## Trig. Equations Examples using CAST Diagrams

These lessons, with videos, examples and step-by-step solutions help A Level Maths students learn to solve trigonometric problems.

Related Pages Trigonometric Functions Trigonometric Graphs Trigonometric Identities Lessons On Trigonometry More Lessons for A Level Maths

What is the CAST diagram? The Cast diagram helps us to remember the signs of the trigonometric functions in each of the quadrants. The CAST diagram is also called the Quadrant Rule or the ASTC diagram.

In the first quadrant, the values are all positive. In the second quadrant, only the values for sin are positive. In the third quadrant, only the values for tan are positive. In the fourth quadrant, only the values for cos are positive.

Note that the mnemonic CAST goes anticlockwise starting from the 4th quadrant.

The following diagram shows how the CAST diagram or Quadrant rule can be used. Scroll down the page for more examples and solutions.

Quadrant Rule or CAST diagram

Using the Quadrant Rule to solve trig. equations

Quadrant Rule for solving trig equations with different ranges

Core (2) Graphs of Trigonometric Functions (4) - CAST Diagram or Unit Circle

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The CAST Diagram

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. 2. About the CAST Diagram

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- stands for Cosine, All, Sine & Tan
- helps us to remember the signs of the trigonometric functions in each of the quadrants
- is also called the Quadrant Rule or the ASTC Diagram
- is used to help solve Trig Equations in a given range

The sine wave below has been divided into 4 quadrants

From the above diagram, the sine wave is:

- Positive in quadrants 1 & 2 (above the x-axis)
- Negative in quadrants 3 & 4 (below the x-axis)

Repeating the above for cosine & tan:

The signs of the three basic trigonometric functions in each quadrant are shown in the table below:

A more convenient way of writing this is to note that all functions are positive in the 1st quadrant, only sine is positive in the 2 nd , only tangent in the 3 rd and only cosine in the 4th. We express this using the CAST diagram as the letters, taken anticlockwise from the bottom right, read C-A-S-T. The letter in each quadrant tells us which trigonometric functions are positive in that quadrant. The `A’ in the 1st quadrant stands for all (meaning sine, cosine and tangent are all positive in this quadrant). ‘S’, `C’ and `T’ ,of course, stand for sine, cosine and tangent. The diagrams below show two different forms of the CAST diagram.

A simpler version of the diagram is shown below:

- A ⇒ All (Sine, Tan & Cosine) Positive 0° – 90°
- S ⇒ Sine Positive 90° – 180°
- T ⇒ Tan Positive 180° – 270°
- C ⇒ Cosine Positive 270° – 360°

The CAST diagram can also be called a ASTC diagram as it keeps the letters in the correct quadrant order. Ways to remember:

- All Sinners Take Care.
- All Students Take Calculus.
- Add Sugar To Coffee.
- A Smart Trig Class.
- All Stations To Central.

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## 6.4 Trigonometric equations

6.4 trigonometric equations (embhm).

Solving trigonometric equations requires that we find the value of the angles that satisfy the equation. If a specific interval for the solution is given, then we need only find the value of the angles within the given interval that satisfy the equation. If no interval is given, then we need to find the general solution. The periodic nature of trigonometric functions means that there are many values that satisfy a given equation, as shown in the diagram below.

## Worked example 12: Solving trigonometric equations

Solve for \(\theta\) (correct to one decimal place), given \(\tan \theta = 5\) and \(\theta \in [\text{0}\text{°};\text{360}\text{°}]\).

## Use a calculator to solve for \(\theta\)

This value of \(\theta\) is an acute angle which lies in the first quadrant and is called the reference angle.

## Use the CAST diagram to determine in which quadrants \(\tan \theta\) is positive

The CAST diagram indicates that \(\tan \theta\) is positive in the first and third quadrants, therefore we must determine the value of \(\theta\) such that \(\text{180}\text{°} < \theta < \text{270}\text{°}\).

Using reduction formulae, we know that \(\tan (\text{180}\text{°} + \theta) = \tan \theta\)

## Use a calculator to check that the solution satisfies the original equation

Write the final answer.

\(\theta = \text{78,7}\text{°}\) or \(\theta = \text{258,7}\text{°}\).

## Worked example 13: Solving trigonometric equations

Solve for \(\alpha\) (correct to one decimal place), given \(\cos \alpha = -\text{0,7}\) and \(\theta \in [\text{0}\text{°};\text{360}\text{°}]\).

## Use a calculator to find the reference angle

To determine the reference angle, we do not include the negative sign. The reference angle must be an acute angle in the first quadrant, where all the trigonometric functions are positive.

## Use the CAST diagram to determine in which quadrants \(\cos \alpha\) is negative

The CAST diagram indicates that \(\cos \alpha\) is negative in the second and third quadrants, therefore we must determine the value of \(\alpha\) such that \(\text{90}\text{°} < \alpha < \text{270}\text{°}\).

Using reduction formulae, we know that \(\cos (\text{180}\text{°} - \alpha) = -\cos \alpha\) and \(\cos (\text{180}\text{°} + \alpha) = -\cos \alpha\)

In the second quadrant:

In the third quadrant:

Note: the reference angle \((\text{45,6}\text{°})\) does not form part of the solution.

\(\alpha = \text{134,4}\text{°}\) or \(\alpha = \text{225,6}\text{°}\).

## Worked example 14: Solving trigonometric equations

Solve for \(\beta\) (correct to one decimal place), given \(\sin \beta = -\text{0,5}\) and \(\beta \in [-\text{360}\text{°};\text{360}\text{°}]\).

To determine the reference angle, we use a positive value.

## Use the CAST diagram to determine in which quadrants \(\sin \beta\) is negative

The CAST diagram indicates that \(\sin \beta\) is negative in the third and fourth quadrants. We also need to find the values of \(\beta\) such that \(-\text{360}\text{°} \leq \beta \leq \text{360}\text{°}\).

