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Double Angle Formulae Simplified Revision Notes

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5.6.2 Double Angle Formulae

Double angle formulae are trigonometric identities that express the trigonometric functions of double angles (i.e., 2θ) in terms of the trigonometric functions of the original angle θ. These formulae are very useful in simplifying expressions, solving trigonometric equations, and analyzing oscillatory motion in physics.

1. Sine of a Double Angle:

The sine of a double angle is given by:

sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta)

  • This formula is derived from the sine sum identity sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B by setting A=B=θ.A = B = \theta.

2. Cosine of a Double Angle:

The cosine of a double angle has three equivalent forms:

cos⁡(2θ)=cos⁡2(θ)−sin⁡2(θ)\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1, the above can also be written as:

cos⁡(2θ)=2cos⁡2(θ)−1\cos(2\theta) = 2\cos^2(\theta) - 1 or cos⁡(2θ)=1−2sin⁡2(θ)\cos(2\theta) = 1 - 2\sin^2(\theta)

  • These forms are useful depending on whether you know or want to simplify the expression in terms of sine or cosine.

3. Tangent of a Double Angle:

The tangent of a double angle is given by:

tan⁡(2θ)=2tan⁡(θ)1−tan⁡2(θ)\tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)}

  • This formula is derived from the tangent sum identity tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡Bby setting A=B=θ.\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} by\ setting\ A = B = \theta.

4. Example Problems Using Double Angle Formulae:

infoNote

Example 1: Calculate sin⁡(2θ) if sin⁡(θ)=35\sin(2\theta)\ if\ \sin(\theta) = \frac{3}{5} and cos⁡(θ)=45\cos(\theta) = \frac{4}{5}.

  • Solution:
  • Use the double angle formula for sine: sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta)
  • Substitute the given values: sin⁡(2θ)=2×35×45=2425\sin(2\theta) = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}
  • So, sin⁡(2θ)=2425\sin(2\theta) = \frac{24}{25}.
infoNote

Example 2: Simplify cos⁡(2θ) if cos⁡(θ)=513\cos(2\theta)\ if\ \cos(\theta) = \frac{5}{13}.

  • Solution:
  • Use the double angle formula for cosine: cos⁡(2θ)=2cos⁡2(θ)−1\cos(2\theta) = 2\cos^2(\theta) - 1
  • First, calculate cos⁡2(θ)\cos^2(\theta): cos⁡2(θ)=(513)2=25169\cos^2(\theta) = \left(\frac{5}{13}\right)^2 = \frac{25}{169}
  • Substitute into the formula: cos⁡(2θ)=2×25169−1=50169−169169=50−169169=−119169\cos(2\theta) = 2 \times \frac{25}{169} - 1 = \frac{50}{169} - \frac{169}{169} = \frac{50 - 169}{169} = \frac{-119}{169}
  • So, cos⁡(2θ)=−119169\cos(2\theta) = \frac{-119}{169}.
infoNote

Example 3: Solve tan⁡\tan(2 θ \theta) = 1 for 0∘≤θ≤180∘.0^\circ \leq \theta \leq 180^\circ.

  • Solution:
  • First, solve for 2θ 2\theta using the basic tangent identity: tan⁡(2θ)=1⇒2θ=45∘ or 2θ=225∘\tan(2\theta) = 1 \quad \Rightarrow \quad 2\theta = 45^\circ \text{ or } 2\theta = 225^\circ
  • Therefore, θ=45∘2=22.5∘\theta = \frac{45^\circ}{2} = 22.5^\circ or θ=225∘2=112.5∘\theta = \frac{225^\circ}{2} = 112.5^\circ.
  • The solutions are θ=22.5∘\theta = 22.5^\circ and θ=112.5∘\theta = 112.5^\circ.

5. Applications of Double Angle Formulae:

  • Simplifying Trigonometric Expressions: Double angle formulae are often used to rewrite trigonometric expressions in a simpler form, especially when dealing with powers of trigonometric functions.
  • Solving Trigonometric Equations: These identities are crucial for solving equations where angles are involved in double forms.
  • Calculus: In integration and differentiation, double angle identities help simplify integrals or derivatives involving trigonometric functions.
  • Physics: Double angle identities are used in wave mechanics, signal processing, and oscillations.

