Double Angle Identities Revision Notes for Grade 12 NSC Matric Mathematics

Double Angle Identities

What are double angle identities?

Double angle identities are special trigonometric formulae that help us find the trigonometric values of twice an angle (2α) using the trigonometric values of the original angle (α). These identities are essential tools for solving complex trigonometric equations and simplifying expressions in your NSC Mathematics exam.

The double angle identities are derived from the compound angle formulae by setting both angles equal to each other.

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Understanding double angle identities is crucial because they appear frequently in NSC Mathematics exams and form the foundation for more advanced trigonometric concepts.

Derivation of sin 2α

To find the formula for $\sin 2α$ , we start with the compound angle formula for sine. We know that $\sin(α + β) = \sin α \cos β + \cos α \sin β$ .

When we let $α = β$ , we can substitute this into the compound angle formula:

$\sin(2α) = \sin(α + α)$
$\sin(2α) = \sin α \cos α + \cos α \sin α$
$\sin(2α) = 2 \sin α \cos α$

Therefore: $\sin 2α = 2 \sin α \cos α$

Derivation of cos 2α

Similarly, for cosine we start with $\cos(α + β) = \cos α \cos β - \sin α \sin β$ .

When we let $α = β$ :

$\cos(2α) = \cos(α + α)$
$\cos(2α) = \cos α \cos α - \sin α \sin α$
$\cos(2α) = \cos² α - \sin² α$

This gives us the first form: $\cos 2α = \cos² α - \sin² α$

Alternative forms of cos 2α

We can derive two additional forms using the Pythagorean identity $\sin² α + \cos² α = 1$ .

Second form: Starting with $\cos 2α = \cos² α - \sin² α$ , we substitute $\cos² α = 1 - \sin² α$ :

$\cos 2α = (1 - \sin² α) - \sin² α$
$\cos 2α = 1 - 2 \sin² α$

Third form: Starting with $\cos 2α = \cos² α - \sin² α$ , we substitute $\sin² α = 1 - \cos² α$ :

$\cos 2α = \cos² α - (1 - \cos² α)$
$\cos 2α = \cos² α - 1 + \cos² α$
$\cos 2α = 2 \cos² α - 1$

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Remember that $\cos 2α$ has three equivalent forms. Choose the form that makes your calculation easiest based on what information you're given in the problem.

Summary of double angle formulae

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Key Double Angle Identities:

$\sin 2α = 2 \sin α \cos α$
$\cos 2α = \cos² α - \sin² α$
$\cos 2α = 1 - 2 \sin² α$
$\cos 2α = 2 \cos² α - 1$

Worked example 1: Finding sin 2α

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Worked Example: Finding sin 2α from a given sin α value

Question: If α is an acute angle and $\sin α = 0.6$ , determine the value of $\sin 2α$ without using a calculator.

Solution:

Step 1: Create a right triangle to find the missing side.

Since $\sin α = 0.6 = \frac{6}{10}$ , we can represent this as a right triangle with opposite side = 6 and hypotenuse = 10.

Using Pythagoras theorem to find the adjacent side:

$x² = r² - y²$
$x² = 10² - 6² = 100 - 36 = 64$
$x = 8$

So $\cos α = \frac{8}{10} = 0.8$

Step 2: Apply the double angle formula.

$\sin 2α = 2 \sin α \cos α$

$\sin 2α = 2 × \frac{6}{10} × \frac{8}{10}$

$\sin 2α = 2 × \frac{48}{100} = \frac{96}{100} = 0.96$

Answer: $\sin 2α = 0.96$

Worked example 2: Proving identities and finding restrictions

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Worked Example: Proving trigonometric identities with restrictions

Question: Prove that $\frac{\sin θ + \sin 2θ}{1 + \cos θ + \cos 2θ} = \tan θ$ . For which values of θ is the identity not valid?

Solution:

Step 1: Simplify the left-hand side using double angle formulae. The right-hand side cannot be simplified, so we work on the left-hand side.

$\text{LHS} = \frac{\sin θ + \sin 2θ}{1 + \cos θ + \cos 2θ}$

Substitute $\sin 2θ = 2 \sin θ \cos θ$ and $\cos 2θ = 2\cos² θ - 1$ :

$\text{LHS} = \frac{\sin θ + 2 \sin θ \cos θ}{1 + \cos θ + 2\cos² θ - 1}$

$= \frac{\sin θ + 2 \sin θ \cos θ}{\cos θ + 2\cos² θ}$

Step 2: Factorise and simplify.

$\text{LHS} = \frac{\sin θ(1 + 2 \cos θ)}{\cos θ(1 + 2 \cos θ)}$

$= \frac{\sin θ}{\cos θ}$

$= \tan θ = \text{RHS}$

The identity is proven.

Step 3: Find restricted values. The identity is undefined when:

$\tan θ$ is undefined: $θ = 90° + k × 180°$ , where $k ∈ ℤ$
The denominator equals zero: $1 + \cos θ + \cos 2θ = 0$

For the denominator: $\cos θ(1 + 2 \cos θ) = 0$ This occurs when:

$\cos θ = 0$ : $θ = 90° + k × 180°$
$1 + 2 \cos θ = 0$ : $\cos θ = -\frac{1}{2}$ , so $θ = 120° + k × 360°$ or $θ = 240° + k × 360°$

Exam tips and common mistakes

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Critical Points to Avoid Common Mistakes:

Choose the right form: Use the most suitable form of $\cos 2α$ for your specific problem
Check restrictions: Always identify where identities are undefined (usually where denominators equal zero)
Watch signs: Be careful with positive and negative signs, especially when substituting

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Helpful Exam Strategies:

Sketch diagrams: For problems involving specific values, draw right triangles to find missing trigonometric ratios
Verify answers: When possible, check your solutions using a calculator to ensure accuracy

Remember!

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Essential Points to Remember:

$\sin 2α = 2 \sin α \cos α$ - the only form for sine double angle
$\cos 2α$ has three equivalent forms - choose the most convenient one for your calculation
Always check for restrictions where identities become undefined
Use Pythagoras theorem to find missing sides when given one trigonometric ratio
Double angle identities are derived from compound angle formulae by setting both angles equal

Double Angle Identities (Grade 12 NSC Matric Mathematics): Revision Notes