Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
8. b. arccot(√3)
Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
8. b. arccot(√3)
Find the values in Exercises 9–12.
9. sin(arccos((√2)/2))
Find the values in Exercises 9–12.
11. tan(arcsin(-1/2))
132. What is special about the functions
f(x) = arcsin((1/√(x²+1)) and g(x)=arctan(1/x)?
Explain.
[Technology Exercise] In Exercises 139–141, find the domain and range of each composite function. Then graph the compositions on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see.
139. a. y=arctan(tan x)
[Technology Exercise] In Exercises 139–141, find the domain and range of each composite function. Then graph the compositions on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see.
141. a. y=arccos(cos x)
Since the hyperbolic functions can be expressed in terms of exponential functions, it is possible to express the inverse hyperbolic functions in terms of logarithms, as shown in the following table.
sinh⁻¹x = ln(x + √(x² + 1)), -∞ < x < ∞
cosh⁻¹x = ln(x + √(x² - 1)), x ≥ 1
tanh⁻¹x = (1/2)ln((1+x)/(1-x)), |x| < 1
sech⁻¹x = ln((1+√(1-x²))/x), 0 < x ≤ 1
csch⁻¹x = ln(1/x + √(1+x²)/|x|), x ≠ 1
coth⁻¹x = (1/2)ln((x+1)/(x-1)), |x| > 1
Use these formulas to express the numbers in Exercises 61–66 in terms of natural logarithms.
63. tanh⁻¹(-1/2)
Since the hyperbolic functions can be expressed in terms of exponential functions, it is possible to express the inverse hyperbolic functions in terms of logarithms, as shown in the following table.
sinh⁻¹x = ln(x + √(x² + 1)), -∞ < x < ∞
cosh⁻¹x = ln(x + √(x² - 1)), x ≥ 1
tanh⁻¹x = (1/2)ln((1+x)/(1-x)), |x| < 1
sech⁻¹x = ln((1+√(1-x²))/x), 0 < x ≤ 1
csch⁻¹x = ln(1/x + √(1+x²)/|x|), x ≠ 1
coth⁻¹x = (1/2)ln((x+1)/(x-1)), |x| > 1
Use these formulas to express the numbers in Exercises 61–66 in terms of natural logarithms.
65. sech⁻¹(3/5)