Edexcel IAL - Pure Maths 3- 2.3 Knowledge and use of double angle formulae- Study notes - New syllabus
Edexcel IAL – Pure Maths 3- 2.3 Knowledge and use of double angle formulae -Study notes- New syllabus
Edexcel IAL – Pure Maths 3- 2.3 Knowledge and use of double angle formulae -Study notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 2.3 Knowledge and use of double angle formulae
Compound Angle Formulae
Compound angle formulae express the trigonometric functions of the sum or difference of two angles in terms of the trigonometric functions of the individual angles.
Compound Angle Formulae
Sine:
\( \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \)
Cosine:
\( \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B \)
Tangent:
\( \tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B} \)
Important sign rule:
- For sine: signs stay the same
- For cosine and tangent: signs change
When Compound Angle Formulae Are Used
- Finding exact trigonometric values
- Simplifying trigonometric expressions
- Proving trigonometric identities
- Solving trigonometric equations
Example
Find the exact value of:
\( \cos 75^\circ \)
▶️ Answer / Explanation
\( \cos 75^\circ = \cos(45^\circ + 30^\circ) \)
Using the cosine compound angle formula:
\( \cos A \cos B – \sin A \sin B \)
\( = \cos45^\circ\cos30^\circ – \sin45^\circ\sin30^\circ \)
\( = \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2} – \dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2} \)
\( = \dfrac{\sqrt{6} – \sqrt{2}}{4} \)
Example
Simplify:
\( \sin(x + 30^\circ)\cos x – \cos(x + 30^\circ)\sin x \)
▶️ Answer / Explanation
Use the identity:
\( \sin A \cos B – \cos A \sin B = \sin(A – B) \)
Here:
\( A = x + 30^\circ,\ B = x \)
So the expression becomes:
\( \sin\!\big((x + 30^\circ) – x\big) = \sin 30^\circ \)
\( = \dfrac{1}{2} \)
Answer: \( \dfrac{1}{2} \)
Example
Prove that:
\( \cos x \cos 2x + \sin x \sin 2x = \cos x \)
▶️ Answer / Explanation
Use the identity:
\( \cos A \cos B + \sin A \sin B = \cos(A – B) \)
Let:
\( A = 2x,\ B = x \)
Then:
\( \cos 2x \cos x + \sin 2x \sin x = \cos(2x – x) \)
\( = \cos x \)
Hence proved.
Double Angle Formulae
Double angle formulae are obtained by applying compound angle formulae with equal angles.
Sine:
\( \sin 2\theta = 2\sin\theta\cos\theta \)
Cosine:
\( \cos 2\theta = \cos^2\theta – \sin^2\theta \)
Alternative forms:
\( \cos 2\theta = 1 – 2\sin^2\theta \)
\( \cos 2\theta = 2\cos^2\theta – 1 \)
Tangent:
\( \tan 2\theta = \dfrac{2\tan\theta}{1 – \tan^2\theta} \)
Half-Angle Applications
Half-angle formulae are derived from the cosine double angle identities.
\( \sin^2\theta = \dfrac{1 – \cos 2\theta}{2} \)
\( \cos^2\theta = \dfrac{1 + \cos 2\theta}{2} \)
Note: The substitution \( t = \tan\dfrac{\theta}{2} \) is not required.
Expressing \( a\cos\theta + b\sin\theta \) in a Single Trigonometric Form
An expression of the form:
\( a\cos\theta + b\sin\theta \)
can be written as:
\( r\cos(\theta – \alpha) \quad \text{or} \quad r\sin(\theta + \alpha) \)
Where:
\( r = \sqrt{a^2 + b^2} \)
\( \cos\alpha = \dfrac{a}{r}, \quad \sin\alpha = \dfrac{b}{r} \)
This method is especially useful for solving equations.
Example
Express:
\( \cos\theta + \sin\theta \)
in the form \( r\cos(\theta – \alpha) \).
▶️ Answer / Explanation
\( r = \sqrt{1^2 + 1^2} = \sqrt{2} \)
\( \cos\alpha = \dfrac{1}{\sqrt{2}}, \quad \sin\alpha = \dfrac{1}{\sqrt{2}} \)
\( \alpha = \dfrac{\pi}{4} \)
Answer:
\( \cos\theta + \sin\theta = \sqrt{2}\cos\!\left(\theta – \dfrac{\pi}{4}\right) \)
Example
Solve:
\( \cos\theta + \sin\theta = 1 \), where \( 0 \le \theta \le 2\pi \).
▶️ Answer / Explanation
From Example 1:
\( \cos\theta + \sin\theta = \sqrt{2}\cos\!\left(\theta – \dfrac{\pi}{4}\right) \)
So:
\( \cos\!\left(\theta – \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}} \)
\( \theta – \dfrac{\pi}{4} = \pm\dfrac{\pi}{4} \)
\( \theta = 0,\ \dfrac{\pi}{2} \)
Example
Simplify:
\( \sin^2 x + \cos^2 2x \)
▶️ Answer / Explanation
Use half-angle identities:
\( \sin^2 x = \dfrac{1 – \cos 2x}{2} \)
\( \cos^2 2x = \dfrac{1 + \cos 4x}{2} \)
So:
\( \sin^2 x + \cos^2 2x = \dfrac{1 – \cos 2x}{2} + \dfrac{1 + \cos 4x}{2} \)
\( = 1 + \dfrac{\cos 4x – \cos 2x}{2} \)
Final simplified form obtained.