5.1 CONCLUSIONS INVOLVING V, BUT NOT 3 OR -, 159
(f) Prove that if A is any square matrix, then A + At is symmetric, whereas A - At
is antisymmetric.
(g) Prove that if A and B are 2 x 2 matrices, then lABl = IAl IB(.
- One approach to the inverse trigonometric functions is to define first the two
functions sin-' and tan-' directly as inverses to sin, restricted to the portion
[ - n/2,42] of its domain, and tan, restricted to ( - 11/2,71/2) respectively. That is,
y = sin-' x if and only if sin y = x, -n/2 I x I n/2, and y = tan-' x if and only
if tan y = x, - 42 < x < 42. Note that domain (sin- ') = range (sin) = [- 1, 11,
and domain (tan- ') = range (tan) = (- co, co).
Having done this, we may then define the other four inverse trigonometric func-
tions cos - l, cot - ', sec - l, and csc - in terms of sin - ' and tan - '. Specifically, we
may let
sec - ' x = cos - '(llx) csc-I x = sin-'(llx)
The first definition is motivated by the fact that the cosine of an acute angle equals
the sine of its complement, so that cos (42 - sin-' x) = sin (sin-' x) = x for any
x E [- 1, 11. The other definitions have similar trigonometric motivations. If you
have not previously done so in your calculus class, calculate the domain and range
of each of the preceding functions and sketch the graph of each. It can be proved,
using the definition of inverse of a function, that sin- and tan - are odd functions;
that is, sin- '(- x) = -sin- 'x and tan- '(- x) = -tan- 'x for all x in the respective
domains. Use these facts to prove:
(a) CO~-~(-X)=~-COS-~X, -1<x11
(b) sec-'(-x)=n-sec-'x, x~ -1 or x2l
(c) csc-'(-x)=-csc-'x, XI-1 or xrl
(d) cot-'(-x)=n-cot-'x, -co <x < co
For a geometric interpretation of these results, see Article 5.2, Exercise 5(c).
- Critique and complete (recall the instructions for this type of exercise, given
in Exercise 11, Article 4.1):
(a) THEOREM For any sets A, B, and C, (A n 6) A (A n C) = A n (6 A C)
"Proof"
(An 6) A (An C) =A n (B A C)
:. [(A n B) n (A n-C)'] u [(A n 6)' n (A n C)] = A n [(B n C') u (6' n C)]
:. [(A n B) n (A' u C)] u [(A' u 6') n (A n C)] = (A n 6 n C) u (A n 6' n C)
.'. [(A n B n A') u (A n B n C)] u [(A'n A n C) u (B'n A n C)]
=(AnBnC)u(AnB'nC)
.'. [a u (A n B n C)] u [a u (6' n A n C)] = (A n B n C) u (A n B' n C)
.'.(An BnC) u(AnB1n C)=(An BnC)u(An B'n C)