# 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 = 180° – α                                                                           θ = α

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 = –( 180° + α)                                                                        θ = –( 360° – α)

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 x-axis.                    Only sine ratio in this                            quadrant is positive θ in 1st Quadrant θ = αsin θ = sin α = PQ/OPcos θ = cos α = –OQ/OP tan θ = tan α = –PQ/OQOQ and PQ are on the positive sides of the axesThe hypotenuse is always positive in each quadrant.All trigonometric ratios in this quadrant are positive.

 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° – αsin θ = sin α = –PQ*/OPcos 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 quadrantis 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.

## Check out our exam guide on other topics here!

Secondary Math Revision Notes

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