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472 Worked-Out Solutions to Exercises: 1-9

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y 3 y = sin 2x 2 1 x 1 3p

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y = sin x

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Illustration for the solution to Problem 6 in Chap 2

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7 The secant is the reciprocal of the cosine The cosine has a range of output values covering the closed interval [ 1,1] That means 1 cos x 1 for all real-number input values x We can break this fact down into the two statements 1 cos x 0 and 0 cos x 1 The reciprocals are the one-ended ranges 1/cos x 1 and 1 1/cos x We can rewrite the above as sec x 1

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and 1 sec x These two inequalities tell us that the secant function never attains any values in the open interval ( 1,1) 8 The cosecant is the reciprocal of the sine The sine has a range of output values covering the closed interval [ 1,1] In other words, no matter what the real-number input x, we always have 1 sin x 1 We can split this into the statements 1 sin x 0 and 0 sin x 1 Therefore, 1/sin x 1 and 1 1/sin x We can rewrite the above as csc x 1 and 1 csc x The output of the cosecant function, like the output of the secant, is never equal to anything in the open interval ( 1,1) 9 We start with the Pythagorean theorem for the sine and cosine, which is sin2 q + cos2 q = 1 When we subtract sin2 q from either side, we get cos2 q = 1 sin2 q

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474 Worked-Out Solutions to Exercises: 1-9

We can divide through by the square of the cosine, as long as we don t allow q to be an odd-integer multiple of p /2 (If it is, then cos q = 0, which means that cos2 q = 0 and we end up dividing by 0) Performing the division, we get cos2 q /cos2 q = 1/cos2 q sin2 q /cos2 q The left-hand side of this equation is equal to 1 regardless of the value of q, as long as it s not one of the forbidden values The first term on the right-hand side is the reciprocal of the cosine squared, which is the same as the secant squared The second term on the right-hand side is the ratio of the sine squared to the cosine squared, which is same as the tangent squared We can therefore simplify the above equation to 1 = sec2 q tan2 q which is, of course, the same as sec2 q tan2 q = 1 10 Again, we start with the Pythagorean theorem for the sine and cosine sin2 q + cos2 q = 1 This derivation goes a lot like the solution to Problem 9 Let s subtract cos2 q from either side to get sin2 q = 1 cos2 q We can divide through by the square of the sine, provided that we don t allow q to be an integer multiple of p (If it is, then we end up dividing by 0) This gives us sin2 q / sin2 q = 1/sin2 q cos2 q /sin2 q The left-hand side of the above equation is always equal to 1, as long as q is not one of the forbidden values The first term on the right-hand side is the reciprocal of the sine squared; that s the same as the cosecant squared The second term on the righthand side is the ratio of the cosine squared to the sine squared That s the same as the cotangent squared We can therefore simplify the above equation to 1 = csc2 q cot2 q which can be rearranged to csc2 q cot2 q = 1