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**Unformatted text preview: **e real number t (or, equivalently, an angle measuring
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t radians) which lies between 0 and π with cos(t) = 2 . We know t = π meets these
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criteria, so arccos 2 = π .
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√ (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
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(d) To ﬁnd arcsin − 1 , we seek the number t in the interval − π , π with sin(t) = − 2 . The
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answer is t = − π so that arcsin − 1 = − π .
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6 (e) Since 0 ≤ π ≤ π , we could simply invoke Theorem 10.26 to get arccos cos π = π .
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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
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√ and cos(t) =
(f) Since 11π...

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