Topic 20 Compound and multiple angles

Theorem 20.1 (Compound angle formulas for sine and cosine) \[ \boxed{ \begin{array}{rcl} \sin(\alpha+\beta) &=& \sin\alpha\cos\beta + \cos\alpha\sin\beta\\ \cos(\alpha+\beta) &=& \cos\alpha\cos\beta - \sin\alpha\sin\beta \end{array} }\]

Proof of Theorem 20.1 (non-examinable):

First, we will arrange points in the plane to arrive at Figure 20.1.

  • Define \(E\) by \(|AE|=1\).
  • Define \(C\) so that \(\angle ACE=90^\circ\).
  • Define \(B\) so that \(\angle ABC=90^\circ\).
  • Form rectangle \(ABDF\), so that \(E\) lies on \(DF\).
Diagram for the proof of compound angle formulas

Figure 20.1: Diagram for the proof of compound angle formulas

Then: \[\begin{align} |CE| &= \sin\beta, & |AC| &= \cos\beta \tag{\(\triangle ACE\)}\\ |EF| &= \cos(\alpha+\beta), & |AF| &=\sin(\alpha+\beta) \tag{\(\triangle AEF\)} \end{align}\] This implies that \[\begin{align} |AB| &= \cos\alpha\cos\beta, & |BC| &= \sin\alpha\cos\beta \tag{\(\triangle ABC\)}\\ |CD| &= \cos\alpha\sin\beta, & |DE| &= \sin\alpha\sin\beta \tag{\(\triangle CDE\)} \end{align}\] Hence \[\begin{align*} \sin(\alpha+\beta) &= |AF| = |BD| = |BC|+|CD| = \sin\alpha\cos\beta+\cos\alpha\sin\beta\\ \cos(\alpha+\beta) &= |EF| = |DF| - |DE| = \cos\alpha\cos\beta - \sin\alpha\sin\beta \end{align*}\] This finishes the proof.

Corollary 20.1 (Compound angle formula for tangent) \[\boxed{ \tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} }\]

Proof of Corollary 20.1 using Theorem 20.1: \[\begin{align*} \tan(\alpha+\beta) &= \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\\ &= \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta} &&\text{(by Theorem)}\\ &= \frac{\frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta}}{1-\frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}} = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}. \end{align*}\] (For the penultimate step, we divide both top and bottom by \(\cos\alpha\cos\beta\).)

Corollary 20.2 (Double angle formulas) \[\boxed{ \begin{array}{rcl} \sin(2\alpha) &=& 2\sin\alpha\cos\alpha\\ \cos(2\alpha) &=& 2\cos^2\alpha - 1 = 1-2\sin^2\alpha\\ \tan(2\alpha) &=& \dfrac{2\tan\alpha}{1-\tan^2\alpha} \end{array} }\]

Proof of double angle cosine formula (Corollary 20.2): \[\begin{align*} \cos(2\alpha) &= \cos\alpha\cos\alpha - \sin\alpha\sin\alpha &&\text{(by Theorem)}\\ &= \cos^2\alpha-\sin^2\alpha\\ &= 2\cos^2\alpha - 1 = 1-2\sin^2\alpha. \end{align*}\] (For the last step, we use Pythagoras’s formula: \(\sin^2\alpha+\cos^2\alpha=1\).)

Example 20.1 Express \(\cos(3\alpha)\) in terms of \(\cos\alpha\).

Solution. \[\begin{align*} \cos(3\alpha) &= \cos(2\alpha+\alpha) \\ &= \cos(2\alpha)\cos\alpha - \sin(2\alpha)\sin\alpha\\ &= (2\cos^2\alpha-1)\cos\alpha - (2\sin\alpha\cos\alpha)\sin\alpha\\ &= 2\cos^3\alpha - \cos\alpha - 2(1-\cos^2\alpha)\cos\alpha\\ &= 4\cos^3\alpha - 3\cos\alpha. \end{align*}\]

Example 20.2 Express \(\tan(3\alpha)\) in terms of \(\tan\alpha\).

