14.1.
9) 3/4. You can use the formula
(14.1) or the picture
10) /2.
See the formula (14.2).
11) Let
= 7 – 2.
Then 0 < <
/2
and 7 =
+ 2.
Therefore, arcsin(sin 7) = arcsin (sin )
=
= 7 – 2.
14.2.
14.3.
1) It can be seen from the picture
2) Similar to 1).
3) Let
= /
2  arc cot x. Then /2
< <
/2
and
tan
= tan(/
2  arccot x) = cot(arccot x) = x. So, tan
= x.
Since /
2 < <
/
2, arctan x =
= /2
 arccot x
4) Let =
arctan x,
= arctan(1/x). Since x > 0, the angles
and
lie in the first quadrant. The solution can be
seen from the picture
14.4.
It follows from more general statement (prove
this statement by yourself): if a function f(x)
is an odd function and is defined on symmetric
interval (–a, a), then the inverse function
f ^{1}(x) is also odd.
14.5.
It can be seen from the graph of arccot x (see
the solution of the problem 13.15).
14.6.
2) Using the formula (7.21), write the equation
as 2 sin x cos2x =0. From here,
sin x = 0 and cos 2x = 0. The solutions are
x = n
and x = /4
+ n /2.
3) Write the equation as sin x = sin x/cos x
or sin x (cos x – 1) = 0. From here,
sin x = 0 and cos x = 1. The solutions are
x = n
4) This is a particular case of the equation
(14.4). Using the general approach, the given
equation can be written as 5(sin x · 3/5
+ cos x · 4/5) = 5 or
cos (x – )
= 1, where
= arcos(4/5). From here, x –
= 2n,
and we get the solutions: x = arcos(4/5)
+ 2n.
14.7
Left part of the identity can be written as a
product of three parts A·B·C, where
A = cos /15·cos
2/15·cos
4/15·cos
7/15,
= cos 3/15·cos
6/15
and
C = cos 5/15
= cos /3
= ½ . To process A, multiply and divide
it by 2^{4}·sin /15,
and apply formula (7.26) four times. Also, use
that sin 8/15
= sin (–
8/15)
= sin /15,
and sin14/15
= sin /15.
As a result, A = 2^{4}. Similar, B =
2^{2}.
14.8.
14.9.
