Chapter 1 | Functions and Graphs
89
⎛ ⎝ cos
⎛ ⎝ 5 π 4
⎞ ⎠
⎞ ⎠
c. cos −1
⎛ ⎝ cos
⎛ ⎝ 2 π 3
⎞ ⎠
⎞ ⎠
d. sin −1
Solution
a. Evaluating sin −1 ⎛
⎝ − 3/2 ⎞ ⎠ is equivalent to finding the angle θ such that sin θ =− 3/2 and
− π /2≤ θ ≤ π /2.
θ =− π /3
The angle
satisfies these two conditions.
Therefore,
sin −1 ⎛
⎝ − 3/2 ⎞
⎠ =− π /3. b. First we use the fact that tan −1 ⎛
⎝ −1/ 3 ⎞
⎠ =− π /6. Then tan( π /6) =−1/ 3. Therefore,
⎛ ⎝ tan −1
⎞ ⎠ =−1/ 3.
⎛ ⎝ −1/ 3 ⎞ ⎠
tan
c. To evaluate cos −1 ⎛ ⎠ , first use the fact that cos(5 π /4) =− 2/2. Then we need to find the angle θ such that cos( θ ) =− 2/2 and 0≤ θ ≤ π . Since 3 π /4 satisfies both these conditions, we have cos ⎛ ⎝ cos −1 (5 π /4) ⎞ ⎠ =cos ⎛ ⎝ cos −1 ⎛ ⎝ − 2/2 ⎞ ⎠ ⎞ ⎠ =3 π /4. ⎝ cos(5 π /4) ⎞ d. Since cos(2 π /3) =−1/2, we need to evaluate sin −1 (−1/2). That is, we need to find the angle θ such that sin( θ ) =−1/2 and − π /2≤ θ ≤ π /2. Since − π /6 satisfies both these conditions, we can conclude that sin −1 (cos(2 π /3)) = sin −1 (−1/2) =− π /6.
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