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Trigonometry
Trigonometry Function Examples
Trigonometry Function Example 1
Question:
Find the value of sin(pi/12)
Solution
sin pi/12 = sin(4pi/12 - 3pi/12)
= sin(pi/3 - pi/4)
= sin pi/3 sin pi/4 - cos pi/3 cos pi/4
= (square root of 3)/2 × (square root of 2)/2 - 1/2 × (square root of 2)/2
= (square root of 2)/2 [(square root of 3)/2 - 1/2]
= (square root of 2)/4 (square root of 3 - 1)
note:
sin pi/3 = (square root of 3)/2
sin pi/4 = (square root of 2)/2
cos pi/3 = 1/2
cos pi/4 = (square root of 2)/2
Trigonometry Function Example 2
Question:
Find the value of an acute angle when 2sin x - square root of 3 = 0
Solution
2 sin x - square root of 3 = 0
2 sin x = square root of 3
sin x = (square root of 3)/2
since sin 60
^{o}
= (square root of 3)/2
therefore, x = 60
^{o}
Trigonometry Function Example 3
Question:
The lengths of the sides of a triangle are 6, 7, and 10. What is the largest angle of this triangle?
Solution
The angle opposite the side with length 10 has the largest angle.
since the three sides have the relation: b < a < c (Given)
then angles will have these relation: angle B < angle A < angle C
law of cosine is:
c
^{2}
= a
^{2}
+ b
^{2}
- 2 a b cos C
10
^{2}
= 7
^{2}
+ 6
^{2}
- (2)(7)(6) cos C
100 = 49 + 36 - 84 cos C
100 = 85 - 84 cos C
15 = - 84 cos C
cos C = - 15/84 = - 0.1786
Since the value of cos C is negative, that is, cos C < 0 , so the angle C is in Quadrant II
in Quadrant II, cos(180
^{o}
- a) = - cos a
If cos C = + 0.1786, angle C = cos
^{-1}
0.1786 = 79.71 degrees
Since cos C = - 0.1786, so angle C = 180
^{o}
- 79.71
^{o}
= 100.29
^{o}
.
Therefore, the angle C is 100.29 degrees.
Trigonometry Function Example 4
Question:
If sin x = 5/13 and x is in the range of [0, pi/2], find tan 2x = ?
Solution
given: sin x = 5/13 and 0 <= x < pi/2
formula: sin
^{2}
x + cos
^{2}
x = 1
then cos
^{2}
x = 1 - sin
^{2}
= 1 -(5/13)
^{2}
= 1 - 25/169 = 169/169 - 25/169 = 144/169 = (12/13)
^{2}
then cos x = +12/13 and cos x = -12/13
since 0 <= x <= pi/2, so x is in Quadrant I, so cos x = 12/13
tan x = (sin x)/(cos x) = 5/13 ÷ 12/13 = 5/
13
×
13
/12 = 5/12
tan 2x = (2 tan x)/(1 - tan
^{2}
x)
= 2 × (5/12) ÷ [1 - (5/12)
^{2}
]
= 5/6 ÷ (1 - 25/144)
= 5/6 ÷ (144/144 - 25/144)
= 5/6 ÷ 119/144
= 5/6 × 144/119
= 1.0084
Or
given: sin x = 5/13 and 0 < x < pi/2
then cos
^{2}
x = 1 - sin
^{2}
= 1 - (5/13)
^{2}
= 1 - 25/169 = 169/169 - 25/169 = 144/169 = 12
^{2}
/13
^{2}
so cos x = +12/13 or cos x = - 12/13
since 0 < x < pi/2
so cos x = 12/13
sin 2x = 2sin x cos x = 2 × 5/13 × 12/13 = 120/169
cos 2x = cos
^{2}
x - sin
^{2}
x = (12/13)
^{2}
- (5/13)
^{2}
= 144/169 - 25/169 = 119/169
tan 2x = sin 2x/cos 2x = 120/169 ÷ 119/169 = 120/
169
×
169
/119 = 120/119 = 1.0084
Trigonometry Function Example 5
Question:
Solution
since cos 30
^{o}
= (square root of 3)/2
so arc cos (square root of 3)/2 = 30
^{o}
since sin 30
^{o}
= 1/2
so sin[ arc cos (square root of 3/2)] = 1/2
Or
let a = arc cos(square root of 3)/2
then cos a = (square root of 3)/2
draw a right triangle with cos a = (square root of 3)/2
cos a = CB/AB, CB = square root of 3, AB = 2
AB
^{2}
= AC
^{2}
+ BC
^{2}
(Pythagorean Theorem)
AC
^{2}
= AB
^{2}
- BC
^{2}
= 2
^{2}
- (square root of 3)
^{2}
= 4 - 3 = 1
AC = square root of 1 = 1
it is a 30
^{o}
, 60
^{o}
, 90
^{o}
right triangle
so the angle a = 30
^{o}
and arc cos(square root of 3)/2 = 30
^{o}
sin a = AC/AB = 1/2
sin (arc cos(square root of 3)/2) = 1/2