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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 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 60o = (square root of 3)/2
therefore, x = 60o
the figure of Sine function when the angle is 60 degree

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.
law of cosine figure
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:
example of a special angles of trigonometry function
Solution
since cos 30o = (square root of 3)/2
so arc cos (square root of 3)/2 = 30o
since sin 30o = 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
example of an inverse trigonometry function
cos a = CB/AB, CB = square root of 3, AB = 2
AB2 = AC2 + BC2 (Pythagorean Theorem)
AC2 = AB2 - BC2 = 22 - (square root of 3)2 = 4 - 3 = 1
AC = square root of 1 = 1
it is a 30o, 60o, 90o right triangle
so the angle a = 30o and arc cos(square root of 3)/2 = 30o
sin a = AC/AB = 1/2
sin (arc cos(square root of 3)/2) = 1/2