5.1.
sin and
cos are
in the range [–1, 1], csc
is in the range (–,
–1]
[1, ).
Therefore, the statements 3), 4) and 5) are incorrect.
5.2.
The value of y is maximum, when the value of sin
x is minimum. Minimum of sin x is –1. Therefore,
maximum of y is 5 + 3 = 8.
The value of y is minimum, when the value of sin
x is maximum. Maximum of sin x is 1. Therefore,
minimum of y is 5 – 3 = 2.
5.3.
5.4.
1) cos (180°– )
= – cos ;
2) sin (270°+ )
= – cos ;
3) cos (90°+ )
= – sin .
5.5.
5.6.
sin 2000° = sin (5·360°+ 180°+
20°) = sin (180°+ 20°) = –
sin 20°.
5.7.
tan(180°+)
= sin (180° + )/
cos(180° + )
= (sin )/(cos
)
= tan.
So, tan(180° + )
= tan.
5.8.
1) sin(– 30°) = – sin(30°)
= – ½
2) tan(– 45°) = – tan(45°)
= –1.
3) cos
= cos (– )
= – 0.4.
4) sin 79° = sin (90°– 11°)
= cos 11° = cos (– 11°) = 0.98.
5.9.
Let h(x) = f(x)·g(x).
1) Let f and g be even functions: f(–x)
= f(x) and g(–x) = g(x). Then
h(–x) = f(–x)·g(–x) =
f(x)·g(x) = h(x). It means, that h is an
even function.
2) Let f and g be odd functions:
f(–x) = –f(x) and g(–x) = –g(x).
Then h(–x) =
f(–x)·g(–x) =[–f(x)]·[–g(x)]
= f(x)·g(x) = h(x). So, h is again an even
function.
3) Let f be an even function,
and g be an odd function. Then h(–x) = f(–x)·g(–x)
=
f(x)·[–g(x)] = –f(x)·g(x)
= –h(x). It means, that h is an odd function.
5.10.
1) even; 2) neither; 3) neither; 4) odd; 5) odd.
5.11.
5.12.
If we accept 0 to be a period of a function, then
any function would be a periodic one with the
period of 0, and, therefore, this term will not
make any difference.
5.13.
Otherwise, the function f would have infinite
many periods: any number T = U·n, where
n is any natural number, would be a period of
f.
5.14.
180°.
5.15.
1) periodic with the period of 180°
2) periodic with the period of 360°.
3) nonperiodic;
4) y= cos(5x+3) = cos[5(x+T)+3] = cos(5x+3+5T).
Hence the cosine has the period of 360°, 5T
= 360°. Therefore, the function is a periodic
one with the period of 360°/5 = 72°.
5.16.
From the cosine theorem, c^{2} = a^{2}
+ b^{2}  2abcos,
where
is the angle between a and b. From here, cos
= (a^{2} + b^{2}  c^{2})/(2ab)
= E/(2ab).
If E > 0, cos >
0 and <
90°. Since c is a biggest side,
is a biggest angle. Therefore, other angels also
are less than 90°, and the triangle is acute.
If E < 0, cos<
0 and >
90°. The triangle is obtuse.
If E = 0, cos
= 0 and =
90°. The triangle is right.
5.17.
Since cot 90° = 0, the entire product also
will be zero.
5.18.
Let’s split the given expression into four
parts: E_{1}· E_{2}·
E_{3}· E_{4}, where
E_{1} = tan 5°·tan 15°·
…·tan
85°,
E_{2} = tan 95°·tan 105°·
…·tan
175°,
E_{3} = tan 185°·tan 195°·
…·tan
265°,
E_{4} = tan 275°·tan 285°·
…·tan
355°.
According to reduction formulas,
E_{2} = tan (90° + 5°)·tan
(90° + 15°)· …·
tan (90° + 85°) = (–1)^{9}·
cot 5°·cot 15°· …·
cot 85°.
Therefore, E_{1}· E_{2}
= (–1)^{9}. Similar,
E_{4} = tan (90° + 185°)·tan
(90° + 195°)· …·
tan (90° + 265°) = (–1)^{9}·
cot 185°·cot 195°· …·
cot 265°.
Therefore, E_{3}· E_{4}
= (–1)^{9} and finally, E_{1}·
E_{2}· E_{3}· E_{4}
= 1.
