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
Best Answer
$$\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)=a$$ 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.