Using reduction formulae, we know that \(\sin (\text{180}\text{°} + \beta) = - \sin \beta\) and \(\sin (\text{360}\text{°} - \beta) = - \sin \beta\)

In the fourth quadrant:

Notice: the reference angle \((\text{30}\text{°})\) does not form part of the solution.

\(\beta = -\text{150}\text{°}\), \(-\text{30}\)\(\text{°}\), \(\text{210}\)\(\text{°}\) or \(\text{330}\)\(\text{°}\).

## Solving trigonometric equations

Determine the values of \(\alpha\) for \(\alpha \in [\text{0}\text{°};\text{360}\text{°}]\) if:

\(4 \cos \alpha = 2\)

\(\alpha = \text{60}\text{°}; \text{300}\text{°}\)

\(\sin \alpha + \text{3,65} = 3\)

\(\alpha = \text{220,5}\text{°}; \text{319,5}\text{°}\)

\(\tan \alpha = 5\frac{1}{4}\)

\(\alpha = \text{79,2}\text{°}; \text{259,2}\text{°}\)

\(\cos \alpha + \text{0,939} = 0\)

\(\alpha = \text{200,1}\text{°}; \text{339,9}\text{°}\)

\(5 \sin \alpha = 3\)

\(\alpha = \text{36,9}\text{°}; \text{143,1}\text{°}\)

\(\frac{1}{2} \tan \alpha = -\text{1,4}\)

\(\alpha = \text{109,7}\text{°}; \text{289,7}\text{°}\)

Determine the values of \(\theta\) for \(\theta \in [-\text{360}\text{°};\text{360}\text{°}]\) if:

\(\sin \theta = \text{0,6}\)

Negative angles:

\(\theta = -\text{323,1}\text{°}; -\text{216,9}\text{°}; \text{36,9}\text{°}; \text{143,1}\text{°}\)

\(\cos \theta + \frac{3}{4} = 0\)

\(\theta = -\text{221,4}\text{°}; -\text{138,6}\text{°}; \text{138,6}\text{°}; \text{221,4}\text{°}\)

\(3 \tan \theta = 20\)

\(\theta = -\text{278,5}\text{°}; -\text{98,5}\text{°}; \text{81,5}\text{°}; \text{261,5}\text{°}\)

\(\sin \theta = \cos \text{180}\text{°}\)

\(\theta = -\text{90}\text{°}; \text{270}\text{°}\)

\(2 \cos \theta = \frac{4}{5}\)

\(\theta = -\text{293,6}\text{°}; -\text{66,4}\text{°}; \text{66,4}\text{°}; \text{293,6}\text{°}\)

## The general solution (EMBHN)

In the previous worked example, the solution was restricted to a certain interval. However, the periodicity of the trigonometric functions means that there are an infinite number of positive and negative angles that satisfy an equation. If we do not restrict the solution, then we need to determine the general solution to the equation. We know that the sine and cosine functions have a period of \(\text{360}\)\(\text{°}\) and the tangent function has a period of \(\text{180}\)\(\text{°}\).

Method for finding the general solution:

- Determine the reference angle (use a positive value).
- Use the CAST diagram to determine where the function is positive or negative (depending on the given equation).
- Find the angles in the interval \([\text{0}\text{°}; \text{360}\text{°}]\) that satisfy the equation and add multiples of the period to each answer.
- Check answers using a calculator.

## Worked example 15: Finding the general solution

Determine the general solution for \(\sin \theta = \text{0,3}\) (give answers correct to one decimal place).

## Use CAST diagram to determine in which quadrants \(\sin \theta\) is positive

The CAST diagram indicates that \(\sin \theta\) is positive in the first and second quadrants.

Using reduction formulae, we know that \(\sin (\text{180}\text{°} - \theta) = \sin \theta\).

In the first quadrant:

where \(k \in \mathbb{Z}\).

## Check that the solution satisfies the original equation

We can select random values of \(k\) to check that the answers satisfy the original equation.

Let \(k = 4\):

This solution is correct.

Similarly, if we let \(k = -2\):

This solution is also correct.

\(\theta = \text{17,5}\text{°} + k \cdot \text{360}\text{°}\) or \(\theta = \text{162,5}\text{°} + k \cdot \text{360}\text{°}\).

## Worked example 16: Finding the general solution

Determine the general solution for \(\cos 2\theta = - \text{0,6427}\) (give answers correct to one decimal place).

## Use CAST diagram to determine in which quadrants \(\cos \theta\) is negative

The CAST diagram shows that \(\cos \theta\) is negative in the second and third quadrants.

Therefore we use the reduction formulae \(\cos (\text{180}\text{°} - \theta) = - \cos \theta\) and \(\cos (\text{180}\text{°} + \theta) = - \cos \theta\).

Remember: also divide the period \((\text{360}\text{°})\) by the coefficient of \(\theta\).

Let \(k = 2\):

Similarly, if we let \(k = -5\):

\(\theta = \text{65}\text{°} + k \cdot \text{180}\text{°}\) or \(\theta = \text{115}\text{°} + k \cdot \text{180}\text{°}\).

## Worked example 17: Finding the general solution

Determine the general solution for \(\tan (2\alpha - \text{10}\text{°}) = \text{2,5}\) such that \(-\text{180}\text{°} \leq \alpha \leq \text{180}\text{°}\) (give answers correct to one decimal place).

## Make a substitution

To solve this equation, it can be useful to make a substitution: let \(x = 2\alpha - \text{10}\text{°}\).

## Use CAST diagram to determine in which quadrants the tangent function is positive

We see that \(\tan x\) is positive in the first and third quadrants, so we use the reduction formula \(\tan (\text{180}\text{°} + x) = \tan x\). It is also important to remember that the period of the tangent function is \(\text{180}\)\(\text{°}\).

Remember: to divide the period \((\text{180}\text{°})\) by the coefficient of \(\alpha\).

## Find the answers within the given interval

Substitute suitable values of \(k\) to determine the values of \(\alpha\) that lie within the interval (\(-\text{180}\text{°} \leq \alpha \leq \text{180}\text{°}\)).

