Thread: trigonometry problem

cos α= -0,007275 α=? (α is in grad ) how is this problem solved ?

you take the arccos/inverse cos of both sides of the equation.

On your calculator the inverse cos button will look some thing like this

Let's start by reviewing what the trig functions mean. The sin(a) = b tells you that the ratio b is made up of taking some angle a and taking its opposite side divided by the hypotenuse of the resultant right triangle. This let's us find a ratio b given some angle a. However, we're going to need to use a slightly different method to find a given some ratio b.

The inverse trig functions, as mentioned above, are sort of like the opposite operation to regular trig functions. sin^-1, cos^-1, and tan^-1 are all inverse functions (they are also sometimes written as arcsin, arccos, and arctan - don't worry, they mean exactly the same thing). The inverse trig functions tell us that, given some ratio b, what the angle would have to be. So, if sin(a) = b, then arcsin(b) = a.

The answer posted above is correct, and I hope this clarifies how to solve the problem.

Originally Posted by blaugrana_paok4

cos α= -0,007275 α=? (α is in grad ) how is this problem solved ?

slove 1-2sin2A=cos3a+sin3a/cos A-sin A