Asked by help!
Find the exact value of each expression, if it exists:
the -1 are representing the inverse functions!
(a) sin -1 (-√2/2)
(b) cos−1 (−1)
(c) sin( sin−1 (π))
(d) cos−1(cos(−4π/ 3))
(e) tan−1 (tan(0.6))
(f) cos−1( cos13π/12))
(g) tan( sin−1 (8 /17))
(h) cot (cos−1 ( 5 /13))
these are what i have. I'm looking to make sure i have done them correctly. If i haven't if you could show me the work to the ones i got wrong that would be great!
a. -π/4
b. π
c. undefined because domain [-1,1]
d. -4π/3
e. .6
f. undefined because domain [0, π]
g. 8/15
h. 5/12
the -1 are representing the inverse functions!
(a) sin -1 (-√2/2)
(b) cos−1 (−1)
(c) sin( sin−1 (π))
(d) cos−1(cos(−4π/ 3))
(e) tan−1 (tan(0.6))
(f) cos−1( cos13π/12))
(g) tan( sin−1 (8 /17))
(h) cot (cos−1 ( 5 /13))
these are what i have. I'm looking to make sure i have done them correctly. If i haven't if you could show me the work to the ones i got wrong that would be great!
a. -π/4
b. π
c. undefined because domain [-1,1]
d. -4π/3
e. .6
f. undefined because domain [0, π]
g. 8/15
h. 5/12
Answers
Answered by
Steve
d: the arccos has principal vales from 0 to π.
cos(-4π/3) = -1/2
cos^-1(-1/2) = 2π/3
f:
cos(13π/12) = -cos(π/12) = -x
arccos(-x) = π-arccos(x) = 11π/12
The others look good. Be careful with the principal values stuff.
cos(-4π/3) = -1/2
cos^-1(-1/2) = 2π/3
f:
cos(13π/12) = -cos(π/12) = -x
arccos(-x) = π-arccos(x) = 11π/12
The others look good. Be careful with the principal values stuff.
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