By Dawkins P.
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Additional resources for Calculus I
E. angles that end at π 6 . Remember that all this says is that we start at π 6 then rotate around in the counter-clockwise direction (n is positive) or clockwise direction (n is negative) for n complete rotations. The same thing can be done for the second solution. So, all together the complete solution to this problem is π + 2π n , n = 0, ± 1, ± 2, ± 3, 6 11π + 2π n , n = 0, ± 1, ± 2, ± 3, 6 As a final thought, notice that we can get − π 6 by using n = −1 in the second solution. Now, in a calculus class this is not a typical trig equation that we’ll be asked to solve.
It’s important to notice that all of these examples used the fact that if you know the first quadrant of the unit circle and can relate all the other angles to “mirror images” of one of the first quadrant angles you don’t really need to know whole unit circle. edu. Another important idea from the last example is that when it comes to evaluating trig functions all that you really need to know is how to evaluate sine and cosine. The other four trig functions are defined in terms of these two so if you know how to evaluate sine and cosine you can also evaluate the remaining four trig functions.
We could just as easily do the work with the original angle our calculator gave us. Now, this is where is seems like we’re just randomly making changes and doing things for no reason. 6345 The fact that we used the calculator answer here seems to contradict the fact that we used a different angle for the first above. The reason for doing this here is to give a second angle that is in the range [ 0, 2π ] . 9177 . 2832) and the general rule of thumb is to keep the initial angles as small as possible.