Calculus I by Dawkins P.

By Dawkins P.

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Extra resources for Calculus I

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Also, we are going to use 4 decimal places of accuracy in the work here. You can use more if you want, but in this class we’ll always use at least 4 decimal places of accuracy. Next, and somewhat more importantly, we need to understand how calculators give answers to inverse trig functions. We didn’t cover inverse trig functions in this review, but they are just inverse functions and we have talked a little bit about inverse functions in a review section. The only real difference is that we are now using trig functions.

Aspx Calculus I 7π 4 (c) Here we should note that = 2π − π 4 so note that this angle will be the mirror image of 7π π and − are in fact the same angle! Also 4 4 π 4 in the fourth quadrant. The unit circle for this angle is Now, if we remember that tan ( x ) = sin ( x ) we can use the unit circle to find the values the cos ( x ) tangent function. So,  7π tan   4   π  sin ( − π 4 ) − 2 2 = = −1 . tan  −  = = 2 2   4  cos ( − π 4 ) π   = 1 and we can see that the tangent function is also called an 4 On a side note, notice that tan  odd function and so for ANY angle we will have tan ( −θ ) = − tan (θ ) .

4π 2π ( 4 ) 28π x=+ = < 2π 15 5 15 π 2π ( 4 ) 29π x= + = < 2π 3 5 15 n = 5. 4π 2π ( 5 ) 34π x= + = > 2π 15 5 15 π 2π ( 5 ) 35π x= + = > 2π 3 5 15 Okay, so we finally got past the right endpoint of our interval so we don’t need any more positive n. Now let’s take a look at the negative n and see what we’ve got. n = –1 . aspx Calculus I n = –2. 4π 2π ( −2 ) 8π + =− > −π 15 5 15 π 2π ( −2 ) 7π =− > −π x= + 3 5 15 x= n = –3. 4π 2π ( −3) 14π + =− > −π 15 5 15 π 2π ( −3) 13π =− > −π x= + 3 5 15 x= n = –4.

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