Summary:

  • Double angle formulae are key trigonometric identities that simplify the analysis of trigonometric functions when dealing with angles that are multiples of other angles.
  • The formulae for sine, cosine, and tangent of double angles are essential tools for simplifying expressions, solving equations, and applying trigonometric concepts in calculus and physics.
  • Mastery of these identities enhances your ability to tackle more complex problems in trigonometry and related fields.

Double Angle Formulae

The compound angle formulae can be used to derive double angle formulae.

infoNote

Example: cos⁡(2θ)\cos(2\theta)

cos⁡(2θ)=cos⁡(θ+θ)=cos⁡θcos⁡θ−sin⁡θsin⁡θ=cos⁡2θ−sin⁡2θ\cos(2\theta) = \cos(\theta + \theta) = \cos\theta\cos\theta - \sin\theta\sin\theta = \cos^2\theta - \sin^2\theta
  • Write as θ+θ\theta + \theta
  • Expand using cos⁡(A+B)\cos(A+B)
∴cos⁡(2θ)≡cos⁡2θ−sin⁡2θ\therefore \cos(2\theta) \equiv \cos^2\theta - \sin^2\thetacos⁡(2θ)≡(1−sin⁡2θ)−sin⁡2θ≡1−2sin⁡2θ\cos(2\theta) \equiv (1 - \sin^2\theta) - \sin^2\theta \equiv 1 - 2\sin^2\thetacos⁡(2θ)≡cos⁡2θ−(1−cos⁡2θ)≡2cos⁡2θ−1\cos(2\theta) \equiv \cos^2\theta - (1 - \cos^2\theta) \equiv 2\cos^2\theta - 1
infoNote
  • Example: sin⁡(2θ)\sin(2\theta)
sin⁡(2θ)=sin⁡(θ+θ)=sin⁡θcos⁡θ+cos⁡θsin⁡θ=2sin⁡θcos⁡θ\sin(2\theta) = \sin(\theta + \theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta
infoNote
  • Example: tan⁡(2θ)\tan(2\theta)
tan⁡(2θ)=tan⁡(θ+θ)=tan⁡θ+tan⁡θ1−tan⁡θtan⁡θ=2tan⁡θ1−tan⁡2θ\tan(2\theta) = \tan(\theta + \theta) = \frac{\tan\theta + \tan\theta}{1 - \tan\theta\tan\theta} = \frac{2\tan\theta}{1 - \tan^2\theta}

Solve each equation for x x in the interval 0≤x≤360∘0 \leq x \leq 360^\circ.

Give your answers to 11 decimal place where appropriate.

a) cos⁡(2x)+cos⁡x=0\cos(2x) + \cos x = 0

  1. cos⁡2x=cos⁡2x−sin⁡2x≡2cos⁡2x−1\cos 2x = \cos^2 x - \sin^2 x \equiv 2\cos^2 x - 1
  2. 2cos⁡2x−1+cos⁡x=0⇒2cos⁡2x+cos⁡x−1=02\cos^2 x - 1 + \cos x = 0 \Rightarrow 2\cos^2 x + \cos x - 1 = 0
  3. Let cos⁡x=12\cos x = \frac{1}{2}
x=arccos⁡(12)=60,300x = \arccos\left(\frac{1}{2}\right) = 60, 300 cos⁡x=−1⇒x=arccos⁡(−1)=180\cos x = -1 \Rightarrow x = \arccos(-1) = 180 image

Thus, the solutions are:

x=60,180,300x = 60, 180, 300

Solve each equation for x in the interval 0≤x≤2πc0 \leq x \leq 2\pi^c.

Give your answers to 1 decimal place where appropriate.

b) sin⁡2x+cos⁡x=0\sin 2x + \cos x = 0

2sin⁡xcos⁡x+cos⁡x=02\sin x \cos x + \cos x = 0

Expand sin⁡2x\sin 2x:

  • Bad Method:
2sin⁡xcos⁡x+1=0(Divide by cos⁡x)2\sin x \cos x + 1 = 0 \quad \text{(Divide by } \cos x\text{)} 2sin⁡x=−12\sin x = -1 sin⁡x=−12\sin x = -\frac{1}{2} x=−16π,π+16π,2π−16πx = -\frac{1}{6}\pi, \pi + \frac{1}{6}\pi, 2\pi - \frac{1}{6}\pi image

