Unformatted text preview: Example
Evaluate the following: √
3
−1 (i) sin
−
2
(ii) tan−1 √ 3 (iii) sec−1(1)
Solution:
√ √
3
3
−
(i) The quantity sin
, equivalently, arcsin −
, is that angle θ ∈ [−π/2, π/2]
2
2
for which
√
3
sin θ = −
.
2
√
But for θ = −π/3 it follows that sin θ = − 3/2, so
√
3
π
−1
sin
−
=− .
2
3
−1 (ii) tan−1 √ 3 is that angle θ ∈ (−π/2, π/2) for which
tan θ = But tan(π/3) = √ √ 3. 3 so it follows that
tan−1 √ 3= π
.
3 (iii) To evaluate sec−1 (1) = arcsec−1 (1), one must determine the angle θ ∈ [0, π/2) ∪ (π/2, π ]
which uniquely satisﬁes
sec(θ) = 1, or cos(θ) = 1
= 1.
1 Since cos(0) = 1, it follows that
sec−1 (1) = 0.
Note: For an easy way to remember the trigonometric functions at standard angles, see
the document Notes on sine and cosine functions which is linked to the statement of the
WeBWorK problem. ...
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.
 Fall '10
 KIHYUNHYUN
 Calculus

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