Notice how some of the values repeat. This is because of the periodic nature of the tangent function. Therefore we need only determine the solution:

\[\alpha = \text{39,1}\text{°} + k \cdot \text{90}\text{°}\] for \(k \in \mathbb{Z}\).

\(\alpha = -\text{140,9}\text{°}\); \(-\text{50,9}\)\(\text{°}\); \(\text{39,1}\)\(\text{°}\) or \(\text{129,1}\)\(\text{°}\).

## Worked example 18: Finding the general solution using co-functions

Determine the general solution for \(\sin (\theta - \text{20}\text{°}) = \cos 2\theta\).

## Use the CAST diagram to decide on the quadrants

The right hand side of the equation is positive, and the sine function is positive in the first and second quadrants. Therefore, we will reduce the equation as if the solution were in the first and second quadrants (using the angle and \(\text{180}\text{°}\) minus the angle).

## Solve the equation

Solution 1:

Solution 2:

For the first solution:

For the second solution:

\(\theta = \text{36,67}\text{°} + k \cdot \text{120}\text{°}\) or \(\theta = \text{250}\text{°} + k \cdot \text{360}\text{°}, \quad k \in \mathbb{Z}\)

## General solution

Find the general solution for each equation.

- Hence, find all the solutions in the interval \([-\text{180}\text{°};\text{180}\text{°}]\).

\(\cos (\theta + \text{25}\text{°}) = \text{0,231}\)

\(\theta = -\text{128,36}\text{°}; -\text{101,64}\text{°}; \text{51,64}\text{°}\)

\(\sin 2\alpha = -\text{0,327}\)

\(\theta = -\text{80,45}\text{°}; -\text{9,54}\text{°}; \text{99,55}\text{°}; \text{170,46}\text{°}\)

\(2 \tan \beta = -\text{2,68}\)

\(\theta = -\text{53,27}\text{°}; \text{126,73}\text{°}\)

\(\cos \alpha = 1\)

\(\alpha = \text{0}\text{°}\)

\(4 \sin \theta = 0\)

\(\theta = -\text{180}\text{°}; \text{0}\text{°}; \text{180}\text{°}\)

\(\cos \theta = -1\)

\(\theta = -\text{180}\text{°}; \text{180}\text{°}\)

\(\tan \frac{\theta}{2} = \text{0,9}\)

\(4 \cos \theta +3 = 1\)

\(\theta = -\text{120}\text{°}; \text{120}\text{°}\)

\(\sin 2\theta = -\frac{\sqrt{3}}{2}\)

\(\theta = -\text{60}\text{°}; -\text{30}\text{°}; \text{120}\text{°}; \text{150}\text{°}\)

\(\cos (\theta + \text{20}\text{°}) = 0\)

\(\theta = - \text{20}\text{°} + n \cdot \text{360}\text{°}\)

\(\sin 3\alpha = -1\)

\(\alpha = \text{30}\text{°} + n \cdot \text{120}\text{°}\)

\(\tan 4\beta = \text{0,866}\)

\(\beta = \text{10,25}\text{°} + n \cdot \text{45}\text{°} \text{ or }\beta = \text{55,25}\text{°} + n \cdot \text{45}\text{°}\)

\(\cos (\alpha - \text{25}\text{°}) = \text{0,707}\)

\(\alpha = \text{70}\text{°} + n \cdot \text{360}\text{°} \text{ or }\alpha = \text{340}\text{°} + n \cdot \text{360}\text{°}\)

\(2 \sin \frac{3\theta}{2} = -1\)

\(\theta = \text{140}\text{°} + n \cdot \text{240}\text{°} \text{ or }\theta = \text{220}\text{°} + n \cdot \text{240}\text{°}\)

\(5 \tan (\beta + \text{15}\text{°}) = \frac{5}{\sqrt{3}}\)

Solving quadratic trigonometric equations

We can use our knowledge of algebraic equations to solve quadratic trigonometric equations.

## Worked example 19: Quadratic trigonometric equations

Find the general solution of \(4 \sin^2 \theta = 3\).

## Simplify the equation and determine the reference angle

Determine in which quadrants the sine function is positive and negative.

The CAST diagram shows that \(\sin \theta\) is positive in the first and second quadrants and negative in the third and fourth quadrants.

Positive in the first and second quadrants:

Negative in the third and fourth quadrants:

\(\theta = \text{60}\text{°} + k \cdot \text{360}\text{°}\) or \(\text{120}\text{°} + k \cdot \text{360}\text{°}\) or \(\text{240}\text{°} + k \cdot \text{360}\text{°}\) or \(\text{300}\text{°} + k \cdot \text{360}\text{°}\)

## Worked example 20: Quadratic trigonometric equations

Find \(\theta\) if \(2 \cos^2 \theta - \cos \theta - 1 = 0\) for \(\theta \in [-\text{180}\text{°};\text{180}\text{°}]\).

## Factorise the equation

Simplify the equations and solve for \(\theta\), substitute suitable values of \(k\).

Determine the values of \(\theta\) that lie within the the given interval \(\theta \in [-\text{180}\text{°};\text{180}\text{°}]\) by substituting suitable values of \(k\).

If \(k = -1\),

If \(k = 0\),

If \(k = 1\),

## Alternative method: substitution

We can simplify the given equation by letting \(y = \cos \theta\) and then factorising as:

We substitute \(y = \cos \theta\) back into these two equations and solve for \(\theta\).

\(\theta = -\text{120}\text{°}\); \(\text{0}\)\(\text{°}\); \(\text{120}\)\(\text{°}\)

## Worked example 21: Quadratic trigonometric equations

Find \(\alpha\) if \(2 \sin^2 \alpha - \sin \alpha \cos \alpha = 0\) for \(\alpha \in [\text{0}\text{°};\text{360}\text{°}]\).

## Factorise the equation by taking out a common factor

Simplify the equations and solve for \(\alpha\).