Graph drawn below the calculations to show solutions.

x=7π6,11π6x = \frac{7\pi}{6}, \frac{11\pi}{6} image

Solution missing for cos⁡x=0\text{Solution missing for } \cos x = 0

  • Good Method:
cos⁡x(2sin⁡x+1)=0\cos x (2\sin x + 1) = 0 2sin⁡x+1=0,cos⁡x=02\sin x + 1 = 0, \quad \cos x = 0 2sin⁡x=−12\sin x = -1 sin⁡x=−12\sin x = -\frac{1}{2} x=−16π,π+16π,2π−16πx = -\frac{1}{6}\pi, \pi + \frac{1}{6}\pi, 2\pi - \frac{1}{6}\pi image

Graph drawn below the calculations to show solutions.

x=7π6,11π6x = \frac{7\pi}{6}, \frac{11\pi}{6} cos⁡x=0⇒x=π2,3π2\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2} image

Graph drawn below the calculations to show solutions.

x=π2,7π6,3π2,11π6\boxed {x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}}

infoNote

Q2 (Jan 2006, Q4) Solve the equation 2sin⁡2θ+cos⁡2θ=12 \sin 2\theta + \cos 2\theta = 1, for 0∘≤θ<360∘0^\circ \leq \theta < 360^\circ.

2(2sin⁡θcos⁡θ)+1−2sin⁡2θ=12(2\sin\theta \cos\theta) + 1-2\sin^2\theta = 1

4sin⁡θcos⁡θ−2sin⁡2θ=04\sin \theta\cos \theta -2\sin ^2\theta = 0

2sin⁡θ(2cos⁡θ−sin⁡θ)=02\sin\theta (2\cos\theta - \sin \theta) = 0

2sin⁡θ=02\sin\theta = 0

sin⁡θ=0\sin\theta = 0

θ=0,360∘,180∘\theta = 0, \xcancel{360^\circ}, 180^\circ

] - 1

cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = \cos^2\theta - \sin^2\theta

=2cos⁡2θ−1=2\cos^2\theta -1

=1−2sin⁡2θ=1 - 2\sin^2\theta

2cos⁡2θ−sin⁡θ=02\cos 2\theta-\sin\theta = 0

2cos⁡2θ=sin⁡θ2\cos 2\theta=\sin\theta

2=sin⁡θcos⁡θ=tan⁡θ2 = \frac {\sin\theta}{\cos\theta} = \tan\theta

θ=arctan⁡(2)=63.435,180+63.433=243.435\theta = \arctan(2) = 63.435, 180 + 63.433 = 243.435

x=0,63.44,180,243.44\boxed {x = 0, 63.44, 180, 243.44}

infoNote

Q8 (Jun 2015, Q2) Express 6cos⁡2θ+sin⁡θ6 \cos 2\theta + \sin \theta in terms of sin⁡θ\sin \theta. Hence solve the equation 6cos⁡2θ+sin⁡θ=06 \cos 2\theta + \sin \theta = 0, for 0∘≤θ<360∘0^\circ \leq \theta < 360^\circ.

  1. Expressing in terms of cos⁡θ\cos\theta : cos⁡2θ=cos⁡2θ−sin⁡2θ⇒2cos⁡2θ−1⇒1−2sin⁡2θ\cos2\theta = \cos^2\theta - \sin^2\theta \\ \Rightarrow 2\cos^2\theta-1 \\ \Rightarrow 1-2\sin^2\theta

  2. Expressing in terms of sin⁡θ\sin\theta: 6(1−2sin⁡2θ)+sin⁡θ=06(1 - 2\sin^2\theta) + \sin\theta = 0

6−12sin⁡2θ+sin⁡θ=06 - 12\sin^2\theta + \sin\theta = 0

Rearrange: 12sin⁡2θ−sin⁡θ−6=012\sin^2\theta - \sin\theta - 6 = 0

  1. Solve the quadratic equation: sin⁡θ=34,θ=41.810∘\sin\theta = \frac{3}{4}, \theta = 41.810^\circ

  2. Use the graph or unit circle to find all solutions: θ=48.59∘,131.4∘,221.8∘,318.2∘\theta = 48.59^\circ, 131.4^\circ, 221.8^\circ, 318.2^\circ

infoNote

Q1 (Jan 2009, Q9)

