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 OLevels AMath 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 rightangled triangle OPQ can be observed, where Q is always on the xaxis 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 2^{nd} Quadrant θ in 1^{st} Quadrant
θ = –( 180° + α) θ = –( 360° – α)
θ in 3^{rd} Quadrant θ in 4^{th} 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.
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 xaxis. Only sine ratio in this quadrant is positive
 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. 
θ in 3^{rd} 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 yaxis. side of the xaxis. Only tangent ratio in this quadrant is positive since both PQ and OQ are negative.  θ in 4^{th} 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 yaxis. Only cosine ratio in this quadrant is positive.

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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