To simplify further, we divide both sides of the equation by \(\cos \alpha\).

Determine the values of \(\alpha\) that lie within the the given interval \(\alpha \in [\text{0}\text{°}; \text{360}\text{°}]\) by substituting suitable values of \(k\).

If \(k = 0\):

If \(k = 1\):

If \(k = 2\):

\(\alpha = \text{0}\text{°}\); \(\text{26,6}\)\(\text{°}\); \(\text{180}\)\(\text{°}\); \(\text{206,6}\)\(\text{°}\); \(\text{360}\)\(\text{°}\)

Find the general solution for each of the following equations:

\(\cos 2\theta = 0\)

\(\theta = \text{45}\text{°} + k \cdot \text{180}\text{°}\) or \(\theta = \text{135}\text{°} + k \cdot \text{180}\text{°}\)

\(\sin (\alpha + \text{10}\text{°}) = \frac{\sqrt{3}}{2}\)

\(\alpha = \text{50}\text{°} + k \cdot \text{360}\text{°}\) or \(\alpha = \text{110}\text{°} + k \cdot \text{360}\text{°}\)

\(2 \cos \frac{\theta}{2} - \sqrt{3} = 0\)

\(\theta = \text{60}\text{°} + k \cdot \text{720}\text{°}\) or \(\theta = \text{660}\text{°} + k \cdot \text{720}\text{°}\)

\(\frac{1}{2} \tan (\beta - \text{30}\text{°}) = -1\)

\(\beta = \text{146,6}\text{°} + k \cdot \text{180}\text{°}\)

\(5 \cos \theta = \tan \text{300}\text{°}\)

\(\theta = \text{110,27}\text{°} + k \cdot \text{360}\text{°}\) or \(\theta = \text{249,73}\text{°} + k \cdot \text{360}\text{°}\)

\(3 \sin \alpha = -\text{1,5}\)

\(\alpha = \text{210}\text{°} + k \cdot \text{360}\text{°}\) or \(\alpha = \text{330}\text{°} + k \cdot \text{360}\text{°}\)

\(\sin 2 \beta = \cos (\beta + \text{20}\text{°})\)

\(\beta = \text{23,3}\text{°} + k \cdot \text{120}\text{°}\)

\(\text{0,5} \tan \theta + \text{2,5}= \text{1,7}\)

\(\theta = \text{122}\text{°} + k \cdot \text{180}\text{°}\)

\(\sin (3 \alpha - \text{10}\text{°}) = \sin (\alpha + \text{32}\text{°})\)

\(\alpha = \text{21}\text{°} + k \cdot \text{180}\text{°}\) or \(\alpha = \text{39,5}\text{°} + k \cdot \text{90}\text{°}\)

\(\sin 2 \beta = \cos 2 \beta\)

\(\beta = \text{22,5}\text{°} + k \cdot \text{90}\text{°}\)

Find \(\theta\) if \(\sin^2 \theta + \frac{1}{2} \sin \theta = 0\) for \(\theta \in [\text{0}\text{°};\text{360}\text{°}]\).

\(\theta = \text{0}\text{°}, \text{180}\text{°}, \text{210}\text{°}, \text{330}\text{°} \text{ or } \text{360}\text{°}\)

Determine the general solution for each of the following:

\(2 \cos^2 \theta - 3 \cos \theta = 2\)

\(\theta = \text{120}\text{°} + k \cdot \text{360}\text{°}\) or \(\theta = \text{240}\text{°} + k \cdot \text{360}\text{°}\)

\(3 \tan^2 \theta + 2 \tan \theta = 0\)

\(\theta = \text{0}\text{°} + k \cdot \text{180}\text{°}\) or \(\theta = \text{146,3}\text{°} + k \cdot \text{180}\text{°}\)

\(\cos^2 \alpha = \text{0,64}\)

\(\alpha = \text{36,9}\text{°} + k \cdot \text{360}\text{°}\) or \(\alpha = \text{143,1}\text{°} + k \cdot \text{360}\text{°}\) or \(\alpha = \text{216,9}\text{°} + k \cdot \text{360}\text{°}\) or \(\alpha = \text{323,1}\text{°} + k \cdot \text{360}\text{°}\)

\(\sin (4\beta + \text{35}\text{°}) = \cos (\text{10}\text{°} - \beta)\)

\(\beta = \text{15}\text{°} + k \cdot \text{120}\text{°}\) or \(\beta = \text{75}\text{°} + k \cdot \text{120}\text{°}\)

\(\sin (\alpha + \text{15}\text{°}) = 2 \cos (\alpha + \text{15}\text{°})\)

\(\alpha = \text{48,4}\text{°} + k \cdot \text{180}\text{°}\)

\(\sin^2 \theta - 4\cos^2 \theta = 0\)

\(\theta = \text{63,4}\text{°} + k \cdot \text{180}\text{°}\) or \(\theta = \text{116,6}\text{°} + k \cdot \text{180}\text{°}\)

\(\dfrac{\cos (2 \theta + \text{30}\text{°})}{2} + \text{0,38} = 0\)

\(\theta = \text{54,8}\text{°} + k \cdot \text{180}\text{°}\) or \(\theta = \text{95,25}\text{°} + k \cdot \text{180}\text{°}\)

Find \(\beta\) if \(\frac{1}{3} \tan \beta = \cos \text{200}\text{°}\) for \(\beta \in [-\text{180}\text{°};\text{180}\text{°}]\).

\(\beta = -\text{70,5}\text{°}\) or \(\beta = \text{109,5}\text{°}\)

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## Course content

- Understand how sin, cos and tan are defined for angles from 0° to 360°
- Understand period and related angles
- Solve basic trigonometric equations.

## Textbook page references

- TeeJay Maths Book N5 pp.196-203
- Zeta National 5+ Maths pp.214-216
- Leckie National 5 Maths pp.272-280

## Related angles

The trig functions sin, cos and tan can be defined for angles greater than 90° by thinking about this diagram:

The four quadrants contain congruent right-angled triangles with the angle \(x^\circ\) at the centre.

