Stitz-Zeager_College_Algebra_e-book

Since cosx and sinx have period 2 its not too dicult

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Unformatted text preview: e real number t (or, equivalently, an angle measuring 2 1 t radians) which lies between 0 and π with cos(t) = 2 . We know t = π meets these 3 1 criteria, so arccos 2 = π . 3 √ (b) The value of arcsin 2 2 is a real number t between − π and 2 number we seek is t = π . Hence, arcsin 4 √ 2 2 π 2 √ with sin(t) = √ is arccos − 2 2 = 2 2 The = π. 4 √ (c) The number t = arccos − 2 2. √ lies in the interval [0, π ] with cos(t) = − 2 2. Our answer 3π 4. 1 (d) To find arcsin − 1 , we seek the number t in the interval − π , π with sin(t) = − 2 . The 2 22 answer is t = − π so that arcsin − 1 = − π . 6 2 6 (e) Since 0 ≤ π ≤ π , we could simply invoke Theorem 10.26 to get arccos cos π = π . 6 6 6 However, in order to make sure we understand why this is the case, we choose to work the example through using the definition of arccosine. Working from the inside out, √ √ arccos cos π = arccos 23 . Now, arccos 23 is the real number t with 0 ≤ t ≤ π 6 √ and cos(t) = (f) Since 11π...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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