3.10 Exercises 89
The relation forcosh
− 1xshows that there exist two values of the function for each
value ofx(> 1 1). These two values differ only in sign, and the positive value is defined
as the principal value.
EXAMPLE 3.30To show that.
Ifx 1 = 1 cosh 1 ythen, becausecosh
21 y 1 − 1 sinh
21 y 1 = 11 , it follows that and
from the definitions (3.47). The result follows sinceln 1 e
y1 = 1 y 1 = 1 cosh
− 1x.
3.10 Exercises
Section 3.2
1.In Figure 3.23, the right-angled triangle ABC has sidesa 1 = 112 and
b 1 = 15. Find cand the sin, cos, tan, cosec, sec and cot of the internal
angles A and B.
2.For the triangle in Exercise 1, find (i)sin
2
A 1 + 1 cos
2
A, (ii)sin
2
B 1 + 1 cos
2
B.
3.Express the following angles in radians:
(i)5° (ii)87° (iii)120° (iv)260° (v)540°
(vi)720°
4.Express the following angles in degrees:
(i)π 210 (ii)π 24 (iii)π 26 (iv)π 23 (v) 3 π 28 (vi) 7 π 28
5.For a circle of radiusr 1 = 14 , find
(i)the angle subtended at the centre of the circle by arc of length 6,
(ii)the length of arc that subtends angleπ 210 at the centre of the circle,
(iii)the length of arc that subtends angleπ 22 at the centre of the circle,
(iv)the circumference of the circle.
6.Use Table 3.2 to find the sine, cosine and tangent of (i) 3 π 24 , (ii) 5 π 24 ,(iii) 7 π 24.
7.By considering the limitθ 1 → 10 of an internal angle of a right-angled triangle, show that
(i)sin 101 = 10 (ii)cos 101 = 11
8.Use the properties of the right-angled isosceles triangle to verify the values of the
trigonometric functions forθ 1 = 1 π 24 in Table 3.2.
9.Sketch diagrams to show that
(i)sin(π 1 − 1 θ) 1 = 1 sin 1 θ (ii)cos(π 1 − 1 θ) 1 = 1 −cosθ (iii)sin(π 1 + 1 θ) 1 = 1 −sinθ
(iv)cos(π 1 + 1 θ) 1 = 1 −cosθ (v)sin(π 221 − 1 θ) 1 = 1 cosθ (vi)cos(π 221 − 1 θ) 1 = 1 sinθ
10.Find the period and sketch the graph (−π 1 ≤ 1 x 1 ≤ 12 π) of (i)sin 12 x, (ii)cos 13 x.
11.Sketch the graph of the harmonic waveφ(x, t) 1 = 1 sin 12 π(x 1 − 1 t) as a function of
x(− 11 ≤ 1 x 1 ≤ 1 2) for values of time t, (i)t 1 = 10 , (ii)t 1 = 1124 , (iii)t 1 = 1122.
lnxx ln cosh sinhy y eln
y±−= +
=
21
sinhyx=± −
21
cosh ln
−=±−
12xxx 1
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ac
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Figure 3.23