Our geometrical definitions of cos θ and sin θ make sense not just for 0 < θ < π/2 as
in the picture but for all real values of θ of either sign. The graphs of these two functions
in the range (¡3π, 3π) are:
By considering the limiting cases of the previously illustrated right-angled triangle
when θ = 0 and when θ = π/2, we immediately deduce (in accordance with the graphs
just drawn) that
sin 0
π
cos
2
cos 0
π
sin
2
= 0,
(2.1)
= 0,
(2.2)
= 1,
(2.3)
= 1.
(2.4)
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2.2.2
Inverse Trigonometric Functions
From the form of the graph of the sine function, we see that it cannot have an inverse
as it stands, since each horizontal line y = c (with ¡1 < c < 1) cuts the graph in
infinitely many points. We restrict it to an interval on which it is an increasing function,
and the
choice for that interval is [¡π/2, π/2]. The increasing
£ πconventional
£ restriction
¤
¤
−1
π
sin : ¡ 2 , 2 ! [¡1, 1] has increasing inverse sin = arcsin : [¡1, 1] ! ¡ π2 , π2 . By
definition, given any y in [¡1, 1], sin−1 y is the unique angle between ¡π/2 and π/2 whose
sine is y.
NB. It is totally different from (sin y)−1 =cosec y.
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NB The graph of sin−1 y has vertical tangents at y = §1, so the slope
these two points.
dx
dy
is infinite at
Question: Over what restriction will cos have an inverse, arccos?
Answer: We restrict cos to be either increasing or decreasing. It is usual to choose the
decreasing restriction cos : [0, π] ! [¡1, 1], with inverse arccos : [¡1, 1] ! [0, π].
As for arcsin, given any y in [¡1, 1], cos−1 (y) = arccos y is the unique angle between 0 and π
whose cosine is y.
tan and arctan
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The function tan is increasing over many intervals, but we conventionally take the interval
containing x = 0: tan : (¡π/2, π/2) ! R with inverse arctan : R ! (¡π/2, π/2).
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