
I know how to get the values of Sin, Cos and Tan until 90 degrees. It'd be easier for me to know how to get the value of these instead of memorizing the huge table. I cant find anything on this except on Wikipedia, but the methods are too complex for my knowledge.
This is in my AS Pure Maths syllabus Edexcel(british examination board)
Any help will be appreciated
Sin is not enough plus i already know how to get those values for Sin Cos and Tan. We can draw the Isosceles triangle and the Equilateral triangle. The we right out the angles 30,60,45 and 90...
Everyone's probably knows this:P But how do you do this if the angles are greater than 90
If you are allowed a calculator it will usually be able to give you a value in surd form so i would just use that.
Edit: You could also try splitting the large angle into smaller angles that you know, e.g. 135 = 90 + 45 both of which you know.
Let's say you had an angle of 135 degrees. A reference angle is measured with respect to the x axis. Since the angle falls in the second quadrant, that means our reference angle is 180-135=45 degrees.
So in evaluating the trig functions there, you need to note that:
If the original angle, , is in the second quadrant [i.e. ], its reference angle is (or in radians, ). Take note that only sine and cosecant have positive values here!
If the original angle, , is in the third quadrant [i.e. ], its reference angle is (or in radians, ). Take note that only tangent and cotangent have positive values here!
If the original angle, , is in the fourth quadrant [i.e. ], its reference angle is (or in radians, ). Tke note that only cosine and secant have positive values here!
Does this make sense?
I think I understand what a quadrant is but we haven't studied radians at all. I haven't seen any questions on angles above 90 degrees so I'm guessing no need to worry.
I'm in AS maths and the unit is C1 so we aren't allowed to use a calculator.
Thanks for your help.