The actual angle is measured anticlockwise from the 0° line .

- 1st quadrant: The actual angle is \(x^\circ.\)
- 2nd quadrant: The actual angle is half a turn and then back \(x^\circ.\) So that's 180–\(x^\circ.\)
- 3rd quadrant: The actual angle is half a turn and then another \(x^\circ\). So that's 180+\(x^\circ.\)
- 4th quadrant: The actual angle is a full turn and then back \(x^\circ.\) So that's 360–\(x^\circ.\)

These four angles are what we call related .

Example 1: \(x\) = 30°

- sin 30° = \(1 \over 2\)
- sin (180–30)° = sin 150° = \(1 \over 2\)
- sin (180+30)° = sin 210° = \(-\frac{1}{2}\)
- sin (360–30)° = sin 330° = \(-\frac{1}{2}\)

Apart from the negatives, they're all equal. That's because these four angles are related.

Example 2: \(x\) = 45°

- cos 45° = \(\frac{\sqrt{2}}{2}\)
- cos (180–45)° = cos 135° = \(-\frac{\sqrt{2}}{2}\)
- cos (180+45)° = cos 225° = \(-\frac{\sqrt{2}}{2}\)
- cos (360–45)° = cos 315° = \(\frac{\sqrt{2}}{2}\)

Again, these are related angles, so they're equal apart from the negatives.

Example 3: \(x\) = 60°

- tan 60° = \(\sqrt{3}\)
- tan (180–60)° = tan 120° = \(-\sqrt{3}\)
- tan (180+60)° = tan 240° = \(\sqrt{3}\)
- tan (360–60)° = tan 300° = \(-\sqrt{3}\)

Same again: they're equal apart from the negatives, because they are related angles.

## ASTC diagram

This is also called a CAST diagram, but we prefer ASTC as it keeps the letters in the correct quadrant order.

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Here is a simpler version of the diagram:

If you want to understand why some of the values for related angles are positive and some are negative, that's easy. Just go back to your study of the basic trig graphs and take a look at when they are above or below the \(x\)-axis. Above is positive and below is negative – simple!

## Trig equations

Trig equations have an infinity of solutions because the wave-like graphs of these functions repeat endlessly.

For this reason, Nat 5 exams ask for such equations to be solved within a given domain , often \(0 \leq x \lt 360.\)

When solving any trig equation, we recommend always finding the related acute angle . This means the 1st quadrant (0° to 90°) angle that is related to the solutions.

We will use the Greek letter alpha (\(\alpha\)) for the related acute angle.

Once we know the related acute angle, we will use it to find the solutions.

## Find a Maths tutor

## Example 1 (calculator)

Solve the equation \( cos\ x^\circ=0.5 \), for \(0 \leq x \lt 360.\)

The related acute angle, \( \alpha\ = cos^{-1}\ 0.5=60^\circ \)

Because \(cos\ x\) > 0, there are solutions in the 1 st (A) and 4 th (C) quadrants.

So \(x=\alpha\) or \(x=360-\alpha\)

\(x=60^\circ\) or \(x=360-60=300^\circ\)

## Example 2 (calculator)

Solve the equation \( cos\ x^\circ=-0.5 \), for \(0 \leq x \lt 360.\)

The only difference here is the negative.

The related acute angle is the same: \( \alpha\ = cos^{-1}\ 0.5=60^\circ \)

Because \(cos\ x\) < 0, there are solutions in the 2 nd (S) and 3 rd (T) quadrants.

So \(x=180-\alpha\) or \(x=180+\alpha\)

\(x=180-60=120^\circ\) or \(x=180+60=240^\circ\)

## Example 3 (calculator)

Solve the equation \( sin\ x^\circ=-0.4 \), for \(0 \leq x \lt 360.\)

The related acute angle \( \alpha\ = sin^{-1}\ 0.4 = 23.6^\circ \) (to 1 d.p.)

Because \(sin\ x\) < 0, there are solutions in the 3 rd (T) and 4 th (C) quadrants.

So \(x=180+\alpha\) or \(x=360-\alpha\)

\(x=180+23.6=203.6^\circ\) or \(x=360-23.6=336.4^\circ\) (to 1 d.p.)

## Recommended student books

## Example 4 (calculator)

Solve the equation \( tan\ x^\circ=5 \), for \(0 \leq x \lt 360.\)

The related acute angle \( \alpha\ = tan^{-1}\ 5 = 78.7^\circ \) (to 1 d.p.)

Because \(tan\ x\) > 0, there are solutions in the 1 st (A) and 3 rd (T) quadrants.

So \(x=\alpha\) or \(x=180+\alpha\)

\(x=78.7^\circ\) or \(x=180+78.7=258.7^\circ\) (to 1 d.p.)

## Example 5 (calculator)

Solve the equation \( 9\tiny\ \normalsize tan\ x^\circ-1=6 \), for \(0 \leq x \lt 360.\)

The previous examples were easier than exam level. This is more like it!

First we rearrange to find a value for \(tan\ x\):

\( 9\tiny\ \normalsize tan\ x^\circ-1=6 \) \( 9\tiny\ \normalsize tan\ x^\circ=7 \) \( tan\ x^\circ=\large \frac{7}{9} \normalsize \)

The related acute angle \( \alpha\ = tan^{-1}\ \large \frac{7}{9} \normalsize = 37.9^\circ \) (to 1 d.p.)

\(x=37.9^\circ\) or \(x=180+37.9=217.9^\circ\) (to 1 d.p.)

## Example 6 (calculator)

Solve the equation \( 3\tiny\ \normalsize sin\ x^\circ+5=4 \), for \(0 \leq x \lt 360.\)

Rearrange to make \(sin\ x\) the subject:

\( 3\tiny\ \normalsize sin\ x^\circ+5=4 \) \( 3\tiny\ \normalsize sin\ x^\circ=-1 \) \( sin\ x^\circ=-\large \frac{1}{3} \normalsize \)

The related acute angle \( \alpha\ = sin^{-1}\ \large \frac{1}{3} \normalsize = 19.5^\circ \) (to 1 d.p.)

