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Unformatted text preview: 132 Chapter 6 Additional Topics 6.1 Example 6.8 2 3 radicallow 13 Figure 6.1.1 Solve the equation 2 sin 3 cos = 1. Solution: We will use the technique which we discussed in Chapter 5 for finding the amplitude of a combination of sine and cosine functions. Take the coefficients 2 and 3 of sin and cos , respectively, in the above equation and make them the legs of a right triangle, as in Figure 6.1.1. Let be the angle shown in the right triangle. The leg with length 3 > 0 means that the angle is above the xaxis, and the leg with length 2 > 0 means that is to the right of the yaxis. Hence, must be in QI. The hypotenuse has length radicallow 13 by the Pythagorean Theorem, and hence cos = 2 radicallow 13 and sin = 3 radicallow 13 . We can use this to transform the equation to solve as follows: 2 sin 3 cos = 1 radicallow 13 parenleftBig 2 radicallow 13 sin 3 radicallow 13 cos parenrightBig = 1 radicallow 13(cos sin...
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This note was uploaded on 01/21/2012 for the course MAC 1130 taught by Professor Dr.cheun during the Fall '11 term at FSU.
 Fall '11
 Dr.Cheun
 Calculus, Addition

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