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- Nov 8th 2012, 08:24 PMEraser147Derivatives
- Nov 8th 2012, 09:24 PMchiroRe: Derivatives
Hey Eraser147.

For your question, you have cos(a-b) = cos(a)cos(b) + sin(a)sin(b) and you also have tan(v) = -4. You also have that cos(a + 2*pi*n) = cos(a) where n is a whole number.

Now you can get v within a given branch by calculating v = arctan(-4) which means cos(v - 12*pi) = cos(v) = cos(arctan(-4)).

Now you have pythagoras' theorem with right angle triangles and you can use this to find cos(arctan(-4)).

Have you covered right angled triangles and how to use them to find things like cos(arcsin(x)) or sin(arctan(x))?