\(x=180+19.5=199.5^\circ\) or \(x=360-19.5=340.5^\circ\) (to 1 d.p.)

## Recommended revision guides

Example 7 (non-calculator).

Write the following in order of size, starting with the smallest.

\(sin\ 200^\circ\) \(sin\ 160^\circ\) \(sin\ 180^\circ\)

Justify your answer.

Obviously this isn't an equation, but a question very like it appeared as Q9 on 2015 Paper 1 .

Answering it without a calculator requires either sketching the graph of \(y=sin\ x\) or thinking about the ASTC diagram.

- \(200^\circ\) is in the 3 rd (T) quadrant, so \(sin\ 200^\circ \lt 0\)
- \(160^\circ\) is in the 2 nd (S) quadrant, so \(sin\ 160^\circ \gt 0\)
- \(180^\circ\) is between the 2 nd and 3 rd quadrants, so \(sin\ 180^\circ=0\)

In order: \(sin\ 200^\circ\) \(sin\ 180^\circ\) \(sin\ 160^\circ\)

## Example 8 (calculator)

SQA National 5 Maths 2014 P2 Q12

Solve the equation \( 11\tiny\ \normalsize cos\ x^\circ-2=3 \), for \(0 \leq x \leq 360.\)

\( 11\tiny\ \normalsize cos\ x^\circ-2=3 \) \( 11\tiny\ \normalsize cos\ x^\circ=5 \) \( cos\ x^\circ=\large \frac{5}{11} \normalsize \)

The related acute angle \( \alpha\ = cos^{-1}\ \large \frac{5}{11} \normalsize = 63.0^\circ \) (to 1 d.p.)

\(x=63.0^\circ\) or \(x=360-63.0=297.0^\circ\) (to 1 d.p.)

## Example 9 (calculator)

SQA National 5 Maths 2016 P2 Q14

Solve the equation \( 2\tiny\ \normalsize tan\ x^\circ+5=-4 \), for \(0 \leq x \leq 360.\)

\( 2\tiny\ \normalsize tan\ x^\circ+5=-4 \) \( 2\tiny\ \normalsize tan\ x^\circ=-9 \) \( tan\ x^\circ=-\large \frac{9}{2} \normalsize \)

The related acute angle \( \alpha\ = tan^{-1}\ \large \frac{9}{2} \normalsize = 77.5^\circ \) (to 1 d.p.)

Because \(tan\ x\) < 0, there are solutions in the 2 nd (S) and 4 th (C) quadrants.

So \(x=180-\alpha\) or \(x=360-\alpha\)

\(x=180-77.5=102.5^\circ\) or \(x=360-77.5=282.5^\circ\) (to 1 d.p.)

## N5 Maths practice papers

Example 10 (calculator).

SQA National 5 Maths 2018 P2 Q8

Solve the equation \( 7\tiny\ \normalsize sin\ x^\circ+2=3 \), for \(0 \leq x \lt 360.\)

\( 7\tiny\ \normalsize sin\ x^\circ+2=3 \) \( 7\tiny\ \normalsize sin\ x^\circ=1 \) \( sin\ x^\circ=\large \frac{1}{7} \normalsize \)

The related acute angle \( \alpha\ = sin^{-1}\ \large \frac{1}{7} \normalsize = 8.2^\circ \) (to 1 d.p.)

Because \(sin\ x\) > 0, there are solutions in the 1 st (A) and 2 nd (S) quadrants.

So \(x=\alpha\) or \(x=180-\alpha\)

\(x=8.2^\circ\) or \(x=180-8.2=171.8^\circ\) (to 1 d.p.)

## Example 11 (calculator)

SQA National 5 Maths 2019 P2 Q14

Solve the equation \( 5\tiny\ \normalsize cos\ x^\circ+2=1 \), for \(0 \leq x \leq 360.\)

\( 5\tiny\ \normalsize cos\ x^\circ+2=1 \) \( 5\tiny\ \normalsize cos\ x^\circ=-1 \) \( cos\ x^\circ=-\large \frac{1}{5} \normalsize \)

The related acute angle \( \alpha\ = cos^{-1}\ \large \frac{1}{5} \normalsize = 78.5^\circ \) (to 1 d.p.)

\(x=180-78.5=101.5^\circ\) or \(x=180+78.5=258.5^\circ\) (to 1 d.p.)

## Example 12 (calculator)

SQA National 5 Maths 2022 P2 Q9

Solve the equation \( 3\tiny\ \normalsize sin\ x^\circ+4=6 \), for \(0 \leq x \leq 360.\)

\( 3\tiny\ \normalsize sin\ x^\circ+4=6 \) \( 3\tiny\ \normalsize sin\ x^\circ=2 \) \( sin\ x^\circ=\large \frac{2}{3} \normalsize \)

The related acute angle \( \alpha\ = sin^{-1}\ \large \frac{2}{3} \normalsize = 41.8^\circ \) (to 1 d.p.)

\(x=41.8^\circ\) or \(x=180-41.8=138.2^\circ\) (to 1 d.p.)

## Example 13 (calculator)

SQA National 5 Maths 2024 P2 Q11

Solve the equation \( 17\tiny\ \normalsize sin\ x^\circ+1=9 \), for \(0 \leq x \lt 360.\)

\( 17\tiny\ \normalsize sin\ x^\circ+1=9 \) \( 17\tiny\ \normalsize sin\ x^\circ=8 \) \( sin\ x^\circ=\large \frac{8}{17} \normalsize \)

The related acute angle \( \alpha\ = sin^{-1}\ \large \frac{8}{17} \normalsize = 28.1^\circ \) (to 1 d.p.)

\(x=28.1^\circ\) or \(x=180-28.1=151.9^\circ\) (to 1 d.p.)

