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 2⁴·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.