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# Trigonometry Function Examples

### Trigonometry Function Example 1

### Trigonometry Function Example 2

### Trigonometry Function Example 3

### Trigonometry Function Example 4

### Trigonometry Function Example 5

- 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

- 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 60
^{o}= Sqrt (3)/2 - therefore, x = 60
^{o}

- 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.

- 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

- Question:

- Solution
- since cos 30
^{o}= Sqrt (3)/2 - so arc cos Sqrt (3)/2 = 30
^{o} - since sin 30
^{o}= 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
- AB
^{2}= AC^{2}+ BC^{2}(Pythagorean Theorem) - AC
^{2}= AB^{2}- BC^{2}= 2^{2}- (Sqrt (3))^{2}= 4 - 3 = 1 - AC = Sqrt (1) = 1
- it is a 30
^{o}, 60^{o}, 90^{o}right triangle - so the angle a = 30
^{o}and arc cos(Sqrt (3))/2 = 30^{o} - sin a = AC/AB = 1/2
- sin (arc cos(Sqrt (3))/2) = 1/2

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