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## Solving trigonometric equations CAST

Subject: Mathematics

Age range: 16+

Resource type: Other

Last updated

10 February 2021

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Presentation with examples on how to solve trigonometric equations using the CAST diagram. AS Maths

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very useful resource

I know not all maths teachers are particularly big fans of using CAST diagrams, but I've found it helpful for many of my students in the past. This is a useful resource to demonstrate and practice using CAST diagrams. I would probably want to demonstrate it for myself, as some of the hand drawn diagrams etc are not always completely clear. Thanks for sharing.

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## Trig Identities: A Crash Course in Complex Math Concepts

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Fundamental trigonometric identities, aka trig identities or trigo identities, are equations involving trigonometric functions that hold true for any value you substitute into their variables.

These identities are essential tools if you want to solve trigonometric equations and perform complex calculations in mathematics, physics or engineering . Understanding all the trigonometric identities can help you simplify seemingly complicated problems, especially in geometry and calculus.

## The Foundation of Trigonometry

Fundamental trigonometric identities, double angle trigonometric identities, triple angle trigonometric identities, half angle identities, sum and difference identities.

Trigonometry is a branch of mathematics . At the heart of trigonometry lie the trigonometric functions, which relate the angles of a triangle to the ratios of its sides.

The most basic trigonometric functions are sine, cosine and tangent, which instructors often teach using the mnemonic SOH-CAH-TOA in right-angled triangles.

From these basic trig functions, we derive other crucial functions, such as secant, cosecant and cotangent, all of which play vital roles in further developing trigonometric theory.

You might hear people refer to sine, cosine, tangent, secant, cosecant and cotangent as the six trigonometric ratios , or trig ratios.

Trigonometric identities form a cornerstone of higher mathematics. They encapsulate all the trigonometric ratios and relationships in a framework that enhances the solving of equations and understanding of geometric and algebraic concepts.

Trigonometric identities encompass a wide range of formulas, but people generally group them into categories based on their specific applications and forms.

There are three main categories comprising eight fundamental trigonometric identities. These categories include reciprocal identities, Pythagorean identities and quotient identities.

## Reciprocal Identities

These identities express the basic trigonometric functions in terms of their reciprocal functions:

- Sine and cosecant : csc( θ ) = 1/sin( θ )
- Cosine and secant : sec( θ ) = 1/cos( θ )
- Tangent and cotangent : cot( θ ) = 1/tan( θ )

## Pythagorean Identities

The Pythagorean trigonometric identities stem from the Pythagorean theorem , also known as the Pythagoras theorem, after the Greek scholar who came up with the mathematical statement.

The trig identities based on the Pythagorean theorem are fundamental to connecting the squares of the primary trigonometric functions:

- Basic Pythagorean identity : sin 2 ( θ ) + cos 2 ( θ ) = 1
- Derived for tangent : 1 + tan 2 ( θ ) = sec 2 ( θ )
- Derived for cotangent : cot 2 ( θ ) + 1 = csc 2 ( θ )

## Quotient Identities

These identities relate the functions through division:

- Tangent as a quotient : tan( θ ) = sin( θ )/cos( θ )
- Cotangent as a quotient : cot( θ ) = cos( θ )/sin( θ )

Of course, there are many more trigonometric identities beyond just these core identities that have applications in specific scenarios, such as double angle, triple angle, half angle and sum and difference identities.

The double angle formulas are trigonometric identities that express trigonometric functions of double angles — that is, angles of the form 2 θ — in terms of trigonometric functions of single angles ( θ ).

These formulas are crucial in various mathematical computations and transformations, particularly in calculus, geometry and solving trigonometric equations.

The primary double angle formulas include those for sine, cosine and tangent.

## Cosine Double Angle Formula

The cosine double angle formula is:

cos(2 θ ) = cos 2 ( θ ) – sin 2 ( θ )

You can also represent this in two alternative forms using the Pythagorean identity sin 2 ( θ ) + cos 2 ( θ ) = 1 :

## Sine Double Angle Formula

The sine double angle formula is:

This formula is derived from the sum identities and is useful for solving problems involving products of sine and cosine.

## Tangent Double Angle Formula

The tangent double angle formula is:

This expression arises from dividing the sine double angle formula by the cosine double angle formula and simplifying using the definition of tangent.

Triple angle formulas, while less commonly used, offer shortcuts in specific scenarios, such as in certain integrals and polynomial equations. These are identities that allow the calculation of the sine, cosine and tangent of three times a given angle (3θ) using the trigonometric functions of the angle itself (θ).

For example, the sine triple angle formula is:

This formula is derived by using the sine double angle formula and the angle sum identity.

Triple angle formulas can be derived from double angle and sum identities and are useful in specific mathematical and engineering contexts, such as simplifying complex trigonometric expressions or solving higher-degree trigonometric equations.

Half angle identities are trigonometric formulas that allow you to prove trigonometric identities for the sine, cosine and tangent of half of a given angle.

Half angle formulas are particularly useful in solving trigonometric equations, integrating trigonometric functions, and simplifying expressions when the angle involved is halved. Half angle formulas are derived from the double angle identities and other fundamental trigonometric identities.

The half angle identities for sine, cosine and tangent use the following half angle formulas:

- Sine half angle identity : sin( θ /2) = ±√((1 – cos θ )/2)
- Cosine half angle identity : cos( θ /2) = ±√((1 + cos θ )/2)
- Tangent half angle identity : tan( θ /2) = sin( θ )/(1 + cos( θ )) = 1 – (cos( θ )/sin( θ ))

In the case of the sine and cosine half angle formulas, the sign depends on the quadrant in which θ /2 resides. The tangent half angle formula you can also express in terms of sine and cosine directly.

These identities are derived by manipulating the double angle identities. For example, the cosine double angle identity cos(2 θ ) = 2cos 2 ( θ ) can be rearranged to express cos 2 ( θ ) in terms of cos(2 θ ) , and then taking the square root (and adjusting for sign based on the angle's quadrant) gives the half angle formula for cosine.