  1. (i) By first expanding cos⁡(2θ+θ)\cos(2\theta + \theta), prove that cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4 \cos^3\theta - 3 \cos\theta. Expressing in terms of cos⁡θ\cos\theta :

cos⁡2θ=cos⁡2θ−sin⁡2θ⇒2cos⁡2θ−1⇒1−2sin⁡2θ\cos2\theta = \cos^2\theta - \sin^2\theta \\ \Rightarrow 2\cos^2\theta-1 \\ \Rightarrow 1-2\sin^2\theta

cos⁡(2θ+θ)=cos⁡(2θ)cos⁡(θ)−sin⁡(2θ)sin⁡(θ)\cos(2\theta + \theta) = \cos(2\theta)\cos(\theta) - \sin(2\theta)\sin(\theta)

Substituting the identities:

=(2cos⁡2θ−1)cos⁡(θ)−2sin⁡θcos⁡θsin⁡(θ)= (2\cos^2\theta - 1)\cos(\theta) - 2\sin\theta\cos\theta\sin(\theta)

Expanding:

=2cos⁡3θ−cos⁡θ−2sin⁡2θcos⁡θ= 2\cos^3\theta - \cos\theta - 2\sin^2\theta\cos\theta

Using sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta:

=2cos⁡3θ−cos⁡θ−2cos⁡θ(1−cos⁡2θ)= 2\cos^3\theta - \cos\theta - 2\cos\theta(1 - \cos^2\theta)=2cos⁡3θ−cos⁡θ−2cos⁡θ+2cos⁡3θ= 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta

=4cos⁡3θ−3cos⁡θ= 4\cos^3\theta - 3\cos\theta

  1. (ii) Hence prove that cos⁡6θ=32cos⁡6θ−48cos⁡4θ+18cos⁡2θ−1\cos 6\theta = 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1.
cos⁡(6θ)=cos⁡(2(3θ))=2cos⁡2(3θ)−1\cos(6\theta) = \cos(2(3\theta)) = 2\cos^2(3\theta) - 1=2cos⁡3(3θ)−cos⁡(3θ)−1= 2\cos^3(3\theta) - \cos(3\theta) - 1

Expanding:

=2[4cos⁡3(3θ)−3cos⁡θ)][4cos⁡3(3θ)−3cos⁡θ)]−1= 2\left [4\cos^3(3\theta) - 3\cos\theta)\right ]\left [4\cos^3(3\theta)-3\cos\theta)\right ] - 1=2[16cos⁡6(3θ)−12cos⁡4(3θ)−12cos⁡4(3θ)−9cos⁡2θ)]−1= 2\left [16\cos^6(3\theta) - 12\cos^4(3\theta)- 12\cos^4(3\theta)-9\cos^2\theta)\right ] - 1

Simplifying:

=32cos⁡6θ−48cos⁡4θ+18cos⁡2θ−1= 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1

  1. (iii) Show that the only solutions of the equation 1+cos⁡6θ=18cos⁡2θ1 + \cos 6\theta = 18\cos^2\theta are odd multiples of 90∘90^\circ. Substitute cos⁡6θ\cos 6\theta from part (ii):
1+(32cos⁡6θ−48cos⁡4θ+18cos⁡2θ−1)=18cos⁡2θ32cos⁡6θ−48cos⁡4θ+18cos⁡2θ=18cos⁡2θ1 + (32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1)\\ = 18\cos^2\theta 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta \\= 18\cos^2\theta

Rearrange:

32cos⁡6θ−48cos⁡4θ=032\cos^6\theta - 48\cos^4\theta = 0

Factor out common terms:

16cos⁡4θ(2cos⁡2θ−3)=016\cos^4\theta (2\cos^2\theta - 3) = 0

cos⁡2θ=32\cos^2\theta = \frac{3}{2} \quad (Not possible, cos⁡2θ\cos^2\theta must be ≤1\leq 1)

Thus, cos⁡2θ=0\cos^2\theta = 0, implying cos⁡θ=0\cos\theta = 0.

Therefore, θ\theta must be odd multiples of 90∘90^\circ.

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