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# 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 cos pi/4 - cos pi/3 sin pi/4
= Sqrt (3)/2 × Sqrt (2)/2 - 1/2 × Sqrt (2)/2
= Sqrt (2)/2 [Sqrt (3)/2 - 1/2]
= Sqrt (2)/4 (Sqrt (3) - 1)
note:
sin pi/3 = Sqrt (3)/2
sin pi/4 = Sqrt (2)/2
cos pi/3 = 1/2
cos pi/4 = Sqrt (2)/2

### Trigonometry Function Example 2

Question:
Find the value of an acute angle when 2sin x - Sqrt (3) = 0
Solution
2 sin x - Sqrt (3) = 0
2 sin x = Sqrt (3)
sin x = Sqrt (3)/2
since sin 60o = Sqrt (3)/2
therefore, x = 60o

### 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:
c2 = a2 + b2 - 2 a b cos C
102 = 72 + 62 - (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(180o - 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 = 180o - 79.71o = 100.29o.
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: sin2x + cos2x = 1
then cos2x = 1 - sin2 = 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 - tan2x)
= 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 cos2x = 1 - sin2 = 1 - (5/13)2 = 1 - 25/169 = 169/169 - 25/169 = 144/169 = 122/132
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 = cos2x - sin2x = (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 30o = Sqrt (3)/2
so arc cos Sqrt (3)/2 = 30o
since sin 30o = 1/2
so sin[ arc cos (Sqrt (3))/2)] = 1/2
Or
let a = arc cos(Sqrt (3))/2
then cos a = Sqrt (3)/2
draw a right triangle with cos a = Sqrt (3)/2
cos a = CB/AB, CB = Sqrt (3), AB = 2
AB2 = AC2 + BC2 (Pythagorean Theorem)
AC2 = AB2 - BC2 = 22 - (Sqrt (3))2 = 4 - 3 = 1
AC = Sqrt (1) = 1
it is a 30o, 60o, 90o right triangle
so the angle a = 30o and arc cos(Sqrt (3))/2 = 30o
sin a = AC/AB = 1/2
sin (arc cos(Sqrt (3))/2) = 1/2

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