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If: $\sin(\alpha) + \sin(\beta) = a$ and $\cos(\alpha) + \cos(\beta) = b$

Determine: $\sin(\alpha + \beta) = ?$ and $\cos(\alpha + \beta) = ?$

The right answers:

$$\sin(\alpha + \beta) = \frac{2ab}{a^2 + b^2} \qquad \cos(\alpha + \beta) = \frac{b^2 - a^2}{a^2+b^2}$$

I started by a*b, which results in

$$\sin\alpha\cos\alpha + \sin\alpha\cos\beta + \sin\beta\cos\alpha + \sin\beta\cos\beta = ab \tag1$$

$$\sin\alpha\cos\alpha + \sin(\alpha + \beta) + \sin\beta\cos\beta = ab \tag2$$

$$\sin(\alpha + \beta) = ab - \frac{1}{2}(\sin2\alpha + \sin0) - \frac{1}{2}(\sin2\beta + \sin0) \tag3$$

$$\sin(\alpha + \beta) = ab - \frac{1}{2}(\sin2\alpha + \sin2\beta) \tag4$$

From this point, I kinda got stuck

Dimitar Kapitanov
  • 73

2 Answers2

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$$\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)=a$$ And $$\cos\alpha+\cos\beta=2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)=b$$

Dividing, $$\tan\left(\frac{\alpha+\beta}{2}\right)=\frac ab=t$$

However, $$\sin(\alpha+\beta)=\frac{2t}{1+t^2}$$ and $$\cos(\alpha+\beta)=\frac{1-t^2}{1+t^2}$$

And it is readily seen that both results follow directly from this.

David Quinn
  • 34,121
1

Here is an alternative method quite fast too using complex exponential:

Let set $\begin{cases}u=e^{i\alpha}\\v=e^{i\beta}\end{cases}\quad$ and $X=e^{i(\alpha+\beta)}=uv\quad $ since we have $\quad\begin{cases}u+v=b+ia\\\frac 1u+\frac 1v=b-ia\end{cases}$

We can develop the second line to get the value of $X$, indeed:

$\dfrac 1u+\dfrac 1v=\dfrac{u+v}{uv}=\dfrac{b+ia}{X}=b-ia\iff X=\dfrac{b+ia}{b-ia}=\underbrace{\dfrac{b^2-a^2}{b^2+a^2}}_{\cos(\alpha+\beta)}+i\,\underbrace{\dfrac{2ab}{b^2+a^2}}_{\sin(\alpha+\beta)}$

zwim
  • 28,563