Calculus I by Dawkins P.

By Dawkins P.

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4π π 2π 11π 16π 17π 22π 23π 28π 29π , , , , , , , , , 15 3 3 15 15 15 15 15 15 15 2π 7π 8π 13π 14π π ,− ,− ,− ,− − ,− 15 15 15 15 15 15 Let’s work another example.  3π 3π  Example 4 Solve sin ( 2 x ) = − cos ( 2 x ) on  − ,   2 2  Solution This problem is a little different from the previous ones. First, we need to do some rearranging and simplification. sin(2 x) = − cos(2 x) sin(2 x) = −1 cos(2 x) tan ( 2 x ) = −1 So, solving sin(2 x) = − cos(2 x) is the same as solving tan(2 x) = −1 . At some level we didn’t need to do this for this problem as all we’re looking for is angles in which sine and cosine have the same value, but opposite signs.

At some level we didn’t need to do this for this problem as all we’re looking for is angles in which sine and cosine have the same value, but opposite signs. However, for other problems this won’t be the case and we’ll want to convert to tangent. aspx Calculus I Looking at our trusty unit circle it appears that the solutions will be, 3π + 2π n, n = 0, ±1, ±2, 4 7π 2x = + 2π n, n = 0, ±1, ±2, 4 2x = Or, upon dividing by the 2 we get all possible solutions. 3π + π n, n = 0, ±1, ±2, 8 7π x= + π n, n = 0, ±1, ±2, 8 x= Now, let’s determine the solutions that lie in the given interval.

Both of these angles along with their coordinates are shown on the following unit circle.  2π  3 From this unit circle we can see that sin  © 2007 Paul Dawkins 24 3   2π and sin  − =  2  3 3  . aspx Calculus I This leads to a nice fact about the sine function. The sine function is called an odd function and so for ANY angle we have sin ( −θ ) = − sin (θ ) [Return to Problems] 7π π π from the = π + so this means we would rotate down 6 6 6 7π π π negative x-axis to get to this angle. Also − =−π − so this means we would rotate up 6 6 6 (b) For this example notice that from the negative x-axis to get to this angle.

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