ASTC Trigo Rule – How to Solve Trigo Equations
In this revision note, you will learn about the ASTC Trigo rule and how to use it to solve Trigonometry (Trigo) equations in O-Levels A-Math exams.
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Cartesian Plane
The cartesian plane can be divided into 4 equal parts. Each part is called a quadrant. We name each of the quadrant as shown in Figure 4 below.
Basic Angle or Reference Angle
Corresponding to each angle formed by the rotation of OP about the origin O, a unique acute angle (also commonly called the basic angle or reference angle or α) and a right-angled triangle OPQ can be observed, where Q is always on the x-axis such that ∡ is 90°. The figure below shows the relationship between the positive angle θ and basic angle α for each quadrant.
θ in 2nd Quadrant θ in 1st Quadrant
θ = 180° – α θ = α
θ in 3rd Quadrant θ in 4th Quadrant
θ = –( 180° + α) θ = – (360° – α)
Negative Angle
The figure below shows the relationship between the negative angle θ and basic angle α for each quadrant
θ in 2nd Quadrant θ in 1st Quadrant
θ = –( 180° + α) θ = –( 360° – α)
θ in 3rd Quadrant θ in 4th Quadrant
θ = –( 180° – α) θ = – α
ASTC Trigo Rule
To remember which trigonometric ratios are positive, your can follow the ASTC Trigo rule: ASTC (Add Sugar To Coffee).
The figure below explains the ASTC Trigo rule and why the trigonometric ratios are positive in each quadrant.
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θ in 2nd Quadrant θ = 180° – α sin θ = sin α = PQ/OP cos θ = cos α = –OQ*/OP tan θ = tan α = –PQ/OQ* *OQ is negative in the value since it is on the negative side of the x-axis. Only sine ratio in this quadrant is positive
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θ in 1st Quadrant θ = α sin θ = sin α = PQ/OP cos θ = cos α = –OQ/OP tan θ = tan α = –PQ/OQ OQ and PQ are on the positive sides of the axes The hypotenuse is always positive in each quadrant. All trigonometric ratios in this quadrant are positive. |
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θ in 3rd Quadrant θ = 180° + α sin θ = sin α = – PQ* /OP cos θ = cos α = –OQ*/OP tan θ = tan α = PQ*/OQ* *OQ is negative in the value since it is on the negative *PQ is negative in value since it is on the negative side of the y-axis. side of the x-axis. Only tangent ratio in this quadrant is positive since both PQ and OQ are negative. |
θ in 4th Quadrant θ = 360° – α sin θ = sin α = –PQ*/OP cos θ = cos α = OQ/OP tan θ = tan α = –PQ*/OQ * PQ is negative in value since it is on the negative side of the y-axis. Only cosine ratio in this quadrant is positive.
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Example of Solving Trigo Equations
Solve the following equations for 0° ≤ x ≤ 360°.
(a) cos x = 0.1256 (b) sin x = −1/2 (c) 2 tan x + 3 = 0
Solution:
(a)
cos x = 0.1256
Basic ∡ = 82.7846°
x = 82.7846°, 360° − 82.7846°
= 82.8°, 277.2° (1 dp) (ans)
Since cos x has a positive value, we infer x must be in the 1st or 4th Quadrant.
We will calculate the value of x for each of those quadrants.
In the first quadrant, x is simply the value of the basic angle α.
(b)
sin x = −1/2
Basic ∡ = 30°
x = 180° + 30°, 360° − 30°
= 210°, 330° (ans)
Since sin x has a negative value, we infer x must be in the 3rd or 4th Quadrant.
We will calculate the value of x for each of those quadrants.
Note that the calculation of α involves sin−1(1/2). Leave out the minus sign in the calculator since α is always acute.
(c)
2 tan x + 3 = 0
tan x = −3/2
Basic ∡ = 56.3100°
x = 180° − 56.3100°, 360 − 56.3100°
= 123.7°, 303.7° (1 dp) (ans)
Since tan x has a negative value, we infer x must be in the 2nd or 4th Quadrant.
We will calculate the value of x for each of those quadrants.
Note that the calculation of α involves tan−1(3/2). Leave out the minus sign in the calculator since α is always acute.
Last Minute Revision for O Level Math?
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