Illustrate local linearity for (x) = x 2 by zooming in on the graph at x =

Illustrate local linearity for ƒ(x) = x² by zooming in on the graph at x = 0.5 (see Example 6).

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Hypotenuse 5 FIGURE 16 FIGURE 17 a Adjacent 2 (o Opposite √21 EXAMPLE 6 For between 0 and 27, the equation cos 0 = has a solution in (0,7) and a solution in (3, 27). Calculate tan in each case. Solution First, using the identity cos² 0+ sin² 0 = 1, we obtain 4 sine = √1 cos² 0 = ₁/1- == 25 √21 5 If 0 <<, then sin is positive and we take the positive square root: tan = sin 0 cos ( = √21/5 2/5 √21 2 To visualize this computation, draw a right triangle with angle such that cos 0 = as in Figure 16. The opposite side then has length √21 = √52-22 by the Pythagorean Theorem. √21 If ³ < 0 < 27, then sin is negative and tan0 = -³ We conclude this section by quoting the Law of Cosines (Figure 17), which is a generalization of the Pythagorean Theorem (see Exercise 62). THEOREM 1 Law of Cosines If a triangle has sides a, b, and c, and is the angle opposite side c, then a² + b² - 2ab cos 0 If 0 = π/2, then cos 0 = 0 and the Law of Cosines reduces to the Pythagorean Theorem.