Half angle identities are crucial for simplifying the integration of trigonometric functions, particularly when integral limits involve pi (π) or when integrating periodic functions. They also play a vital role in various fields of science and engineering where wave functions and oscillations are analyzed.

Sum identities in trigonometry are essential formulas that allow for the calculation of the sine, cosine and tangent of the sum of two angles. Conversely, difference formulas allow you to calculate the sine, cosine and tangent of the difference between two angles.

These identities are incredibly useful for simplifying expressions, solving trigonometric equations and performing complex calculations.

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The Cast diagram helps us to remember the signs of the trigonometric functions in each of the quadrants. The CAST diagram is also called the Quadrant Rule or the ASTC diagram. In the first quadrant, the values are all positive. In the second quadrant, only the values for sin are positive.

Trigonometry at AS/A Level includes solving trigonometric equations, and finding multiple solutions using a CAST diagram (or graph). In this video I introduc...

How to solve trigonometric equations using the CAST diagram? This video provides a step-by-step guide with examples and tips to help you master this topic. You can also watch other related videos ...

This video explains how the CAST diagram can be used to solve trig equations.Solving Trig Equations 1:https://youtu.be/cILaBqbmPX0Practice Questions: https:/...

Prove the identities: θ. θ. tan θ sin θ 1. − cos θ ≡ 1 + cos θ. Find the exact value of the sin θ and tan θ given cos θ = − 3 4 and is θ obtuse. Notes. When solving trigonometric equations, the CAST diagram can be used to help calculate all the required angles. There are two important points to remember about using the CAST diagram:

Trigonometry: The CAST Diagram. Angles can be measured from 0 degrees on a set of x-y axes, in a counterclockwise direction. Angles in standard position like this have trigonometric fractions associated with them, in the following way: If the angle is less than 90 degrees, the trig fractions for the angle are made in the ordinary way.

Next: Solving Trig Equations 1 Video. A video introducing how to solve trig equations using the CAST diagram.

Solving linear trigonometric equations. You should have already seen two ways to solve linear trig equations By sketching the graph (see Graphs of Trigonometric Functions) you can read off all the solutions in a given range (or interval); By using trigonometric identities you can simplify harder equations; Another way to find solutions is by using the CAST diagram which shows where each ...

2. About the CAST Diagram To learn about the CAST Diagram please click on the Trigonometry Theory Guide (HSN) link and read from page 2. Please also find in Sections 2 & 3 below trig equations videos 1 & 2, mind maps (see under Trigonometry) and worksheets on this topic to help your understanding. The Essential Skills 13 worksheet, along with worksheets including actual SQA Exam Questions, are ...

Solving trigonometric equations Solving trigonometric equations in degrees. Trigonometric equations can be solved in degrees or radians using CAST and its period to find other solutions within the ...

The CASt diagram provides a method to find multiple solutions to trigonometric equations. It indicates the quadrants in which the elementary trigonometric functions are positive (or negative). It is normally shown as below. The significance of each eletter is shown in the diagram below. Example: To solve in the range 0° - 360°, find Since ...

A full explanation of how to use the CAST diagram to find all the solutions of a trigonometric equation.

Trigonometric equations can be solved in degrees or radians using CAST and its period to find other solutions within the range, including multiple or compound angles and the wave function. Part of ...

Solving trigonometric equations requires that we find the value of the angles that satisfy the equation. If a specific interval for the solution is given, then we need only find the value of the angles within the given interval that satisfy the equation. ... Use the CAST diagram to determine in which quadrants \(\tan \theta\) is positive. The ...

Solving trig equations using a CAST diagram. Recaps where a CAST diagram comes from, relating to the graphs. Then moves onto solving more complex equations using the diagram as this was not the first lesson on solving trig equations. to let us know if it violates our terms and conditions. Our customer service team will review your report and ...

ASTC/CAST diagrams. Solving trigonometic equations. Notes, videos, examples and other great resources. ... Solve basic trigonometric equations. Textbook page references. TeeJay Maths Book N5 pp.196-203; Zeta National 5+ Maths pp.214-216; Leckie National 5 Maths pp.272-280; Recommended revision course.

Another way to find solutions is by using the CAST diagram which shows where each function has positive solutions. You may be asked to use degrees or radians to solve trigonometric equations. Make sure your calculator is in the correct mode. Remember common angles. 90° is ½π radians. 180° is π radians. 270° is 3π/2 radians.

Trig equations using CAST diagram. Subject: Mathematics. Age range: 5-7. Resource type: Worksheet/Activity. File previews. docx, 177.47 KB. pdf, 398.55 KB. I'm a little bit torn when it comes to teaching trig equations using the cast diagram. It seems like teaching without the understanding, however I find they get the right answer more often ...

Here I show you how to use the quadrant rule or CAST diagram to solve trigonometric equations in different ranges. It saves having to use trigonometric graph...

Solving trigonometric equations CAST. Subject: Mathematics. Age range: 16+. Resource type: Other. File previews. pptx, 646.32 KB. Presentation with examples on how to solve trigonometric equations using the CAST diagram. AS Maths. Creative Commons "Sharealike".

Solve each equation for X X in the interval 0 ≤ X ≤ 360 0 ≤ X ≤ 360. Here is my attempt: sin sin is negative and so tan tan and cos cos are positive. This would mean that 17.5 17.5 is added to 180 180 and subtracted from 360 360. So the final answers for X X should be 227.5 227.5 and 372.5 372.5 degrees. However, the solutions say that ...

Searching for a comprehensive guide to conquer trigonometric equations at the A-Level? Look no further! In this tutorial, we delve into the intricacies of so...

These are identities that allow the calculation of the sine, cosine and tangent of three times a given angle (3θ) using the trigonometric functions of the angle itself (θ). For example, the sine triple angle formula is: sin (3 θ) = 3sin ( θ) - 4sin 3 (θ) This formula is derived by using the sine double angle formula and the angle sum ...

There's this question where we are using CAST diagrams to solve trig equations, I'm able to get the positive values however I don't know the…