Solution. \[\begin{align*} \tan(3\alpha) &= \tan(2\alpha+\alpha)\\ &=\frac{\tan(2\alpha)+\tan\alpha}{1-\tan(2\alpha)\tan\alpha}\\ &=\frac{\frac{2\tan\alpha}{1-\tan^2\alpha}+\tan\alpha}{1-\frac{2\tan\alpha}{1-\tan^2\alpha}\cdot\tan\alpha}\\ &=\frac{2\tan\alpha+\tan\alpha(1-\tan^2\alpha)}{1-\tan^2\alpha - 2\tan\alpha\cdot\tan\alpha}\\ &=\frac{3\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha}. \end{align*}\]

Corollary 20.3 (Changing sums of (co)sines into products) \[\boxed{ \begin{array}{rcl} \sin\alpha + \sin\beta &=& 2\sin\tfrac{\alpha+\beta}2\cos\tfrac{\alpha-\beta}2\\ \sin\alpha - \sin\beta &=& 2\cos\tfrac{\alpha+\beta}2\sin\tfrac{\alpha-\beta}2\\ \cos\alpha + \cos\beta &=& 2\cos\tfrac{\alpha+\beta}2\cos\tfrac{\alpha-\beta}2\\ \cos\alpha - \cos\beta &=& - 2\sin\tfrac{\alpha+\beta}2\sin\tfrac{\alpha-\beta}2\\ \end{array} }\]

Proof of Corollary 20.3 (non-examinable):

Note that \(\frac{\alpha+\beta}2+\frac{\alpha-\beta}2 = \alpha\) and \(\frac{\alpha+\beta}2-\frac{\alpha+\beta}2=\beta\). Then \[\begin{align*} \sin\alpha+\sin\beta &= \sin\left(\tfrac{\alpha+\beta}2+\tfrac{\alpha-\beta}2\right) + \sin\left(\tfrac{\alpha+\beta}2-\tfrac{\alpha-\beta}2\right)\notag\\ &= \sin\tfrac{\alpha+\beta}2\cos\tfrac{\alpha-\beta}2 + \cos\tfrac{\alpha+\beta}2\sin\tfrac{\alpha-\beta}2 \notag\\ &\phantom{=}\,\, + \sin\tfrac{\alpha+\beta}2\cos\left(-\tfrac{\alpha-\beta}2\right) + \cos\tfrac{\alpha+\beta}2\sin\left(-\tfrac{\alpha-\beta}2\right) \notag\\ &= 2\sin\tfrac{\alpha+\beta}2\cos\tfrac{\alpha-\beta}2, \end{align*}\] since \(\cos(-x)=\cos(x)\) and \(\sin(-x)=-\sin(x)\). The rest are similar.

Example 20.3 Prove that \(\dfrac{\sin(7\theta)+\sin(5\theta)}{\cos(7\theta) - \cos(5\theta)} = -\cot\theta.\)

Solution. \[\begin{align*} \mathrm{LHS} &= \frac{2\sin\frac{7\theta+5\theta}{2}\cos\frac{7\theta-5\theta}{2}}{-2\sin\frac{7\theta+5\theta}{2}\sin\frac{7\theta-5\theta}{2}}\\ &= \frac{\sin6\theta\cos\theta}{-\sin6\theta\sin\theta} = \mathrm{RHS}. \end{align*}\]

Example 20.4 Express \(\sin50^\circ + \sin30^\circ\) as a product of (co)sines.

Solution. \(\sin50^\circ+\sin30^\circ = 2\sin\dfrac{50^\circ+30^\circ}{2}\cos\dfrac{50^\circ-30^\circ}{2} = 2\sin 40^\circ\cos 10^\circ\).

Example 20.5 Without a calculator, evaluate \(16\cdot\sin75^\circ\cdot \cos15^\circ\).

Solution. First, we want to find \(\alpha\) and \(\beta\) so that \[ 75^\circ=\frac{\alpha+\beta}{2}, \quad\quad\quad 15^\circ=\frac{\alpha-\beta}{2}. \] Solving the equations, we get \(\alpha=90^\circ\) and \(\beta=60^\circ\). Hence \[\begin{align*} 16\sin 75^\circ\cos15^\circ &= 16\sin\frac{90^\circ+60^\circ}{2}\cos\frac{90^\circ-60^\circ}{2}\\ &= 8(\sin90^\circ+\sin60^\circ)\\ &= 8\left(1+\frac{\sqrt3}{2}\right) = 8 + 4\sqrt{3}. \end{align*}\]

Example 20.6 Prove that \(\dfrac{\sin2\alpha}{1+\cos2\alpha}=\tan\alpha\).

Solution. \[\begin{align*} \mathrm{LHS} &= \frac{2\sin\alpha\cos\alpha}{1+2\cos^2\alpha -1}\\ &= \frac{2\sin\alpha\cos\alpha}{2\cos^2\alpha} = \frac{\sin\alpha}{\cos\alpha} = \mathrm{RHS}. \end{align*}\]