I'm trying to solve this problem:
If $\sin\theta+\sin\phi=a$ and $\cos\theta+ \cos\phi=b$, then find $\tan \dfrac{\theta-\phi}2$.
So seeing $\dfrac{\theta-\phi}2$ in the argument of the tangent function, I first thought of converting the left-hand sides of the givens to products which gave me:
$$2\sin\frac{\theta+\phi}2\cos\frac{\theta-\phi}2=a\quad,\quad2\cos\frac{\theta+\phi}2\cos\frac{\theta-\phi}2=b$$
But then, on dividing the two equations (assuming $b\ne0$), I just get the value of $\tan\dfrac{\theta+\phi}2$.
I don't know how else to proceed.
Any help would be appreciated!
Best Answer
It may be a long way to find it by mine, but it works. As $$\sin\theta+\sin\phi=a,~~~\cos\theta+ \cos\phi=b$$ then $$\cos(\theta-\phi)=\cos\phi\cos\theta+\sin\phi\sin\theta=\frac{a^2+b^2-2}2$$ Now use: $$\cos x=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$$