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Answers To The Exercises: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14
Lesson 5
 


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) non-periodic;

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, c2 = a2 + b2 - 2abcos, where is the angle between a and b. From here, cos = (a2 + b2 - c2)/(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: E1· E2· E3· E4, where
E1 = tan 5°·tan 15°· …·tan 85°,
E2 = tan 95°·tan 105°· ·tan 175°,
E3 = tan 185°·tan 195°· ·tan 265°,
E4 = tan 275°·tan 285°· ·tan 355°.

According to reduction formulas,
E2 = tan (90° + 5°)·tan (90° + 15°)· · tan (90° + 85°) = (–1)9· cot 5°·cot 15°· · cot 85°.
Therefore, E1· E2 = (–1)9. Similar,
E4 = tan (90° + 185°)·tan (90° + 195°)· · tan (90° + 265°) = (–1)9· cot 185°·cot 195°· · cot 265°.

Therefore, E3· E4 = (–1)9 and finally, E1· E2· E3· E4 = 1.

 

 

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